= \sum\limits^{k}_{i=0} (-1)^i \binom{k}{i} x^{k-i} x^{2i} {x^D}^i =  \sum\limits^{k+1}_{i=0} (-1)^i \binom{k}{i} x^{k-i+2i} {x^D}^i \\
= \sum\limits^{k}_{i=0} (-1)^i\binom{k}{i} x^{k+i} {x^D}^i = \sum\limits^{k}_{i=0} (-1)^i\binom{k}{i} x^{k} = (1-1)^{k} x^{k} = 0
\end{gather*}

So $n_x$ has nilpotent index less than or equal to $k$.

* Using [D.2] and [D.3], we have:

\begin{gather*}
c_x n_x = c_x(x - c_x) = c_x x - c_x c_x = x x^D x x - x x^D x x x^D x\\
= x^3 x^D - x^3 x^D x x^D =  x^3 x^D - x^3 x^D = 0
\end{gather*}

So $c_x n_x = 0$, and similarly  $n_x c_x = 0$.

* From the definition of $n_x$ it follows that $x = c_x + (x - c_x) = c_x + n_x$.

Suppose that $x$ has a core-nilpotent decomposition $(c,n)$. Let $k \geq 1$ be the nilpotent index of $n$ so that $n^{k} = 0$. Using [CND.1], we set $x^D := c^D$ and show that $x^D$ satisfies the three Drazin inverse axioms.

* First note that, for $m > 0$, $x^m = (c + n)^m = c^m + n^m$ as the cross-terms vanish using [CND.3]. So, in particular, for $j \geq k$, we have that $x^{j} = c^{j}$ using [CND.2]. Now since $\mathsf{ind}(c) \leq 1 \leq k$, by Lemma <ref>.(<ref>) we compute that:

\begin{gather*}
x^{k+1} x^D = (c+n)^{k+1} c^D = (c^{k+1} + n^{k+1}) c^D = c^{k+1} c^D \\ = c^{k} = c^{k} + n^{k} = (c+n)^{k} = x^{k}
\end{gather*}

So ${\sf ind}(x) \leq k$.

* By [CND.3], we have that $cn = 0 = 0c$. So by Lemma <ref>, we have that $c^Dn = 0c^D$, so $c^D n =0$. So using this and [D.2], we compute that:

\[x^D x x^D = c^D (c + n) c^D = c^D c c^D + c^D n c^D = c^D + 0 = c^D = x^D \]

* By [CND.3], we have that $cn= nc$. So by Lemma <ref>, we have that $c^Dn = nc^D$. So using this and [D.3], we have:

\[x x^D = (c+n)c^D = cc^D + nc^D = c^Dc +c^Dn = c^D(c+n) = x^D x\]

So $x$ is Drazin with Drazin inverse $x^D := c^D$ and ${\sf ind}(x) \leq k$.

It remains to show that a core-nilpotent decomposition is unique. So let $x$ be Drazin. Suppose that $x$ has another core-nilpotent decomposition $(c,n)$. Then we have that $c_x^D = x^D = c^D$. Since $\mathsf{ind}(c) \leq 1$ and $\mathsf{ind}(c_x) \leq 1$, by Corollary <ref> we have that $c = (c^D)^D = (c^D_x)^D = c_x$, so $c = c_x$. By [CND.4], $c + n = x = c_x + n_x$, so subtracting $c=c_x$ from both sides gives us $n=n_x$. So we conclude that $(c_x, n_x)$ is the unique core-nilpotent decomposition of $x$.

Therefore, we obtain another characterization of when an additive category is Drazin:

An additive category $\mathbb{X}$ is Drazin if and only if every endomorphism has a core-nilpotent decomposition.

§.§ Kernel-cokernel coincidence
In <cit.>, Robinson and Puystjens provide conditions for being Drazin in an additive category in terms of kernels and cokernels. Here we revisit their result and show how their setup is in fact equivalent to being Drazin such that the complement of the induced idempotent is split.

In an additive category, the complement of an idempotent $e: A \to A$ is the endomorphism $e^c: A \to A$ defined as $e^c: =1_A - e$. Not that since $e$ is an idempotent, we have that $ee^c =0 = e^c e$. Moreover, the complement of an idempotent is again an idempotent, so we may consider when it splits. In an additive category, we say that an idempotent $e$ is complement-split if $e^c$ is split. In particular, we may consider a Drazin endomorphism whose induced idempotent is complement-split.

In an additive category $\mathbb{X}$, $x: A\to A$ is Drazin complement-split if $x$ is Drazin and its induced impotent $e_x: A \to A$ (i.e. $e_x = xx^D = x^Dx$) is complement-split.

We now show an endomorphism $x$ is Drazin complement-split precisely when some power of $x$ has a kernel and a cokernel which are isomorphic. In particular, this means that we can describe the Drazin inverse of $x$ in terms of the splitting of the complement of the induced idempotent.

In an additive category $\mathbb{X}$, $x: A \to A$ is Drazin complement-split if and only if there is a $k \in \mathbb{N}$ such that $x^{k+1}$ has a kernel and cokernel:

\[ \xymatrixcolsep{5pc}\xymatrix{ \mathsf{ker}(x^{k+1}) ~\ar@{>->}[r]^-{\kappa} & A \ar@<1ex>[r]^-{x^{k+1}} \ar@<-1ex>[r]_0 & A \ar@{->>}[r]^-{\lambda} & \mathsf{coker}(x^{k+1}) } \]

such that $\kappa \lambda$ is an isomorphism and $x^{k+1}: (A, e^c_{\lambda,\kappa}) \to (A, e^c_{\lambda,\kappa})$ is an isomorphism in $\mathsf{Split}(\mathbb{X})$, where $e^c_{\lambda,\kappa}$ is the complement of the idempotent $e_{\lambda,\kappa} = \lambda (\kappa\lambda)^{-1} \kappa$.

Suppose that $x$ is Drazin complement-split where $e^c_x$ splits via $r^c: A \to K$ and $s^c: K \to A$. Let $\mathsf{ind}(x) = k$. We first show that $s^c: K \to A$ is the kernel of $x^{k+1}$ and that  $r^c: A \to K$ is the cokernel of $x^{k+2}$. Starting with the kernel, we first compute using Lemma <ref>.(<ref>) that:

\[ x^{k+1} e^c_x = x^{k+1}(1_A- e_x) = x^{k+2} - x^{k+1}e_x =   x^{k+1} - x^{k+1}x x^D = x^{k+1} - x^{k+1+1} x^D = x^{k+1} - x^{k+1} = 0 \]

So $x^{k+1} e^c_x = 0$. Since $e^c_x r^c = s^c$, post-composing both since by $r^c$ gives us that $x^{k+1} s^c =0$. Now suppose that we had a map $f: B \to A$ such that $f x^{k+1} =0$ as well. Now observe that using idempotency and [D.3], we have that:

\[ f e_x = f (e_x)^{k+1} = f (x x^D)^{k+1} = f x^{k+2} (x^D)^{k+1} = 0 (x^D)^{k+1} = 0   \]

So $f e_x =0$, and so it follows that $fe^c_x = f$. Then define $f^\sharp: B \to K$ as $f^\sharp \colon= f r^c$. Since $r^c s^c = e^c_x$, it follows that $f^\sharp r^c =f$. Since $s^c$ is monic (since $s^c r^c =1_K$), it follows that $f^\sharp$ must be unique. So we conclude that $s^c: K \to A$ is the kernel of $x^{k+1}$ as desired. By similar arguments, one can also show that $r^c: A \to K$ is the cokernel of $x^{k+1}$. Now setting $\kappa = s^c$ and $\lambda= r^c$, trivially $\kappa \lambda = 1_K$ is an isomorphism. Moreover:

\[e_{\lambda,\kappa} =  \lambda (\kappa\lambda)^{-1} \kappa = \lambda \kappa = r^c s^c = e^c_x\]

So $e_{\lambda,\kappa} = e^c_x$ and $e^c_{\lambda,\kappa} = e_x$. Now as was shown in the proof of Theorem <ref>, $x^{k+1}: (A, e_x) \to (A, e_x)$ is an isomorphism in the idempotent splitting, or in other words, $x^{k+1}: (A, e^c_{\lambda,\kappa}) \to (A, e^c_{\lambda,\kappa})$ is an isomorphism in $\mathsf{Split}(\mathbb{X})$ as desired.

By Theorem <ref>, since $x^{k+1}: (A, e^c_{\lambda,\kappa}) \to (A, e^c_{\lambda,\kappa})$ is an isomorphism in $\mathsf{Split}(\mathbb{X})$, we get that $x$ is Drazin. So it remains to explain why its induced idempotent is complement-split. Recall that in the proof of Theorem <ref> we showed that the Drazin inverse of $x$ is $x^D := v x^k = x^k v$, where $v: (A, e^c_{\lambda,\kappa}) \to (A, e^c_{\lambda,\kappa})$ is the inverse of $x^{k+1}: (A, e^c_{\lambda,\kappa}) \to (A, e^c_{\lambda,\kappa})$ in $\mathsf{Split}(\mathbb{X})$. So its induced idempotent is $e_x = x^{k+1} v$. However $x^{k+1} v= 1_{(A, e^c_{\lambda,\kappa})} = e^c_{\lambda,\kappa}$. So $e_x = e^c_{\lambda,\kappa}$ and thus $e_x^c = e_{\lambda,\kappa}$. However, clearly $e^c_x = e_{\lambda,\kappa}$ splits via $\lambda: A \to \mathsf{coker}(x^{k+1})$ and $(\kappa\lambda)^{-1} \kappa: \mathsf{coker}(x^{k+1}) \to A$. So we conclude that $x$ is Drazin complement-split.

In <cit.>, rather than asking that $x^{k+1}$ is an isomorphism in the idempotent splitting, Robinson and Puystjens ask that $x^{k+1}+e_{\lambda,\kappa}$ be an actual isomorphism. However, the following lemma shows that these statements are equivalent.

In an additive category $\mathbb{X}$, if $f: (A,e) \to (A,e)$ is a map in $\mathsf{Split}(\mathbb{X})$, then $f: (A,e) \to (A,e)$ is an isomorphism in $\mathsf{Split}(\mathbb{X})$ if and only if $f + e^c$ is an isomorphism in $\mathbb{X}$.

For both the directions, assume that $f: (A,e) \to (A,e)$ is a map in $\mathsf{Split}(\mathbb{X})$, which recall means that $ef = f =fe$.

($\Rightarrow$)Suppose that $f: (A,e) \to (A,e)$ is an isomorphism in $\mathsf{Split}(\mathbb{X})$ with inverse $g: (A,e) \to (A,e)$. Recall that this means that: $eg=g=ge$ and also that $fg=e=gf$. Now note that this implies that $fe^c=0=fe^c$ and $ge^c =0 = e^cg$. Then define $(f + e^c)^{-1} := g + e^c$. So we compute that:

\[ (f +e^c)(f + e^c)^{-1} = (f + e^c) (g + e^c) = fg + fe^c + e^c g + e^c e^c = e + 0 + 0 + e^c = e + e^c = 1_A \]

\[ (f + e^c)^{-1}(f +e^c) = (g + e^c) (f + e^c) = gf + ge^c + e^c f + e^c e^c = e + 0 + 0 + e^c = e + e^c = 1_A  \]

So $f + e^c$ is an isomorphism.

$(\Leftarrow)$ Suppose that $f + e^c$ is an isomorphism. We must find an inverse for $f: (A,e) \to (A,e)$ in $\mathsf{Split}(\mathbb{X})$. So consider the map $e(f + e^c)^{-1})e$. We clearly have that $e(f + e^c)^{-1})e: (A,e) \to (A,e)$ is a map in $\mathsf{Split}(\mathbb{X})$, so it remains to show that it is the inverse of $f: (A,e) \to (A,e)$. So we compute:

\begin{gather*}
f e(f + e^c)^{-1})e = e f (f + e^c)^{-1})e = (ef + 0) (f + e^c)^{-1})e \\
=  (ef + ee^c) (f + e^c)^{-1})e = e (f + e^c)  (f + e^c)^{-1})e = e e = e
\end{gather*}

So $f e(f + e^c)^{-1})e = e$, and similarly we can show that $e(f + e^c)^{-1})e f = e$ as well.

A keen-eyed reader may note that in the statement of the lemma, we started with a map in the idempotent splitting, while in <cit.> this is not stated. However, this is not an issue. Indeed, if $x^{k+1}$ has a kernel and cokernel as above, such that $\kappa \lambda$ is an isomorphism, then we can compute that:

\[ x^{k+1}e^c_{\lambda,\kappa} = x^{k+1}(1_A - \lambda (\kappa\lambda)^{-1} \kappa) = x^{k+1} - x^{k+1}  \lambda (\kappa\lambda)^{-1} \kappa = x^{k+1} - 0 = x^{k+1}  \]

So $x^{k+1}e^c_{\lambda,\kappa} = x^{k+1}$ and similarly $x^{k+1}e^c_{\lambda,\kappa} = x^{k+1}$. So $x^{k+1}: (A,e^c_{\lambda,\kappa}) \to (A,e^c_{\lambda,\kappa})$ is a map in the idempotent splitting. Then under the other conditions of the above theorem or <cit.>, by applying the above lemma, the statement that $x^{k+1}: (A, e^c_{\lambda,\kappa}) \to (A, e^c_{\lambda,\kappa})$ is an isomorphism is equivalent to saying that $x^{k+1} + e_{\lambda,\kappa}$ is an isomorphism. Finally, we also recapture the formula for the Drazin inverse in <cit.> in terms of the complement of the induced idempotent (recalling that we showed that $e_{\lambda,\kappa}=e^c_x$ and $e^c_{\lambda,\kappa} = e_x$):

In an additive category $\mathbb{X}$, if $x: A \to A$ is Drazin complement-split and $\mathsf{ind}(x) = k$, then the following equality holds:

\[ x^D = x^k (x^{k+1} + e^c_x)^{-1} = (x^{k+1} + e^c_x)^{-1} x^k \]

As shown in the proof of Theorem <ref>, $x^{k+1}: (A, e_x) \to (A, e_x)$ is an isomorphism in $\mathsf{Split}(\mathbb{X})$. In the same proof, we also showed that $x^D = v x^k = x^k v$, where $v: (A, e_x) \to (A, e_x)$ is the inverse of $x^{k+1}: (A, e_x) \to (A, e_x)$ in $\mathsf{Split}(\mathbb{X})$. However, by Lemma <ref>, we have that $x^{k+1} + e^c_x$ is an isomorphism and that $v = e_x (x^{k+1} + e^c_x ) e_x$. Therefore, we get that $x^D = x^k (x^{k+1} + e^c_x)^{-1} = (x^{k+1} + e^c_x)^{-1} x^k$ as desired.

§.§ Fitting Decomposition
In Section <ref>, we reviewed how to in order to compute the Drazin inverse of a matrix, we first decomposed that matrix as a block diagonal matrix, involving an invertible matrix and a nilpotent matrix, conjugated by another invertible matrix. This approach generalizes to finite-dimensional vector spaces thanks to Fitting's Decomposition Theorem <cit.>. Here, we generalize this decomposition to additive categories with finite biproducts, and call it a Fitting decomposition. Moreover, we show that having a Fitting decomposition is equivalent to being both Drazin split and Drazin complement-split, which we call being Drazin decomposable.

In an additive category $\mathbb{X}$, $x: A\to A$ is Drazin decomposable if $x$ is both Drazin split and Drazin complement-split.

We now work in an additive category with finite biproducts: we denote the biproduct by $\oplus$. To help understand Fitting decompositions, it will be useful to recall the matrix representation for maps between biproducts. Indeed, recall that a map of type $F: A_1 \oplus \hdots \oplus A_n \to B_1 \oplus \hdots \oplus B_m$ is uniquely determined by a family of maps $f_{i,j}: A_i \to B_j$. As such, $F$ can be represented as an $n\times m$ matrix:

\[ F := \begin{bmatrix} f_{1,1} & f_{1,2} & \hdots & f_{1,m} \\
f_{2,1} & f_{2,2} & \hdots & f_{2,m} \\
\vdots & \vdots & \ddots & \vdots \\
f_{n,1} & f_{n,2} & \hdots & f_{n,m}
\end{bmatrix}
\]

Moreover, composition corresponds to matrix multiplication, and identities correspond to the identity matrix. See <cit.> for details of the matrix representation of biproducts.

We now define a Fitting decomposition of an endomorphism by turning <cit.> into a definition:

In an additive category $\mathbb{X}$ with finite biproducts, a Fitting decomposition of $x: A \to A$ is a triple $(p, \alpha, \eta)$ consisting of an isomorphism $p: A \to I \oplus K$, an isomorphism $\alpha: I \to I$, and a nilpotent endomorphism $\eta: K \to K$ such that the following equality holds:

\[ x = p \begin{bmatrix} \alpha & 0 \\ 0 & \eta \end{bmatrix} p^{-1} \]

In an additive category $\mathbb{X}$ with finite biproducts, $x: A \to A$ is Drazin decomposable if and only if $x$ has a Fitting's decomposition. Explicitly:

* If $x$ is Drazin decomposable, where $e_x$ splits via $r: A \to I$ and $s: I \to A$, and $e^c_x$ splits via $r^c: A \to K$ and $s^c: K \to A$, then define the maps $p: A \to I \oplus K$, $\alpha: I \to I$, and $\eta: K \to K$ as follows:

\begin{align*}
p = \begin{bmatrix}
r & r^c
\end{bmatrix} && \alpha = s x r && \gamma_{x^k} = s^c n_x r^c
\end{align*}

Then $(p, \alpha, \eta)$ is a Fitting's decomposition of $x$.

* If $x$ has a Fitting's decomposition $(p, \alpha, \eta)$, then $x$ is Drazin complement-split with Drazin inverse defined as follows:

\[ x^D = p \begin{bmatrix} \alpha^{-1} & 0 \\ 0 & 0 \end{bmatrix} p^{-1} \]

and moreover, writing $p: A \to I \oplus K$ and $p^{-1}: I \oplus K \to I$ in matrix form:

\begin{align*}
p = \begin{bmatrix}
r & r^c
\end{bmatrix} &&  p^{-1} = \begin{bmatrix}
s \\ s^c
\end{bmatrix}
\end{align*}

the induced idempotent $e_x: A \to A$ splits via $r: A \to I$ and $s: I \to A$, and $e_x^c: A \to A$ splits via $r^c: A \to K$ and $s^c: K \to A$.

Suppose that $x$ is Drazin decomposable, where $e_x$ splits via $r: A \to I$ and $s: I \to A$, and $e^c_x$ splits via $r^c: A \to K$ and $s^c: K \to A$. We first show that $p$ is an isomorphism. So define $p^{-1}: I \oplus K \to X$ as follows:

\begin{align*}
p^{-1} = \begin{bmatrix}
s \\ s^c
\end{bmatrix}
\end{align*}

Now using that $e_x = rs$ and $e_x^c = r^c s^c$, we first compute that:

\[  pp^{-1} = \begin{bmatrix}
r & r^c
\end{bmatrix} \begin{bmatrix}
s \\ s^c
\end{bmatrix} = rs + r^c s^c = e_x + e_x^c = e_x + 1_A - e_x = 1_A \]

On the other hand, since $ee^c = 0 = e^c e$, it follows that $s r^c=0$ and $s^c r=0$. So by using that $sr=1_I$ and $s^c r^c = 1_K$, we compute that:

\[  p^{-1} p =\begin{bmatrix}
s \\ s^c
\end{bmatrix} \begin{bmatrix}
r & r^c
\end{bmatrix} =  \begin{bmatrix}
sr & s r^c \\ s^c r & s^c r^c
\end{bmatrix} =  \begin{bmatrix}
1_I & 0 \\ 0 & 1_K
\end{bmatrix} = 1_{I \oplus K} \]

So $p$ is indeed an isomorphism. On the other hand, in the proof of Lemma <ref> we have already shown that $\alpha$ is an isomorphism as well. Now if $\mathsf{ind}(x) =k$, then in the proof of Theorem <ref> we showed that $n_x$ was indeed nilpotent and that $n^{k+1}_x =0$. Moreover, recall that in the proof of Theorem <ref> we also showed that $x^D = c^D_x$ and $n_x c^D_x =0$. So it follows that $e_x n_x =0$ and similarly we also have $n_x e_x =0$. Therefore, we get that $e^c_x n_x = n_x = n_x e^c_x$. Then using this and $e_x^c = r^c s^c$, we compute that:

\[ \eta^{k+1} =  s^c n_x r^c  s^c n_x r^c \hdots  s^c n_x r^c = s^c n_x e^c_x n_x \hdots e^c n_x r^c = s^c n^{k+1}_x r^c = s^c 0 r^c = 0 \]

So $\eta$ is nilpotent as desired. Now recall that by Lemma <ref>, it follows that $r \alpha s = x e_x = x x^D x$. In other words, $r \alpha s = c_x$. On the other hand, from $e^c_x n_x = n_x = n_x e^c_x$, it also follows that $r^c \eta s^c = n_x$. Therefore, by the core-nilpotent decomposition, we finally compute that

\[ p \begin{bmatrix} \alpha & 0 \\ 0 & \eta \end{bmatrix} p^{-1} = \begin{bmatrix}
r & r^c
\end{bmatrix} \begin{bmatrix} \alpha & 0 \\ 0 & \eta \end{bmatrix} \begin{bmatrix}
s \\ s^c
\end{bmatrix} =  \begin{bmatrix}
r \alpha & r^c \eta
\end{bmatrix}  \begin{bmatrix}
s \\ s^c
\end{bmatrix} = r \alpha s + r^c \eta s^c = c_x + n_x = x  \]

and conclude that that $(p, \alpha, \eta)$ is a Fitting's decomposition of $x$ as desired.

We first show that $x$ is Drazin by showing that $x^D$ satisfies the three Drazin inverse axioms.

* First note that for all $m \in \mathbb{N}$, we have that:

\[ x^m = p \begin{bmatrix} \alpha^m & 0 \\ 0 & \eta^m \end{bmatrix} p^{-1}  \]

Now let $k$ be the smallest natural number such that $\eta^{k+1} =0$. Then we compute:

\begin{gather*}
x^{k+1+1} x^d =  p \begin{bmatrix} \alpha^{k+1+1} & 0 \\ 0 & \eta^{k+1+1} \end{bmatrix} p^{-1} p \begin{bmatrix} \alpha^{-1} & 0 \\ 0 & 0 \end{bmatrix} p^{-1} =  p \begin{bmatrix} \alpha^{k+1} \alpha & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} \alpha^{-1} & 0 \\ 0 & 0 \end{bmatrix} p^{-1} \\
=  p \begin{bmatrix} \alpha^{k+1} \alpha \alpha^{-1} & 0 \\ 0 & 0 \end{bmatrix} p^{-1} =  p \begin{bmatrix} \alpha^{k+1} & 0 \\ 0 & \eta^{k+1} \end{bmatrix} p^{-1} = x^{k+1}
\end{gather*}

* We easily compute that:

\begin{gather*} x^D x x^D =  p \begin{bmatrix} \alpha^{-1} & 0 \\ 0 & 0 \end{bmatrix} p^{-1} p \begin{bmatrix} \alpha & 0 \\ 0 & \eta \end{bmatrix} p^{-1}  p \begin{bmatrix} \alpha^{-1} & 0 \\ 0 & 0 \end{bmatrix} p^{-1} \\ = p \begin{bmatrix} \alpha^{-1} & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} \alpha & 0 \\ 0 & \eta \end{bmatrix}  \begin{bmatrix} \alpha^{-1} & 0 \\ 0 & 0 \end{bmatrix} p^{-1} = p \begin{bmatrix} \alpha^{-1} \alpha \alpha^{-1} & 0 \\ 0 & 0 \eta 0 \end{bmatrix} p^{-1} = p \begin{bmatrix} \alpha^{-1} & 0 \\ 0 & 0 \end{bmatrix} p^{-1} = x^D \end{gather*}

* We easily compute that:

\begin{gather*} x x^D =   p \begin{bmatrix} \alpha & 0 \\ 0 & \eta \end{bmatrix} p^{-1}  p \begin{bmatrix} \alpha^{-1} & 0 \\ 0 & 0 \end{bmatrix} p^{-1}  = p \begin{bmatrix} \alpha & 0 \\ 0 & \eta \end{bmatrix}  \begin{bmatrix} \alpha^{-1} & 0 \\ 0 & 0 \end{bmatrix} p^{-1}\\
= p \begin{bmatrix} \alpha^{-1} \alpha  & 0 \\ 0 & \eta 0 \end{bmatrix} p^{-1} = p \begin{bmatrix} 1_I & 0 \\ 0 & 0 \end{bmatrix} p^{-1} = p \begin{bmatrix}  \alpha \alpha^{-1} & 0 \\ 0 & 0 \eta  \end{bmatrix} p^{-1} \\
=  p \begin{bmatrix} \alpha^{-1} & 0 \\ 0 & 0 \end{bmatrix}  \begin{bmatrix} \alpha & 0 \\ 0 & \eta \end{bmatrix} p^{-1} =    p \begin{bmatrix} \alpha^{-1} & 0 \\ 0 & 0 \end{bmatrix} p^{-1} p \begin{bmatrix} \alpha & 0 \\ 0 & \eta \end{bmatrix} p^{-1} = x^D x
\end{gather*}

So $x$ is Drazin. Next, we compute its induced idempotent:

\[ e_x = x x^D =  p \begin{bmatrix} 1_I & 0 \\ 0 & 0 \end{bmatrix} p^{-1} = \begin{bmatrix}
r & r^c
\end{bmatrix} \begin{bmatrix} 1_I & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix}
s \\ s^c
\end{bmatrix} =  \begin{bmatrix}
r & 0
\end{bmatrix} \begin{bmatrix}
s \\ s^c
\end{bmatrix} = rs   \]

So $e_x =rs$. We compute its complement to be:

\[ e^c_x = 1_A - e_x = p p^{-1} =   \begin{bmatrix}
r & r^c
\end{bmatrix} \begin{bmatrix}
s \\ s^c
\end{bmatrix} - e_x = rs + r^c s^c - e_x = e_x + r^c s^c - e_x = r^c s^c \]

So $e^c_x = r^c s^c$. Now note that we also have that:

\[  \begin{bmatrix}
1_I & 0 \\ 0 & 1_K
\end{bmatrix} = 1_{I \oplus K}=  p^{-1} p =\begin{bmatrix}
s \\ s^c
\end{bmatrix} \begin{bmatrix}
r & r^c
\end{bmatrix} =  \begin{bmatrix}
sr & s r^c \\ s^c r & s^c r^c
\end{bmatrix}  \]

In particular, this gives us that $sr = 1_I$ and $s^c r^c = 1_K$. So both $e_x$ and $e^c_x$ split, and we conclude that $x$ is Drazin decomposable as desired.

Theorem <ref> implies that if an additive category with finite biproducts is also idempotent complete, then an endomorphism is Drazin if and only if it has a Fitting decomposition. As such:

An additive category $\mathbb{X}$ with finite biproducts is Drazin complete if and only if every endomorphism has a Fitting decomposition.

Moreover, we can also use the Fitting decomposition to describe the core-nilpotent decomposition.

In an additive category $\mathbb{X}$ with finite biproducts, if $x: A \to A$ has a Fitting's decomposition $(p, \alpha, \eta)$, and is therefore Drazin, then the core and nilpotent-part of $x$ are determined by:

\begin{align*}
c_x :=  p \begin{bmatrix} \alpha & 0 \\ 0 & 0 \end{bmatrix} p^{-1} &&  n_x := p \begin{bmatrix} 0 & 0 \\ 0 & \eta \end{bmatrix} p^{-1}
\end{align*}

We compute:

\begin{gather*} c_X = x x^D x =  p \begin{bmatrix} \alpha & 0 \\ 0 & \eta \end{bmatrix} p^{-1} p \begin{bmatrix} \alpha^{-1} & 0 \\ 0 & 0 \end{bmatrix} p^{-1}  p \begin{bmatrix} \alpha & 0 \\ 0 & \eta \end{bmatrix} p^{-1} = p \begin{bmatrix} \alpha & 0 \\ 0 & \eta \end{bmatrix} \begin{bmatrix} \alpha^{-1} & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} \alpha & 0 \\ 0 & \eta \end{bmatrix} p^{-1}\\
=  p \begin{bmatrix} \alpha \alpha^{-1} \alpha & 0 \\ 0 & \eta 0 \eta \end{bmatrix} p^{-1} = p \begin{bmatrix} \alpha & 0 \\ 0 & 0\end{bmatrix} p^{-1} \end{gather*}

So the desired equality holds. Then by the definition of $n_x$, the other equality holds as well.

§.§ Image-Kernel Decomposition
In Section <ref>, we mentioned that for a ring $R$ and a $R$-module $M$, an $R$-linear endomorphism $f: M \to M$ was Drazin if and only if $f$ had a “Fitting decomposition”, which recall was that $M = \mathsf{im}(f^k) \oplus \mathsf{ker}(f^k)$ for some $k \geq 1$ <cit.>. We generalize this sort of decomposition in an Abelian category and show that it is indeed equivalent to being Drazin.

In an Abelian category $\mathbb{X}$, denote the kernel and image of an endomorphism $x: A \to A$, as follows:

\begin{align*}
\xymatrixcolsep{5pc}\xymatrix{ \mathsf{ker}(x) \ar@{>->}[r]^-{\kappa}\ar@/^3pc/[rrr]^-{0} & A \ar[rr]^-{f}  \ar@{->>}[dr]_-{\epsilon}  && A   \\
& & \mathsf{im}(f) \ar@{>->}[ur]_-{\iota} }
\end{align*}

where recall that $\kappa$ and $\iota$ are monic, and $\epsilon$ is epic. Now define $\psi$ as the canonical map:

\[ \psi = \begin{bmatrix}
\iota \\
\kappa
\end{bmatrix}: \mathsf{im}(x) \oplus \mathsf{ker}(x) \to A  \]

In an Abelian category $\mathbb{X}$, a map $x: A \to A$ has an image-kernel decomposition in case the map $\psi$ as defined above is an isomorphism.

We will now prove that, in an Abelian category, an endomorphism is Drazin if and only if it has a power which has an image-kernel decomposition.

In an Abelian category $\mathbb{X}$, $x: A \to A$ is Drazin if and only if there is some $k \in \mathbb{N}$ such that $x^{k+1}$ has an image-kernel decomposition.

Suppose that $x$ is Drazin with $\mathsf{ind}(x) = k$. Our objective is to show that $\psi: \mathsf{im}(x^{k+1}) \oplus \mathsf{ker}(x^{k+1}) \to A$ is an isomorphism. To construct its inverse, we need to first construct maps $A \to \mathsf{im}(x^{k+1})$ and $A \to \mathsf{ker}(x^{k+1})$. First note that by idempotency and [D.3], we get that $e_x= e_x^{k+1} = (x^D)^{k+1} x^{k+1}$. So by using the universal property of the image, there is a monic map $\mathsf{im}(e_x) \to \mathsf{im}(x^{k+1})$, which then allows us to build a map $\phi_1: A \to \mathsf{im}(x^{k+1})$ such that $\phi_1 \iota = e_x$. On the other hand, recall that in the proof of Theorem <ref>, we showed that $e_x x^{k+1} = x^{k+1}$. Then it follows that $e^c_x x^{k+1} = 0$. So by the universal property of the kernel, let $\phi_2: \mathsf{ker}(x^{k+1}) \to A$ be the unique map such that $\phi_2 \kappa^\circ_{x^{k+1}} = e^c_x$. Then define $\phi$ as:

\[ \phi = \begin{bmatrix}
\phi_1 &
\phi_2
\end{bmatrix}  : A \to \mathsf{im}(x^{k+1}) \oplus \mathsf{ker}(x^{k+1})\]

So we first compute that:

\[ \phi \psi = \begin{bmatrix}
\phi_1 &
\phi_2
\end{bmatrix} \begin{bmatrix}
\iota \\
\kappa
\end{bmatrix}  = \phi_1 \iota + \phi_2 \kappa = e_x + e^c_x = 1_A \]

To prove that $\phi \psi$ is the identity, we must work out what $\iota \phi_1$, $\iota \phi_2$, $\kappa \phi_1$, and $\kappa \phi_2$ are. First, recall again in the proof of Theorem <ref>, we showed that $x^{k+1} e_x =x^{k+1}$, which also implies that $x^{k+1} e^c_x = 0$. So we first compute that:

\[ \epsilon \iota \phi_1 \iota = x^{k+1} e_x = x^{k+1} = \epsilon \iota \]

\[ \epsilon \iota \phi_2 \kappa = x^{k+1} e^c_x = 0 = \epsilon 0 \iota  \]

Since $\epsilon$ is epic and $\iota$ is monic, it follows that $\iota \phi_1 = 1_{\mathsf{im}(x^{k+1})}$ and $\iota \phi_2 =0$. Next, using idempotency and [D.3], $e_x = x^{k+1} (x^D)^{k+1}$, then we have that $\kappa e_x =0$. So we compute:

\[ \kappa \phi_1 \iota = \kappa e_x = 0 = 0 \iota  \]

\[ \kappa \phi_2 \kappa = \kappa e^c_x = \kappa (1_A -e_x) = \kappa - \kappa e_x = \kappa - 0 = \kappa   \]

Since $\iota$ and $\kappa$ are monic, it follows that $\kappa \phi_1 = 0$ and $\kappa \phi_2 = 1_{\mathsf{ker}(x^{k+1})}$. So finally we get that:

\[ \psi \phi = \begin{bmatrix}
\iota \\
\kappa
\end{bmatrix}  \begin{bmatrix}
\phi_1 &
\phi_2
\end{bmatrix}  =  \begin{bmatrix}
\iota \phi_1 &\iota \phi_2
\kappa \phi_1 & \kappa \phi_2
\end{bmatrix} =   \begin{bmatrix}
1_{\mathsf{im}(x^{k+1})} & 0\\
0 & 1_{\mathsf{ker}(x^{k+1})}
\end{bmatrix} = 1_{\mathsf{im}(x^{k+1}) \oplus \mathsf{ker}(x^{k+1})} \]

So we conclude that $\psi: \mathsf{im}(x^{k+1}) \oplus \mathsf{ker}(x^{k+1}) \to A$ is an isomorphism.

Suppose that $\psi: \mathsf{im}(x^{k+1}) \oplus \mathsf{ker}(x^{k+1}) \to A$ is an isomorphism. To show that $x$ is Drazin, we will show that $\psi$ is part of a Fitting decomposition of $x$. So let $\phi_1: A \to \mathsf{im}(x^{k+1})$ and $\phi_2: A \to \mathsf{ker}(x^{k+1})$ be the components of $\psi^{-1}: A \to \mathsf{im}(x^{k+1}) \oplus \mathsf{ker}(x^{k+1})$. Then from $\psi \psi^{-1} = 1_{\mathsf{im}(x^{k+1}) \oplus \mathsf{ker}(x^{k+1})}$ and $\psi^{-1} \psi = 1_A$, we get the following equalities:

\begin{gather*}
\iota \phi_1 = 1_{\mathsf{im}(x^{k+1})} \qquad \iota \phi_2 =0 \qquad \kappa \phi_1 = 0 \qquad  \kappa \phi_2 = 1_{\mathsf{ker}(x^{k+1})} \qquad \phi_1 \iota + \phi_2 \kappa = 1_A
\end{gather*}

Now for the isomorphism part of our decomposition, define $\alpha: \mathsf{im}(x^{k+1}) \to \mathsf{im}(x^{k+1})$ to be the composite $\alpha \colon = \iota x \phi_1$. To show that $\alpha$ is an isomorphism, we will use the fact that in an Abelian category, a map is an isomorphism if and only if it is monic and epic. To show that $\alpha$ is monic and epic, we will need to first compute some useful identities. To start with, $\epsilon \iota = x^{k+1}$ and $\iota \phi_1 = 1_{\mathsf{im}(x^{k+1})}$ together imply that $x^{k+1} \phi_1 = \epsilon$. Moreover, from $\epsilon \iota = x^{k+1}$ and $\kappa x^{k+1} =0$, since $\iota$ is monic, it follows that $\kappa \epsilon = 0$. From this and $\phi_1 \iota + \phi_2 \kappa = 1_A$ we also get that $\phi_1 \iota \epsilon = \epsilon$. On the other hand, we also get:

\[ x \epsilon = x x^{k+1} \phi_1 = x^{k+1} x \phi_1 = \epsilon \iota x \phi_1 = \epsilon \alpha\]

So $x \epsilon = \epsilon \alpha$. From this, we have:

\[ \epsilon \alpha \iota = x \epsilon \iota = x x^{k+1} = x^{k+1} x = \epsilon \iota x \]

Then since $\epsilon$ is epic, we get that $\alpha \iota = \iota x$, which also tells us that $\iota x^n = \alpha^n \iota$ for all $n \in \mathbb{N}$. Then from $\iota \phi_1 = 1_{\mathsf{im}(x^{k+1})}$, we get that $\alpha^n = \iota x^n \phi_1$.

Now since we are in an Abelian category, to show that $\alpha$ is monic, it suffices to show that if $f \alpha =0$ then $f=0$. So suppose that $f \alpha =0$. Then we have that:

\[ f \iota x^{k+1} = f \iota x x^k = f \alpha \iota x^k = 0 \iota x^k = 0 \]

So $f \iota x^{k+1} =0$, and so by the universal property of the kernel, there exists a unique $h$ such that $f \iota = h \kappa$. Post-composing both sides by $\phi_2$, since $\kappa \phi_2 = 1_{\mathsf{ker}(x^{k+1})}$ and $\iota \phi_2 =0$, we get that $h=0$ and so $f \iota =0$. However, $\iota$ is monic, therefore $f=0$ as desired. So $\alpha$ is monic. Dually, since we are in an Abelian category, to show that $\alpha$ is epic, it suffices to show that if $\alpha g= 0$ then $g=0$. So suppose that $\alpha g=0$. Then

\[ \epsilon g = \phi_1 \iota \epsilon g = \phi_1 \iota x^{k+1} \phi_1 g = \phi_1 \iota \alpha^{k+1} g = \phi_1 \iota \alpha^k \alpha g = \phi_1 \iota \alpha^k  0 = 0 \]

Since $\epsilon$ is epic, it follows that $g=0$, as desired. So $\alpha$ is epic. Therefore, we conclude that $\alpha$ is an isomorphism.

For the nilpotent part of the decomposition, note that since $\kappa x^{k+1} =0$, we also have that $\kappa x x^{k+1} =0$. Therefore, by the universal property of the kernel, there exists a unique map $\eta: \mathsf{ker}(x^{k+1}) \to \mathsf{ker}(x^{k+1})$ such that $\eta \kappa = \kappa x$. However, since $\kappa \phi_2 = 1_{\mathsf{ker}(x^{k+1})}$, we get that $\eta = \kappa x \phi_2$. Moreover, we compute that:

\[ \kappa x = \kappa x  \left(\phi_1 \iota + \phi_2 \kappa \right) = \kappa x \phi_1 \iota +  \kappa x \phi_2 \kappa = \eta \kappa \phi_1 \iota + \kappa x \phi_2 \kappa = 0 + \kappa x \phi_2 \kappa = \kappa x \phi_2 \kappa \]

So $\kappa x = \kappa x \phi_2 \kappa$. From this, we can prove by induction that $\eta^{m} = \kappa x^m \phi_2$. The base case $m=0$ follows from $\kappa \phi_2 = 1_{\mathsf{ker}(x^{k+1})}$. So suppose that for all $0 \leq j \leq n$, $\eta^{n} = \kappa x^n \phi_2$.

\[ \eta^{n+1} = \eta \eta^n = \eta \kappa x^n \phi_2 = \kappa x \phi_2 \kappa x^n \phi_2 = \kappa x x^n \phi_2 = \kappa x^{n+1} \phi_2 \]

So we do have that $\eta^{m} = \kappa x^m \phi_2$ for all $m \in \mathbb{N}$. Now for $m=k+1$, we get that:

\[ \eta^{k+1} = \kappa x^{k+1} \phi_2 = 0 \phi_2 = 0    \]

So $\eta$ is nilpotent as desired.

Lastly, we finally compute that:

\begin{gather*}
\psi^{-1} \begin{bmatrix}
\alpha & 0 \\
0 & \eta
\end{bmatrix} \psi = \begin{bmatrix}
\phi_1 &
\phi_2
\end{bmatrix} \begin{bmatrix}
\alpha & 0 \\
0 & \eta
\end{bmatrix} \begin{bmatrix}
\iota \\
\kappa
\end{bmatrix} = \begin{bmatrix}
\phi_1 \alpha &
\phi_2 \eta
\end{bmatrix} \begin{bmatrix}
\iota \\
\kappa
\end{bmatrix} \\
= \phi_1 \alpha \iota + \phi_2 \eta \kappa = \phi_1 \iota x + \phi_2 \kappa x =  \left(\phi_1 \iota + \phi_2 \kappa \right) x = x
\end{gather*}

So we conclude that $(\psi^{-1}, \alpha, \eta)$ is a Fitting's decomposition of $x$. So by Theorem <ref>, we get that $x$ is Drazin.

§.§ Fitting's Results

In an Abelian category, if every endomorphism of an object $A$ has some non-trivial power which has an image-kernel decomposition then, by Theorem <ref>, $A$ is a Drazin object. From Section <ref>, we know that finite length modules of any ring are Drazin. This means that every endomorphism of a finite length module is Drazin, and so, using Theorem <ref>, has a non-trivial power with an image-kernel decomposition. This observation is often called Fitting's Decomposition Theorem <cit.>:

For any ring $R$ and any finite length $R$-module $M$, every $R$-linear endomorphism $f:M \to M$ has a power $f^k$, for some $k \geq 1$, with an image-kernel decomposition.

Fitting also made another striking observation, often called Fitting's Lemma: namely that every endomorphism of an indecomposible finite length module is either an isomorphism or a nilpotent:

For any ring $R$ and any finite length indecomposable $R$-module $M$, every endomorphism is either an automorphism or nilpotent.

We may see this immediately: we know that any Drazin endomorphism has a Fitting decomposition and, thus, as the module is indecomposable, only one of the nilpotent part or the core part can be non-zero.  This has the consequence that the endomorphism ring of an indecomposable finite length module is local.  Being local means there is a maximal ideal outside of which every element is invertible.  Of course, the maximal ideal in this case is the collection of all the nilpotent elements, in other words, it is the nilpotent radical. It is worth recalling that the nilpotent elements always form a two-sided ideal in a commutative ring but do not necessarily do so in a noncommutative ring.  However, in a noncommutative ring in which every element is either a unit (an automorphism) or nilpotent, the nilpotent elements do form a two-sided ideal. The main difficulty is to show that a nilpotent element, $n$, when multiplied on the left (or right) by an element $r$, still has the result nilpotent. Of course, it suffices to show that $r n$ (and $n r$) cannot be a unit. For this note that $r n n^{k-1} = r 0 = 0$ (and $n^{k-1} n r = 0 r = 0$), where $k$ is the nilpotent index of $n$, so that neither $r n$ nor $n r$ are units which means they are nilpotent.

§ DRAZIN INVERSES OF OPPOSING PAIRS OF MAPS

Arriving with categorical eyes to the subject of Drazin inverses it is natural to want to have a Drazin inverse of an arbitrary map. However, to have a Drazin inverse of a map $f: A \to B$, one really needs an opposing map $g: B \to A$ to allow for the iteration which is at the heart of the notion of a Drazin inverse. The Drazin inverse of an opposing pair, $(f,g)$, is itself an opposing pair whose properties we develop in this section.

§.§ The Drazin inverse of an opposing pair
In a category $\mathbb{X}$, we denote a pair of maps of dual type $f: A \to B$ and $g: B \to A$ by $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ and refer to it as an opposing pair.

In a category $\mathbb{X}$, a Drazin inverse of $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ is an opposing pair $\xymatrixcolsep{1.5pc}\xymatrix{(g^{\frac{D}{f}},f^{\frac{D}{g}}): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ satisfying the following properties:

* There is a $k \in \mathbb{N}$ such that $(fg)^k f f^{\frac{D}{g}} = (fg)^k$ and $(gf)^k g g^{\frac{D}{f}} = (gf)^k$.

* $f^{\frac{D}{g}} f f^{\frac{D}{g}} = f^{\frac{D}{g}}$ and $g^{\frac{D}{f}} g g^{\frac{D}{f}} = g^{\frac{D}{f}}$;

* $ f f^{\frac{D}{g}} =  g^{\frac{D}{f}} g$ and $ f^{\frac{D}{g}} f =  g g^{\frac{D}{f}}$.

The map $f^{\frac{D}{g}}: B \to A$ is called the Drazin inverse of $f$ over $g$, while the map $g^{\frac{D}{f}}: A \to B$ is called the Drazin inverse of $g$ over $f$. The Drazin index of $(f,g)$ is the smallest $k \in \mathbb{N}$ such that [DV.1] holds, which we denote by $\mathsf{ind}(f,g)=k$.

There is a slight redundancy in the above definition in that, thanks to [DV.3], we do not need to assume both equalities for [DV.1].

In a category $\mathbb{X}$, if for $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ there is a $\xymatrixcolsep{1.5pc}\xymatrix{(g^{\frac{D}{f}},f^{\frac{D}{g}}): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ which satisfies [DV.3], then the following are equivalent:

* [DV.1] holds;

* There is a $p \in \mathbb{N}$ such that $(fg)^p f f^{\frac{D}{g}} = (fg)^p$;

* There is a $q \in \mathbb{N}$ such that $(gf)^q g g^{\frac{D}{g}} = (gf)^q$.

We prove that $(i) \Rightarrow (ii) \Rightarrow (iii)$. Clearly, we already have that $(i) \Rightarrow (ii)$. For $(ii) \Rightarrow (iii)$, suppose that there is a $p \in \mathbb{N}$ such that $(fg)^p f f^{\frac{D}{g}} = (fg)^p$. Then using [DV.3] and that $(gf)^{p+1} = g(fg)^p f$, we compute:

\[ (gf)^{p+1} g g^{\frac{D}{g}} = g(fg)^p f g g^{\frac{D}{f}} = g(fg)^p f f^{\frac{D}{g}} f = g(fg)^p f  = (gf)^{p+1} \]

So $(iii)$ holds. For $(iii) \Rightarrow (i)$, suppose that there is a $q \in \mathbb{N}$ such that $(gf)^q g g^{\frac{D}{g}} = (gf)^q$. Now that we also have that $(gf)^{q+1} g g^{\frac{D}{g}} = (gf)^{q+1}$. On the othr hand, using [DV.3] and that $(fg)^{q+1} = f(gf)^q g$, we compute:

\[ (fg)^{q+1} f f^{\frac{D}{g}} = f(gf)^q g f f^{\frac{D}{g}} = f(gf)^q g g^{\frac{D}{f}} g = f (gf)^q g = (gf)^{q+1} \]

So [DV.1] holds. So we conclude that $(i) \Leftrightarrow (ii) \Leftrightarrow (iii)$ as desired.

Drazin inverses of opposing pairs share many of the same properties as Drazin inverses of endomorphisms, such as the fact that they are unique (if they exist) and also absolute. Before establishing these properties, we will show how Drazin inverses of opposing pairs are linked to Drazin inverses of endomorphisms.

§.§ Drazin Opposing Pairs
From an opposing pair $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$, we get two endomorphisms $fg: A \to A$ and $gf: B \to B$. Our objective is to show that $(f,g)$ has a Drazin inverse if and only if the composites $fg$ and $gf$ both have Drazin inverses. To do so, we first show that, surprisingly, if one of these composites is Drazin, then so is the other one: this is sometimes known as Cline's formula [11].

In a category $\mathbb{X}$, for an opposing pair $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$, $fg: A \to A$ is Drazin if and only if $gf: B \to B$ is Drazin. Explicitly,

* If $fg$ is Drazin, then $gf$ is Drazin where $(gf)^D \colon = g (fg)^D (fg)^D f$;

* If $gf$ is Drazin, then $fg$ is Drazin where $(fg)^D \colon = f (gf)^D (gf)^D g$.

($\Rightarrow$) Suppose that $fg$ is Drazin. We must show that $(gf)^D \colon = g (fg)^D (fg)^D f$ satisfies the three axioms of a Drazin inverse.

* Suppose that $\mathsf{ind}(fg)=k$. Then using Lemma <ref>.(<ref>) for $fg$, we compute:

\begin{gather*}
(gf)^{k+2} (gf)^D = g (fg)^{k+1} f (gf)^D = g (fg)^{k+1} f g (fg)^D (fg)^D f \\
= g (fg)^{k+2} ((fg)^D)^2 f = g (fg)^k f = (gf)^{k+1}
\end{gather*}

* Using [D.2] and [D.3] for $fg$, we compute that:

\begin{gather*} (gf)^D gf (gf)^D = g (fg)^D (fg)^D f gf g (fg)^D (fg)^D f = \\
g (fg)^D f g (fg)^D  (fg)^D f g (fg)^D f = g (fg)^D (fg)^D f = (gf)^D     \end{gather*}

* Using [D.3] for $fg$, we compute that:

\begin{gather*} (gf)^D gf = g (fg)^D (fg)^D f g f = g (fg)^D f g (fg)^D  f \\
= g f g (fg)^D  (fg)^D  f = gf (gf)^D  \end{gather*}

So we conclude that $gf$ is Drazin.

($\Leftarrow$) Suppose that $gf$ is Drazin. By similar calculations as above, we can show that $(fg)^D \colon = f (gf)^D (gf)^D g$ satisfies the three axioms of a Drazin inverse. So we conclude that $fg$ is Drazin.

This suggests the following definition:

In a category $\mathbb{X}$, a Drazin opposing pair is an opposing pair

$\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ such that $fg$ or $gf$ is Drazin (so, by Lemma <ref>,  both are Drazin).

Below we will show that being a Drazin opposing pair is equivalent to having a Drazin inverse for the pair. In order to prove this, it will be useful to establish the following identities for Drazin opposing pairs:

In a category $\mathbb{X}$, if $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ is a  Drazin opposing pair, then we have that $(fg)^D f = f (gf)^D$ and $(gf)^D g = g (fg)^D$.

Suppose that $(f,g)$ is a Drazin opposing pair. Then by Lemma <ref> we have that $(gf)^D = g (fg)^D (fg)^D f$ and $(fg)^D = f (gf)^D (gf)^D g$. Then using [D.2] and [D.3] for $gf$, we compute that:

\[ (fg)^D f = f (gf)^D (gf)^D g f = f (gf)^D g f (gf)^D = f (gf)^D \]

So $(fg)^D f = f (gf)^D$. Similarly, we can also show that $(gf)^D g = g (fg)^D$.

We may now prove the promised equivalence between a Drazin opposing pair and an opposing pair which has a Drazin inverse:

In a category $\mathbb{X}$, $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ has a Drazin inverse if and only if $(f,g)$ is a Drazin opposing pair. Explicitly:

* If $(f,g)$ has a Drazin inverse $(g^{\frac{D}{f}},f^{\frac{D}{g}})$
then $(f,g)$ is a Drazin opposing pair where $(fg)^D \colon = g^{\frac{D}{f}} f^{\frac{D}{g}}$ with $\mathsf{ind}(fg) \leq \mathsf{ind}(f,g)$ and $(gf)^D \colon = f^{\frac{D}{g}}g^{\frac{D}{f}}$ with $\mathsf{ind}(gf) \leq \mathsf{ind}(f,g)$;

* If $(f,g)$ is a Drazin opposing pair then $(f,g)$ has a Drazin inverse $(g^{\frac{D}{f}},f^{\frac{D}{g}})$ where $f^{\frac{D}{g}} := g (fg)^D = (gf)^D g$ and $g^{\frac{D}{f}} := f (gf)^D = (fg)^D f$, and we also have that $\mathsf{ind}(fg) \leq \mathsf{max}(\mathsf{ind}(f,g), \mathsf{ind}(gf))$.

($\Rightarrow$) Suppose that $(f,g)$ has a Drazin inverse $(g^{\frac{D}{f}},f^{\frac{D}{g}})$ with $\mathsf{ind}(f,g)=k$. We first show that $(fg)^D \colon = g^{\frac{D}{f}} f^{\frac{D}{g}}$ satisfies the three axioms of a Drazin inverse.

* Note that $(fg)^{k+1}=f (gf)^k g$ and also that $f (gf)^{k}= (fg)^{k} f$. Then using [DV.1], we compute that:

\begin{gather*}
(fg)^{k+1} (fg)^D \!=\! f (gf)^k g (fg)^D \!=\! f (gf)^{k} g g^{\frac{D}{f}} f^{\frac{D}{g}} = f (gf)^{k}  f^{\frac{D}{g}} = (fg)^{k} f f^{\frac{D}{g}} \!=\! (fg)^{k}
\end{gather*}

* Using [DV.2] and [DV.3], we compute that:

\begin{gather*}
(fg)^D fg (fg)^D \!=\! g^{\frac{D}{f}} f^{\frac{D}{g}} fg g^{\frac{D}{f}} f^{\frac{D}{g}} = g^{\frac{D}{f}} g g^{\frac{D}{f}} g g^{\frac{D}{f}} f^{\frac{D}{g}}  = g^{\frac{D}{f}} g g^{\frac{D}{f}} f^{\frac{D}{g}} = g^{\frac{D}{f}} f^{\frac{D}{g}} = (fg)^D
\end{gather*}

* Using [D.3] for $fg$, we compute that:

\begin{gather*}
(fg)^D fg  \!=\! g^{\frac{D}{f}} f^{\frac{D}{g}} fg = g^{\frac{D}{f}} g g^{\frac{D}{f}} g = g^{\frac{D}{f}} g = f f^{\frac{D}{g}} = f f^{\frac{D}{g}} f f^{\frac{D}{g}} = f g g^{\frac{D}{f}} f^{\frac{D}{g}} = fg (fg)^D
\end{gather*}

So we conclude that $fg$ is Drazin with $\mathsf{ind}(fg) \leq \mathsf{ind}(f,g)$. Then by Lemma <ref>, we also get that $gf$ is Drazin, and it is easy to check that $\mathsf{ind}(gf) \leq \mathsf{ind}(f,g)$ as well. Now using the formula from Lemma <ref>, we can also check using [DV.2] and [DV.3] that:

\begin{gather*}
(gf)^D = g (fg)^D (fg)^D f = g g^{\frac{D}{f}} f^{\frac{D}{g}} g^{\frac{D}{f}} g g^{\frac{D}{f}}= f^{\frac{D}{g}} g^{\frac{D}{f}}
\end{gather*}

So we conclude that $(f,g)$ is a Drazin opposing pair.

($\Leftarrow$) Suppose that $(f,g)$ is a Drazin opposing pair. Note by Lemma <ref> that setting $f^{\frac{D}{g}} := g (fg)^D = (gf)^D g$ and $g^{\frac{D}{f}} := f (gf)^D = (fg)^D f$ is well-defined. So now we must show that $(g^{\frac{D}{f}},f^{\frac{D}{g}})$ satisfies the three axioms of a Drazin inverse. We start with the third one:

* Using [D.3] for $fg$, we compute that:

\begin{gather*}
f f^{\frac{D}{g}} = f g (fg)^D = (fg)^D fg = g^{\frac{D}{f}} g
\end{gather*}

So $ f f^{\frac{D}{g}} =  g^{\frac{D}{f}} g$, and similarly we can show that $f^{\frac{D}{g}} f =  g g^{\frac{D}{f}}$.

* Let $k = \mathsf{max}(\mathsf{ind}(fg), \mathsf{ind}(gf))$. Then by Lemma <ref>.(<ref>) for $fg$, we compute that:

\begin{gather*}
(fg)^k f f^{\frac{D}{g}} = (fg)^k f g (fg)^D = (fg)^{k+1} (fg)^D = (fg)^k
\end{gather*}

As such, since [DV.3] holds, by Lemma <ref> we then get that [DV.1] holds.

* Using [D.2] for $gf$, we compute that:

\begin{gather*}
f^{\frac{D}{g}}  f f^{\frac{D}{g}} =  (gf)^D g f (gf)^D g = (gf)^D g = f^{\frac{D}{g}}
\end{gather*}

So $ f^{\frac{D}{g}}  f f^{\frac{D}{g}} = f^{\frac{D}{g}}$, and similarly we can show that $g^{\frac{D}{f}}  g g^{\frac{D}{f}}  =g$ as well.

So we conclude that $(f,g)$ has a Drazin inverse and $\mathsf{ind}(fg) \!\leq\! \mathsf{max}(\mathsf{ind}(f,g), \mathsf{ind}(gf))$.

Therefore, from now on we will simply say that:

In a category $\mathbb{X}$, an opposing pair $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ is Drazin if $(f,g)$ is a Drazin opposing pair or equivalently if $(f,g)$ has a Drazin inverse.

§.§ Properties of Drazin Opposing Pairs
Now that we have related Drazin inverses of opposing pairs to usual Drazin inverses of endomorphisms, we can use this to access properties of Drazin opposing pairs which are analogues for those of Drazin endomorphisms.

We begin with the all-important fact that Drazin inverses of opposing pairs, if they exist, are unique.

In a category $\mathbb{X}$, if $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ has a Drazin inverse, then it is unique.

Suppose that $(f,g)$ has two Drazin inverse $(h,k)$ and $(p,q)$. By Theorem <ref>, $(f,g)$ is also Drazin, so $fg$ and $gf$ are both Drazin, and we may use $(h,k)$ and $(p,q)$ to build their Drazin inverses. Since Drazin inverses are unique, we must have that $pq = (fg)^D = hk$ and $qp = (gf)^D = kh$. Now using [DV.2] and [DV.3], we compute:

\begin{gather*}
h = h g h = h k f = p q f = p g p = p
\end{gather*}

So $h=p$, and similarly we can also show that $k=q$. So we conclude that Drazin inverses of opposing pairs are unique.

This allows us to talk of the Drazin inverse of  $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$
and to write it as:

\[ \xymatrixcolsep{1.5pc}\xymatrix{(f,g)^D := (g^{\frac{D}{f}},f^{\frac{D}{g}}): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}\]

Moreover, implies that the formulas of the Drazin inverses in Theorem <ref> are canonical.

Drazin inverses of opposing pairs are also absolute:

Let $\mathsf{F}: \mathbb{X} \to \mathbb{Y}$ be a functor and let $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ be Drazin in $\mathbb{X}$, then the opposing pair $\xymatrixcolsep{1.5pc}\xymatrix{(\mathsf{F}(f),\mathsf{F}(g)): \mathsf{F}(A<EMAIL_ADDRESS>& \mathsf{F}(B<EMAIL_ADDRESS>}$ is Drazin where in particular $\mathsf{F}(f)^{\frac{D}{\mathsf{F}(g)}} = \mathsf{F}(f^{\frac{D}{g}})$ and $\mathsf{F}(g)^{\frac{D}{\mathsf{F}(f)}} = \mathsf{F}(g^{\frac{D}{f}})$.

Suppose that $(f,g)$ is Drazin, which in particular means that $fg$ and $gf$ are Drazin. By Proposition <ref>, Drazin inverses are absolute, so $\mathsf{F}(fg)= \mathsf{F}(f) \mathsf{F}(g)$ and $\mathsf{F}(gf)= \mathsf{F}(g) \mathsf{F}(f)$ are also Drazin. As such, we get that $(\mathsf{F}(f),\mathsf{F}(g))$ is Drazin. Moreover, applying the formulas from Theorem <ref>, we compute that:

\begin{gather*}
\mathsf{F}(f)^{\frac{D}{\mathsf{F}(g)}} = \mathsf{F}(g) \mathsf{F}(fg)^D =  \mathsf{F}(g) \mathsf{F}( (fg)^D ) = \mathsf{F}( g (fg)^D) = \mathsf{F}(\mathsf{F}(f^{\frac{D}{g}}))
\end{gather*}

So $\mathsf{F}(f)^{\frac{D}{\mathsf{F}(g)}} = \mathsf{F}(f^{\frac{D}{g}})$, and similarly we can also compute that $\mathsf{F}(g)^{\frac{D}{\mathsf{F}(f)}} = \mathsf{F}(g^{\frac{D}{f}})$.

Much as before, the Drazin inverse of an opposing pair commutes with everything with which the opposing pair commutes:

In a category $\mathbb{X}$, suppose that $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ and $\xymatrixcolsep{1.5pc}\xymatrix{(u,v): A^\prime<EMAIL_ADDRESS>& B^\prime<EMAIL_ADDRESS>}$ are Drazin. Then if the diagram on the left commutes, then the diagram on the right commutes:

\[\begin{array}[c]{c}
\xymatrixcolsep{5pc}\xymatrix{A \ar[d]_-f \ar[r]^-a & A' \ar[d]^-v \\ B \ar[d]_-g \ar[r]^-b	 & B' \ar[d]^-w  \\ A \ar[r]_-a & A' }
\end{array}  \Rightarrow \begin{array}[c]{c}    \xymatrixcolsep{5pc}\xymatrix{A \ar[d]_-{g^{\frac{D}{f}}} \ar[r]^-a & A' \ar[d]^-{w^{\frac{D}{v}}} \\ B \ar[d]_-{f^{\frac{D}{g}}} \ar[r]^-b	 & B' \ar[d]^-{v^{\frac{D}{w}}}  \\ A \ar[r]_-a & A' }
\end{array}   \]

If $(f,g)$ and $(u,v)$ are Drazin, we have that $fg$ and $vw$ are Drazin. So by Lemma <ref>, the outer square of the diagram on the left implies that $(fg)^D a = a (vw)^D$. Now by using this and the formulas in Theorem <ref>, we compute that:

\[ g^{\frac{D}{f}} b = (fg)^D f b = (fg)^Da v = a (vw)^D v = a w^{\frac{D}{v}}\]

So $g^{\frac{D}{f}} b = a w^{\frac{D}{v}}$, and similarly we can also show that $f^{\frac{D}{g}} a = b v^{\frac{D}{w}}$. So the diagram on the right commutes as desired.

It is straightforward to see that if an opposing pair is Drazin, then its symmetric pair will also be Drazin:

In a category $\mathbb{X}$, $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ is Drazin if and only if $\xymatrixcolsep{1.5pc}\xymatrix{(g,f): B<EMAIL_ADDRESS>& A<EMAIL_ADDRESS>}$ is Drazin.

By definition, $(f,g)$ is Drazin if and only if $fg$ is Drazin  or $gf$ is Drazin if and only if $(g,f)$ is Drazin.

We can also recover the Drazin inverses of an endomorphism from the Drazin inverse of the opposing pair consisting of the endomorphism paired with the identity: the Drazin inverse of an endomorphism over the identity is the usual Drazin inverse.

In a category $\mathbb{X}$, $x: A \to A$ is Drazin if and only if $\xymatrixcolsep{1.5pc}\xymatrix{(x,1_A): A<EMAIL_ADDRESS>& A<EMAIL_ADDRESS>}$ is Drazin, or equivalently if $\xymatrixcolsep{1.5pc}\xymatrix{(1_A,x): A<EMAIL_ADDRESS>& A<EMAIL_ADDRESS>}$ is Drazin. Explicitly:

* If $x$ is Drazin then $(x,1_A)$ has a Drazin inverse with $x^{\frac{D}{1_A}} = x^D$ and $1^{\frac{D}{x}} = x x^D$;

* If $(x,1_A)$ is Drazin, then $x$ is Drazin where $x^D = x^{\frac{D}{1_A}} 1^{\frac{D}{x}} = 1^{\frac{D}{x}} x^{\frac{D}{1_A}}$.

Suppose that $x$ is Drazin. Then trivially $x 1_A = x = 1_A x$ is Drazin. So $(x,1_A)$ (or equivalent $(1_A, x)$) is Drazin. Applying the formulas from Theorem <ref>, we get that:

\begin{gather*}
x^{\frac{D}{1_A}} = 1_A (1_A x)^D = x^D \qquad 1^{\frac{D}{x}} = x (1_A x)^D = x x^D
\end{gather*}

($\Leftarrow$) Suppose that $(x,1_A)$ is Drazin. So $x 1_A= x = 1_A x$ is Drazin. Applying the formulas from Theorem <ref>, we compute that:

\[ x^D = (1_A x)^D = x^{\frac{D}{1_A}} 1^{\frac{D}{x}}  \qquad x^D = (x1_A)^D = 1^{\frac{D}{x}} x^{\frac{D}{1_A}} \]

Thus, we have yet another equivalent characterization of when a category is Drazin.

A category $\mathbb{X}$ is Drazin if and only if every opposing pair in $\mathbb{X}$ is Drazin.

If $\mathbb{X}$ is Drazin, then for an opposing pair $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$, $fg$ and $gf$ are both Drazin, so $(f,g)$ is Drazin. Conversely, if every opposing pair is Drazin, then by Lemma <ref>, every endomorphism must be Drazin.

§.§ Opposing Pairs of Isomorphisms
We now show that an opposing pair has a Drazin index of zero precisely when it is a pair of opposing isomorphisms. The Drazin inverse of an opposing pair of isomorphisms is the opposing pair of their inverses.

In a category $\mathbb{X}$, $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ is Drazin with $\mathsf{ind}(f,g) =0$ if and only if $f$ and $g$ are isomorphism. Explicitly:

* If $(f,g)$ is Drazin with $\mathsf{ind}(f,g) =0$, then $f$ and $g$ are isomorphisms where $f^{-1} := f^{\frac{D}{g}}$ and $g^{-1} := g^{\frac{D}{f}}$;

* If $f$ and $g$ are isomorphisms, then $(f,g)$ is Drazin with $f^{\frac{D}{g}} :=  f^{-1}$ and $g^{\frac{D}{f}} :=  g^{-1}$, and also $\mathsf{ind}(f,g) =0$.

$(\Rightarrow)$ Suppose that $(f,g)$ is Drazin with $\mathsf{ind}(f,g) =0$. Then [DV.1] tells us that $f f^{\frac{D}{g}} = 1_A$ and $g g^{\frac{D}{f}} = 1_B$. Now [DV.3] gives us that $f^{\frac{D}{g}} f = g g^{\frac{D}{f}} = 1_B$ and $g^{\frac{D}{f}} g= f f^{\frac{D}{g}} = 1_A$. So we get that $f$ and $g$ are indeed isomorphisms with respective inverses $f^{-1} = f^{\frac{D}{g}}$ and $g^{\frac{D}{f}}= g^{-1}$.

$(\Leftarrow)$ Suppose that $f$ and $g$ are isomorphisms. Then the composites $fg$ and $gf$ are also isomorphisms. By Lemma <ref>, we get that $fg$ and $gf$ are Drazin with $\mathsf{ind}(fg) = 0$ and $\mathsf{ind}(gf) = 0$. As such, $(f,g)$ is Drazin. Applying the formulas from Theorem <ref> and Lemma <ref>, we compute that:

\[ f^{\frac{D}{g}} = g (fg)^D = g (fg)^{-1} = g g^{-1} f^{-1} = f^{-1} \]

So $f^{\frac{D}{g}} =  f^{-1}$, and similarly we also get that $g^{\frac{D}{f}} =  g^{-1}$. Moreover, we also have that $\mathsf{ind}(f,g) =0$.

§.§ Group Inverses of Opposing Pairs
Recall that for endomorphisms, having a Drazin index less than or equal to one was equivalent to having a group inverse. So here we introduce the analogue notion of a group inverse for opposing pairs. We then show that for opposing pairs, having a group inverse is again equivalent to having a Drazin index less than or equal to one.

In a category $\mathbb{X}$, a group inverse of $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ is an opposing pair $\xymatrixcolsep{1.5pc}\xymatrix{(g^{\frac{D}{f}},f^{\frac{D}{g}}): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ satisfying the following properties:

* $fg g^{\frac{D}{f}} g = fg$ and $g f f^{\frac{D}{g}} f = gf$.

* $f^{\frac{D}{g}} f f^{\frac{D}{g}} = f^{\frac{D}{g}}$ and $g^{\frac{D}{f}} g g^{\frac{D}{f}} = g^{\frac{D}{f}}$;

* $ f f^{\frac{D}{g}} =  g^{\frac{D}{f}} g$ and $ f^{\frac{D}{g}} f =  g g^{\frac{D}{f}}$.

In a category $\mathbb{X}$, for an opposing pair $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$, the following are equivalent:

* $(f,g)$ is Drazin with $\mathsf{ind}(f,g) \leq 1$;

* $fg$ and $gf$ have group inverses;

* $(f,g)$ has a group inverse.

($(i) \Rightarrow (ii)$) Suppose that $(f,g)$ is Drazin with $\mathsf{ind}(f,g) \leq 1$. So by Theorem <ref>, we have that $fg$ and $gf$ are Drazin, and $\mathsf{ind}(fg) \leq 1$ and $\mathsf{ind}(gf) \leq 1$. However by Lemma <ref>, this is the same as saying that $fg$ and $gf$ have group inverses.

($(ii) \Rightarrow (iii)$) Suppose that $fg$ and $gf$ have group inverses. Then by Lemma <ref>, $fg$ and $gf$ are both Drazin with $\mathsf{ind}(fg) \leq 1$ and $\mathsf{ind}(gf) \leq 1$. So by Theorem <ref>, $(f,g)$ is Drazin with $\mathsf{ind}(f,g) \leq \mathsf{max}(\mathsf{ind}(fg) , \mathsf{ind}(gf) ) \leq 1$. We show that its Drazin inverse $(g^{\frac{D}{f}},f^{\frac{D}{g}})$ is also its group inverse. Note that [GV.2] and [GV.3] are the same as [DV.2] and [DV.3]. So it remains to show [GV.1].
Then using that $\mathsf{ind}(f,g) \leq 1$ and [DV.3], we compute that:

\[ fg g^{\frac{D}{f}} g = fg f f^{\frac{D}{g}} = fg \]

So $fg g^{\frac{D}{f}} g = fg$, and similarly we can also get that $gf = g f f^{\frac{D}{g}} f = f$. So we conclude that $(g^{\frac{D}{f}},f^{\frac{D}{g}})$ is a group inverse of $(f,g)$.

($(iii) \Rightarrow (i)$) Suppose that $(f,g)$ has a group inverse $(g^{\frac{D}{f}},f^{\frac{D}{g}})$. We will show that it is also a Drazin inverse. Note that [DV.2] and [DV.3] are the same as [GV.2] and [GV.3]. So it remains to show [DV.1]. So using [GV.1] and [GV.3], we compute that:

\[ (fg)^1 f f^{\frac{D}{g}} = fg f f^{\frac{D}{g}} = fg g^{\frac{D}{f}} g = fg = (fg)^1 \]

So $(fg)^1 f f^{\frac{D}{g}} = (fg)^1$, and similarly we can also show that $(gf)^1 g g^{\frac{D}{f}} = (gf)^1$. So we conclude that $(f,g)$ is Drazin with $\mathsf{ind}(f,g) \leq 1$.

§.§ Drazin Opposing Pairs and Idempotents
In Section <ref> we saw how to equivalently characterize having a Drazin inverse in terms of isomorphisms in the idempotent completion. The objective of this section is to do the same for the Drazin inverse of an opposing pair.

We first recall from Lemma <ref> that every Drazin endomorphism $x$ induces an idempotent $e_x = xx^D = x^D x$. Then for a Drazin opposing pair $(f,g)$, we get two idempotents $e_{fg} = fg(fg)^D$ and $e_{gf} = gf(gf)^D = (gf)^D gf$. Moreover, we can also use the Drazin inverse to get isomorphisms between these two idempotents.

In a category $\mathbb{X}$, if $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ is Drazin, then:

*   For the induced idempotents, $e_{fg} = f f^{\frac{D}{g}} = g^{\frac{D}{f}}g$ and $e_{gf} = f^{\frac{D}{g}} f = g g^{\frac{D}{f}}$;

* $f f^{\frac{D}{g}} f: (A, e_{fg}) \to (B, e_{gf})$ is an isomorphism in $\mathsf{Split}(\mathbb{X})$ whose inverse is given by $f^{\frac{D}{g}}: (B, e_{gf}) \to (A, e_{fg})$;

* $g g^{\frac{D}{f}} g: (B, e_{gf}) \to (A, e_{fg})$ is an isomorphism in $\mathsf{Split}(\mathbb{X})$ whose inverse is given by $g^{\frac{D}{f}}: (A, e_{gf}) \to  (B, e_{fg}) $.

* Recall the formulas from Theorem <ref> that $f^{\frac{D}{g}} = g (fg)^D = (gf)^D g$ and $g^{\frac{D}{f}} = f (gf)^D = (fg)^D f$. Then combining these with the definition of the induced idempotents, we immediately get that $e_{fg} = f f^{\frac{D}{g}} = g^{\frac{D}{f}}g$ and $e_{gf} = f^{\frac{D}{g}} f = g g^{\frac{D}{f}}$ as desired.

* We must first show that $f f^{\frac{D}{g}} f$ and $f^{\frac{D}{g}}$ are maps in the idempotent completion. First by using (<ref>) and [DV.2] we compute that:

\[ e_{gf} f^{\frac{D}{g}} e_{fg} = f^{\frac{D}{g}} f   f^{\frac{D}{g}} f f^{\frac{D}{g}} =  f^{\frac{D}{g}} f f^{\frac{D}{g}} = f^{\frac{D}{g}} \]

So $e_{gf} f^{\frac{D}{g}} e_{fg} = f^{\frac{D}{g}}$. On the other hand

\[ e_{fg} f f^{\frac{D}{g}} f e_{gf} =  f f^{\frac{D}{g}}  f f^{\frac{D}{g}} f  f^{\frac{D}{g}} f =  f f^{\frac{D}{g}} f  f^{\frac{D}{g}} f =  f f^{\frac{D}{g}} f   \]

So $e_{fg} f f^{\frac{D}{g}} f e_{gf} = f f^{\frac{D}{g}} f$ as well. As such, we get  $f f^{\frac{D}{g}} f: (A, e_{fg}) \to (B, e_{gf})$ and $f^{\frac{D}{g}}: (B, e_{gf}) \to (A, e_{fg})$  are indeed maps in  $\mathsf{Split}(\mathbb{X})$. Now we must show that they are inverses of each other in $\mathsf{Split}(\mathbb{X})$. However, using (<ref>) again, we easily see that:

\[  f f^{\frac{D}{g}} f  f^{\frac{D}{g}} = e_{fg} e_{fg} = e_{fg} =1_{(A, e_{fg})}\]

\[ f^{\frac{D}{g}} f f^{\frac{D}{g}} f = e_{gf} e_{gf} = e_{gf} = 1_{(B, e_{gf})} \]

So we conclude that $f f^{\frac{D}{g}} f: (A, e_{fg}) \to (B, e_{gf})$ is an isomorphism in $\mathsf{Split}(\mathbb{X})$ with inverse $f^{\frac{D}{g}}: (B, e_{gf}) \to (A, e_{fg})$, as desired.

* This is shown by a similar argument.

We now prove the analogue of Theorem <ref> for Drazin inverses of opposing pairs:

In a category $\mathbb{X}$, $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ is Drazin if and only if there are idempotents $e_A: A \to A$ and $e_B: B \to B$, and a $k \in \mathbb{N}$, such that both of the maps $(fg)^kf: (A,e_A) \to (B,e_B)$ and $(gf)^kg: (B,e_B) \to (A,e_A)$ are isomorphisms[It is important to note that this statement does not say that $(fg)^kf$ and $(gf)^kg$ are inverses in $\mathsf{Split}(\mathbb{X})$] in $\mathsf{Split}(\mathbb{X})$.

($\Rightarrow$)  Suppose that $(f,g)$ is Drazin. So $fg$ and $gf$ are Drazin. As our idempotents, we take the induced ones $e_{fg}: A \to A$ and $e_{gf}: B \to B$. Now we first note that by Lemma <ref>.(<ref>) and using the same arguments as in the first half of the proof of Theorem <ref>, we have that for a $x: A \to A$ which is Drazin, for all $\mathsf{ind}(x) \leq j$ that $x^{j+1}: (A,e_x) \to (A,e_x)$ is an isomorphism in $\mathsf{Split}(\mathbb{X})$. So let $k = \mathsf{max}\left( \mathsf{ind}(fg), \mathsf{ind}(gf) \right)$. Then we get that $(fg)^{k+1}: (A,e_{fg}) \to (A,e_{fg})$ and $(gf)^{k+1}: (A,e_{gf}) \to (A,e_{gf})$ are isomorphisms. Now by Lemma <ref>, we also have that $f^{\frac{D}{g}}: (B, e_{gf}) \to (A, e_{fg})$ and $g^{\frac{D}{f}}: (A, e_{gf}) \to  (B, e_{fg})$ are isomorphisms. So the composites, $(fg)^{k+1} g^{\frac{D}{f}}: (A, e_{gf}) \to  (B, e_{fg})$ and $(gf)^{k+1} f^{\frac{D}{g}}: (A,e_{gf}) \to (A,e_{gf})$ are also isomorphisms. However, using the formulas from Theorem <ref> and Lemma <ref>.(<ref>), we get that:

\[ (fg)^{k+1} g^{\frac{D}{f}} = (fg)^{k+1} (fg)^D f = (fg)^k f   \]

So $(fg)^{k+1} g^{\frac{D}{f}} = (fg)^k f$, and similarly we can also show that $(gf)^{k+1} f^{\frac{D}{g}} =  (gf)^kg$. So we conclude that $(fg)^kf: (A,e_A) \to (B,e_B)$ and $(gf)^kg: (B,e_B) \to (A,e_A)$ are isomorphisms in $\mathsf{Split}(\mathbb{X})$ as desired.

($\Leftarrow$) Suppose for some $e_A: A \to A$ and $e_B: B \to B$, and a $k \in \mathbb{N}$, that both  $(fg)^kf: (A,e_A) \to (B,e_B)$ and $(gf)^kg: (B,e_B) \to (A,e_A)$ are isomorphisms in $\mathsf{Split}(\mathbb{X})$, with respective inverses $v: (B, e_B) \to (A,e_A)$ and $u: (A,e_A) \to (B,e_B)$. Explicitly this means we have the following identity:

\begin{align*}
(fg)^kf e_B = (fg)^kf = e_A (fg)^kf && e_B (gf)^kg = (gf)^kg = (gf)^kg e_A \\
v e_A = v = e_B v && e_A u = u = u e_B \\
(fg)^kf v = e_A = u (gf)^kg && v (fg)^kf  = e_B = (gf)^kg u
\end{align*}

We will now prove that $fg$ is Drazin. To do so, we will first show that $(fg)^{k+1}$ is an isomorphism in the idempotent splitting. Using the above identity, we first compute:

\[ e_A (fg)^{k+1} = e_A (fg)^k fg = (fg)^k fg = (fg)^{k+1} \]

\[ (fg)^{k+1} e_A = f (gf)^k g e_A =  f (gf)^k g = (fg)^{k+1} \]

So we get that $(fg)^{k+1}: (A, e_A) \to (A,e_A)$ is indeed a map in $\mathsf{Split}(\mathbb{X})$. On the other hand, since $u: (A,e_A) \to (B,e_B)$ and $v: (B, e_B) \to (A,e_A)$ are maps in $\mathsf{Split}(\mathbb{X})$, it follows that $u (gf)^k v: (A, e_A) \to (A, e_A)$ is also a map $\mathsf{Split}(\mathbb{X})$. To show that this map is the inverse, we first need to do some calculations:

\begin{gather*}
(gf)^k v = (gf)^k e_B v = (gf)^k (gf)^k g u v = (gf)^k g (fg)^k u v = e_B  (gf)^k g (fg)^k u v \\
= e_B (gf)^k (gf)^k g u v = (gf)^k e_B v =  (gf)^k v
\end{gather*}

So $(gf)^k v = e_B (gf)^k v$, and similarly we can also show that $u (gf)^k =u (gf)^k e_A$. With these identities in hand, we can observe that:

\begin{gather*}
(fg)^{k+1} u (gf)^k v = f (gf)^k g u (gf)^k v = f e_B (gf)^k v = f(gf)^k v = (fg)^k f v = e_A = 1_{(A,e_A)}
\end{gather*}

\begin{gather*}
u (gf)^k v  (fg)^{k+1} =  u  (gf)^k v (fg)^k f g = u (gf)^k e_A g = u (gf)^k g = e_A = 1_{(A,e_A)}
\end{gather*}

So we get that $(fg)^{k+1}: (A, e_A) \to (A,e_A)$ is an isomorphism in $\mathsf{Split}(\mathbb{X})$. Then by Theorem <ref>, $fg$ is Drazin. Therefore we conclude that $(f,g)$ is Drazin.

§.§ Drazin Opposing Pairs and Binary Idempotents
Recall from Section <ref> that a binary idempotent is, using the terminology from this section, an opposing pair $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ such that $fgf=f$ and $gfg= g$, and this implied that $fg$ and $gf$ are idempotents. Here we explain how binary idempotents are linked to Drazin opposing pairs. Moreover, binary idempotents will play a key role in the following sections.

We first observe that in the same way that an idempotent is its own Drazin inverse, a binary idempotent regarded as an opposing pair is its own Drazin inverse.

In any category $\mathbb{X}$, $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ is a binary idempotent if and only if $(f,g)$ its own Drazin inverse, so $(f,g)^D = (f,g)$ (so that $f^{\frac{D}{g}} = g$ and $g^{\frac{D}{f}} = f$).

Since $(f,g)$ is a binary idempotent, $fg$ and $gf$ are idempotents, and so by Lemma <ref>, are Drazin and their own Drazin inverses. Therefore $(f,g)$ is Drazin. Applying the formulas from Theorem <ref>, we compute:

\[ f^{\frac{D}{g}} = g(fg)^D = gfg = g \]

So $f^{\frac{D}{g}} = f$, and similarly, we can also show that $g^{\frac{D}{g}} = f$.

$(\Leftarrow)$ Conversely if $(f,g)$ is Drazin with itself as a Drazin inverse, then [DV.2] gives us that
$gfg = g$ and $fgf =f$.  Thus, $(f,g)$ is a binary idempotent.

From any Drazin opposing pair we get two binary idempotents. We can see this directly but it is instructive to see how this is related to taking, for opposing pairs, the Drazin inverse of a Drazin inverse.  The implicit binary idempotents can then be seen as “mixing” the Drazin inverse and the Drazin inverse of the Drazin inverse of an opposing pair.

In a category $\mathbb{X}$, if $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ is Drazin then:

*  $(f,g)^D := (g^{\frac{D}{f}}, f^{\frac{D}{g}})$ is Drazin with Drazin inverse $(f,g)^{DD}:=(f f^{\frac{D}{g}}f, g g^{\frac{D}{f}}g)$ and $\mathsf{ind}((f,g)^D) \leq 1$;

*  $(f,g)^{DD}:=(f f^{\frac{D}{g}}f, g g^{\frac{D}{f}}g)$ is Drazin with Drazin inverse $(f,g)^D$;

*  $(f f^{\frac{D}{g}}f,f^{\frac{D}{g}})$ and
$(g g^{\frac{D}{f}}g, g^{\frac{D}{f}})$ are binary idempotents.

* Recall by the formulas from Theorem <ref>, we have that $g^{\frac{D}{f}} f^{\frac{D}{g}} = (gf)^D$ and $f^{\frac{D}{g}} g^{\frac{D}{f}} = (fg)^D$. However by Lemma <ref>.(<ref>), a Drazin inverse is always Drazin, so $g^{\frac{D}{f}} f^{\frac{D}{g}}$ and $f^{\frac{D}{g}} g^{\frac{D}{f}}$ are Drazin. As such, $(g^{\frac{D}{f}}, f^{\frac{D}{g}})$ is Drazin. Now applying the formulas from Lemma <ref>.(<ref>) and Theorem <ref>, as well as using [DV.2] and [DV.3], we compute:

\begin{gather*}
g^{\frac{D}{f}} ( f^{\frac{D}{g}} g^{\frac{D}{f}})^D = g^{\frac{D}{f}} ((gf)^D)^D =  g^{\frac{D}{f}} gf (gf)^D gf = g^{\frac{D}{f}} gf f^{\frac{D}{g}} g^{\frac{D}{f}} gf \\
= f f^{\frac{D}{g}} f f^{\frac{D}{g}} f f^{\frac{D}{g}} f = f f^{\frac{D}{g}} f f^{\frac{D}{g}} f = f f^{\frac{D}{g}} f f^{\frac{D}{g}} f = f f^{\frac{D}{g}} f
\end{gather*}

So the Drazin inverse of $f^{\frac{D}{g}}$ over $g^{\frac{D}{f}}$ is $f f^{\frac{D}{g}} f$. Similarly, we can also compute that the Drazin inverse of $g^{\frac{D}{f}}$ over $f^{\frac{D}{g}}$ is $g g^{\frac{D}{f}}g$. In fact, since $g^{\frac{D}{f}} f^{\frac{D}{g}}$ and $f^{\frac{D}{g}} g^{\frac{D}{f}}$ have group inverses, then by Lemma <ref>, $(f f^{\frac{D}{g}}f, g g^{\frac{D}{f}}g)$ is a group inverse of $(g^{\frac{D}{f}}, f^{\frac{D}{g}})$, and so $\mathsf{ind}(g^{\frac{D}{f}}, f^{\frac{D}{g}}) \leq 1$.

* By (<ref>), we know that $(f f^{\frac{D}{g}}f, g g^{\frac{D}{f}}g)$ is a Drazin inverse. So by (<ref>) again, we get that $(f f^{\frac{D}{g}}f, g g^{\frac{D}{f}}g)$ has a Drazin inverse, which by [DV.2] is precisely $(g^{\frac{D}{f}}, f^{\frac{D}{g}})$.

* Using [DV.2], we compute that:

\begin{gather*}
(ff^{\frac{D}{g}}f) f^{\frac{D}{g}} (ff^{\frac{D}{g}}f) = f(f^{\frac{D}{g}}f f^{\frac{D}{g}}) f (f^{\frac{D}{g}}ff^{\frac{D}{g}})f = f(f^{\frac{D}{g}} f f^{\frac{D}{g}}) f = f f^{\frac{D}{g}}f \end{gather*}

\begin{gather*}
f^{\frac{D}{g}} (f (f^{\frac{D}{g}} f) f^{\frac{D}{g}} = (f^{\frac{D}{g}} f f^{\frac{D}{g}}) f f^{\frac{D}{g}} = f^{\frac{D}{g}} f f^{\frac{D}{g}} = f^{\frac{D}{g}}
\end{gather*}

So $(ff^{\frac{D}{g}}f,f^{\frac{D}{g}})$ is a binary idempotent. Similarly we can also show that $(g g^{\frac{D}{f}}g, g^{\frac{D}{f}})$ is a binary idempotent.

Using binary idempotents, we can also precisely characterize when an opposing pair is the Drazin inverse of another opposing pair. To do so, it will be first useful to observe the following:

In a category $\mathbb{X}$, $\xymatrixcolsep{1.5pc}\xymatrix{(f,g): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ is a Drazin inverse if and only if $(f,g)$ is Drazin with $\mathsf{ind}(f,g) \leq 1$, and $f f^{\frac{D}{g}} f = f$ and $g g^{\frac{D}{f}} g = g$. Explicitly:

* If $(f,g)$ is the Drazin inverse of $(h,k)$, then $(f,g)$ is Drazin with $\mathsf{ind}(f,g) \leq 1$ where $f^{\frac{D}{g}} = k g k$ and $g^{\frac{D}{f}} = h f h$, and $(f, hfh)$ and $(g, kgk)$ are binary idempotents.

* If $(f,g)$ is Drazin with $\mathsf{ind}(f,g) \leq 1$, and $f f^{\frac{D}{g}} f = f$ and $g g^{\frac{D}{f}} g = g$, then $(f,g)$ is the Drazin inverse of $(g^{\frac{D}{f}}, f^{\frac{D}{g}})$.

($\Rightarrow$) Suppose that $(f,g)$ is the Drazin inverse of $(h,k)$, so $(f,g) = (k^{\frac{D}{h}}, h^{\frac{D}{k}})$. Then by Proposition <ref>.(<ref>), we get that $(f,g)$ is Drazin with Drazin inverse $(g^{\frac{D}{f}}, f^{\frac{D}{g}})= ( h h^{\frac{D}{k}} h ,  k k^{\frac{D}{h}} k ) = (h f h, k g k)$ and $\mathsf{ind}(f,g) \leq 1$. Also by Proposition <ref>.(<ref>), we get that $(f,f^{\frac{D}{g}})$ and $(g, g^{\frac{D}{f}})$ are binary idempotents, which gives us that $f f^{\frac{D}{g}} f = f$ and $g g^{\frac{D}{f}} g = g$.

($\Leftarrow$) By assumption we have that $(f,g) = (f f^{\frac{D}{g}}f, g g^{\frac{D}{f}}g)$. However, Proposition <ref>.(<ref>) tells us that $(f,g) = (f f^{\frac{D}{g}}f, g g^{\frac{D}{f}}g)$ is the Drazin inverse of $(g^{\frac{D}{f}}, f^{\frac{D}{g}})$.

In a category $\mathbb{X}$, $(f,g)$ is a Drazin inverse if and only if $(f,g)$ is Drazin and both
$(f,f^{\frac{D}{g}})$ and $(g^{\frac{D}{f}},g)$ are binary idempotents.

From Lemma <ref> we know the forward implication is true.  For the reverse implication, it remains to show that the conditions of Lemma <ref> hold. Note that the two required equalities  $f f^{\frac{D}{g}} f = f$ and $g g^{\frac{D}{f}} g = g$ are immediate as we ask for binary idempotents. As such, it remains to show why ${\sf ind}((f,g)) \leq 1$ (so that it is a group inverse). For this, we compute that:

\[ (fg)ff^{\frac{D}{g}} = (fg)g^{\frac{D}{f}} g
= f (g g^{\frac{D}{f}} g) = fg \]

and similarly we can show that $(gf)g g^{\frac{D}{f}} = gf$. So we get that ${\sf ind}((f,g)) \leq 1$. Thus we can apply Lemma <ref> to get that $(f,g)$ is a Drazin inverse.

§.§ Being Drazin in a Dagger Category
In this section, we consider Drazin inverses in dagger categories. In particular, in the next section we will relate Drazin inverses to another sort of generalized inverse that arises in dagger categories, called the Moore-Penrose inverse.

Recall that a dagger category is a category $\mathbb{X}$ equipped with a functor $\dagger: \mathbb{X}^{op} \to \mathbb{X}$ which is the identity on objects and involutive. Explicitly, this means that for every map $f: A \to B$, there is a map of dual type $f^\dagger: B \to A$ called the adjoint of $f$ such that $(fg)^\dagger = g^\dagger f^\dagger$, $1_A^\dagger = 1_A$, and $f^{\dagger \dagger} = f$. For a more in-depth introduction to dagger categories, we refer the reader to [7].

Since the adjoint of an endomorphism is an endomorphism, it makes sense to ask whether the adjoint of a Drazin endomorphism is Drazin. It turns out that the Drazin inverse of the adjoint is the adjoint of the Drazin inverse:

In a dagger category $\mathbb{X}$, if $x: A \to A$ is Drazin then $x^\dagger: A \to A$ is also Drazin where $(x^\dagger)^D = (x^D)^\dagger$.

Clearly, if $x$ is Drazin in $\mathbb{X}$ then $x$ is Drazin in $\mathbb{X}^{op}$ with $x^D$ as its Drazin inverse. Since Drazin inverses are absolute by Proposition <ref>, it follows that $x^\dagger$ is Drazin and $(x^\dagger)^D = (x^D)^\dagger$ as desired.

An important source of a pair of opposing maps in a dagger category is the pair of a map and its adjoint. So for any map $f: A \to B$, we get the opposing pair $\xymatrixcolsep{1.5pc}\xymatrix{(f,f^\dagger): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$, which we call an adjoint opposing pair. When an adjoint opposing pair is Drazin, it turns out that the Drazin inverse of a map over its adjoint is the adjoint of the Drazin inverse of the adjoint over the map:

In a dagger category $\mathbb{X}$, if $\xymatrixcolsep{1.5pc}\xymatrix{(f,f^\dagger): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ has a Drazin inverse then:

*  $f^{\frac{D}{f^\dagger}}= ((f^\dagger)^{\frac{D}{f}})^\dagger$ and so $(f,f^\dagger)^D = ((f^\dagger)^{\frac{D}{f}},(f^{\frac{D}{f^\dagger}})^\dagger)$ is an adjoint pair;

*  The induced idempotents are $\dagger$-idempotents, that is, $e_{ff^\dagger}^\dagger = e_{ff^\dagger}$ and $e_{f^\dagger f}^\dagger = e_{f^\dagger f}$.

* If $(f,f^\dagger)$ has a Drazin inverse, by Theorem <ref> $(f,f^\dagger)$ is Drazin, so $ff^\dagger$ and $f^\dagger f$ are Drazin. Then using the formulas from Theorem <ref> and Lemma <ref>, we compute:

\[ ((f^\dagger)^{\frac{D}{f}})^\dagger = ( f(f^\dagger f)^D )^\dagger = ((f^\dagger f)^D)^\dagger f^\dagger = ((f^\dagger f)^\dagger)^D f^\dagger = (f^\dagger f)^D f^\dagger = f^{\frac{D}{f^\dagger}} \]

So $f^{\frac{D}{f^\dagger}}= ((f^\dagger)^{\frac{D}{f}})^\dagger$ as desired.

* By Lemma <ref>.(<ref>), we know that $e_{ff^\dagger}= (f^\dagger)^{\frac{D}{f}} f^{\frac{D}{f^\dagger}}$ and $e_{f^\dagger f} =  f^{\frac{D}{f^\dagger}} (f^\dagger)^{\frac{D}{f}}$. However by (<ref>), we then get that $e_{ff^\dagger}= \left(f^{\frac{D}{f^\dagger}} \right)^\dagger f^{\frac{D}{f^\dagger}}$ and $e_{f^\dagger f} =  f^{\frac{D}{f^\dagger}} \left(f^{\frac{D}{f^\dagger}} \right)^\dagger $ – which implies that $e_{ff^\dagger}^\dagger$ and $e_{f^\dagger f}^\dagger$ are $\dagger$-idempotents.

We can also precisely say when an adjoint opposing pair is the Drazin inverse of another adjoint opposing pair.

In a dagger category $\mathbb{X}$, $\xymatrixcolsep{1.5pc}\xymatrix{(f,f^\dagger): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ is a Drazin inverse if and only if $(f,f^\dagger)$ is Drazin with $\mathsf{ind}(f,f^\dagger) \leq 1$ and $f f^{\frac{D}{f^\dagger}} f = f$. Explicitly:

* If $(f,f^\dagger)$ is the Drazin inverse of $(g,g^\dagger)$, then $(f,f^\dagger)$ is Drazin with $\mathsf{ind}(f,f^\dagger) \leq 1$ where $f^{\frac{D}{f^\dagger}} = g^\dagger f g^\dagger$ and $(f, f^{\frac{D}{f^\dagger}})$ is a binary idempotent.

* If $(f,f^\dagger)$ is Drazin with $\mathsf{ind}(f,f^\dagger) \leq 1$ and $f f^{\frac{D}{f^\dagger}} f = f$, then $(f,f^\dagger)$ is the Drazin inverse of the adjoint opposing pair $( (f^{\frac{D}{f^\dagger}})^\dagger , f^{\frac{D}{f^\dagger}})$.

($\Rightarrow$) This is immediate from Lemma <ref>.

($\Leftarrow$) If $f f^{\frac{D}{f^\dagger}} f = f$, then Lemma <ref>.(<ref>) tell us that the dagger of this identity is precisely $f^\dagger (f^{\frac{D}{f^\dagger}})^\dagger f^\dagger = f^\dagger$. Then by applying Lemma <ref>, we get that $(f,f^\dagger)$ is the Drazin inverse of $( (f^{\frac{D}{f^\dagger}})^\dagger , f^{\frac{D}{f^\dagger}})$.

§.§ The Moore-Penrose Inverse

The remainder of this section is dedicated to discussing the relationship between Drazin opposing pairs and Moore-Penrose inverses. For a more in-depth introduction to Moore-Penrose inverses and examples, we invite the reader to see [8, 12].

In a dagger category $\mathbb{X}$, a Moore-Penrose inverse <cit.> of a map $f: A \to B$ is a map of dual type $f^\circ: B \to A$ such that:

* $f f^\circ f = f$

* $f^\circ f f^\circ = f^\circ$

* $(ff^\circ)^\dagger = ff^\circ$

* $(f^\circ f)^\dagger = f^\circ f$

Like Drazin inverses, Moore-Penrose inverses are unique <cit.>. Moreover, if $f$ has a Moore-Penrose inverse, then so does $f^\dagger$, where $f^{\dagger \circ} = f^{\circ \dagger}$ <cit.>. We also observe that having a Moore-Penrose inverse gives a binary idempotent:

In a dagger category $\mathbb{X}$, if $f: A \to B$ has a Moore-Penrose inverse then $(f,f^\circ)$ is a binary idempotent
whose induced idempotents are $\dagger$-idempotents.

[MP.1] and [MP.2] ensure that it is a binary idempotent while [MP.3] and [MP.4] ensure that the induced idempotents are $\dagger$-idempotents.

We now give an equivalent characterization of having a Moore-Penrose inverse in terms of Drazin inverses.

In a dagger category $\mathbb{X}$, $f: A \to B$ has a Moore-Penrose inverse if and only if $\xymatrixcolsep{1.5pc}\xymatrix{(f,f^\dagger): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ is Drazin with $\mathsf{ind}(f,f^\dagger)\leq 1$ and $f f^{\frac{D}{f^\dagger}} f = f$. Explicitly:

* If $f$ has a Moore-Penrose inverse, then $(f,f^\dagger)$ is Drazin with $\mathsf{ind}(f,g) \leq 1$ where:

* $f^{\frac{D}{f^\dagger}} = f^\circ$ and $(f^\dagger)^{\frac{D}{f}} = f^{\circ \dagger}$;

* $(ff^\dagger)^D = f^{\circ \dagger} f^\circ $ and $(f^\dagger f)^D = f^\circ  f^{\circ \dagger}$.

* If $(f,f^\dagger)$ is Drazin with $\mathsf{ind}(f,f^\dagger)\leq 1$ and $f f^{\frac{D}{f^\dagger}} f = f$, then $f$ has a Moore-Penrose inverse where $f^\circ = f^{\frac{D}{f^\dagger}}$.

$(\Rightarrow)$ We will first show that $((f^\dagger)^{\frac{D}{f}}, f^{\frac{D}{f^\dagger}}) = (f^{\circ \dagger}, f^\circ)$ satisfies the three axioms of a group inverse.

* Using [MP.1], we compute that:

\begin{gather*}
f f^\dagger (f^\dagger)^{\frac{D}{f}} f^\dagger  = f f^\dagger f^{\circ \dagger}  f^\dagger = f ( f f^\circ f)^\dagger = f f^\dagger
\end{gather*}

\begin{gather*}
f^\dagger f f^{\frac{D}{f^\dagger}} f = f^\dagger f f^\circ f = f^\dagger f
\end{gather*}

* Using [MP.2], we compute that:

\begin{gather*}
f^{\frac{D}{f^\dagger}} f f^{\frac{D}{f^\dagger}}f^{\frac{D}{g}} = f^\circ f f^\circ = f^\circ = f^{\frac{D}{f^\dagger}}
\end{gather*}

\begin{gather*}
(f^\dagger)^{\frac{D}{f}} f^\dagger (f^\dagger)^{\frac{D}{f}} = f^{\circ \dagger}  f^\dagger f^{\circ \dagger} = (f^\circ f f^\circ)^\dagger =   f^{\circ \dagger} =  (f^\dagger)^{\frac{D}{f}}
\end{gather*}

* Using [MP.3] and [MP.4], we compute that:

\begin{gather*}
f f^{\frac{D}{f^\dagger}} = f f^\circ = (f f^\circ)^\dagger =  f^{\circ \dagger} f^\dagger =  (f^\dagger)^{\frac{D}{f}} f^\dagger
\end{gather*}

\begin{gather*}
f^{\frac{D}{f^\dagger}} f =  f^\circ f = ( f^\circ f )^\dagger =  f^\dagger f^{\circ \dagger} =  f^\dagger(f^\dagger)^{\frac{D}{f}}
\end{gather*}

So $(f^{\circ \dagger}, f^\circ)$ is a group inverse of a $(f,f^\dagger)$, and so by Lemma <ref>, $(f,f^\dagger)$ is Drazin with $\mathsf{ind}(f,g) \leq 1$. Now applying the formulas from Theorem <ref>, we compute that:

\begin{align*}
(ff^\dagger)^D = (f^\dagger)^{\frac{D}{f}} f^{\frac{D}{f^\dagger}} =  f^{\circ \dagger} f^\circ && (f^\dagger f)^D = f^{\frac{D}{f^\dagger}} (f^\dagger)^{\frac{D}{f}} =   f^\circ  f^{\circ \dagger}
\end{align*}

Lastly, [MP.1] is precisely the statement that $f f^{\frac{D}{f^\dagger}} f = f$.

$(\Leftarrow)$ We will show that $f^\circ = f^{\frac{D}{f^\dagger}}$ satisfies the four axioms of a Moore-Penrose inverse. However note that [MP.1] is precisely the assumption that $f f^{\frac{D}{f^\dagger}} f = f$, while [MP.2] is precisely [DV.3]. Now for [MP.3], using Lemma <ref> and [DV.3], we compute that:

\begin{gather*}
(f f^\circ)^\dagger = (f^\circ)^\dagger f^\dagger = (f^{\frac{D}{f^\dagger}})^\dagger f^\dagger = (f^\dagger)^{\frac{D}{f}} f^\dagger = f f^{\frac{D}{f^\dagger}} = f f^\circ
\end{gather*}

While [MP.4] is shown using similar calculations. So we conclude that $f$ has a Moore-Penrose inverse.

By Lemma <ref>, we can re-express Theorem <ref> to say that a map has a Moore-Penrose inverse if and only if its induced adjoint opposing pair is a Drazin inverse.

In a dagger category $\mathbb{X}$, $f: A \to B$ has a Moore-Penrose inverse if and only if $\xymatrixcolsep{1.5pc}\xymatrix{(f,f^\dagger): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ is a Drazin inverse. Explicitly:

* If $f$ has a Moore-Penrose inverse, then $(f,f^\dagger)$ is the Drazin inverse of $(f^{\circ \dagger}, f^\circ)$.

* If $(f,f^\dagger)$ is a Drazin inverse of an adjoint opposing pair $(g, g^\dagger)$, then $f$ has a Moore-Penrose inverse with $f^\circ = g^\dagger f g^\dagger$.

$(\Rightarrow)$ By Theorem <ref>, we know that $(f,f^\dagger)$ is Drazin with Drazin inverse $(f^{\circ \dagger}, f^{\circ})$ and $\mathsf{ind}(f,f^\dagger) \leq 1$, and also that $f f^\circ f = f$. Then by Lemma <ref>, $(f,f^\dagger)$ is the Drazin inverse of $(f^{\circ \dagger}, f^{\circ})$.

$(\Leftarrow)$ Suppose that $(f,f^\dagger)$ is a Drazin inverse. Then by Proposition <ref>, $(f,f^\dagger)$ is Drazin, where by Lemma <ref>, we may set $(f,f^\dagger)^{D} = (g, g^\dagger)$, and so we get that $(g, g^\dagger)^{D} = (f,f^\dagger)$. So $(f,f^\dagger)$ is a Drazin inverse of an adjoint opposing pair. Then Lemma <ref> tells us that in fact $(f,f^\dagger)^{D} = (g f^\dagger g, g^\dagger f g^\dagger)$ and $\mathsf{ind}(f,f^\dagger) \leq 1$, and also that $f g^\dagger f g^\dagger f = f$. Then by Theorem <ref>, $f$ has a Moore-Penrose inverse given by $f^\circ = g^\dagger f g^\dagger$.

Now a Moore-Penrose dagger category <cit.> is a dagger category such that every map has a Moore-Penrose inverse. Then by Proposition <ref>, we obtain a novel equivalent description of when a dagger category is Moore-Penrose.

A dagger $\mathbb{X}$ is Moore-Penrose if and only if every adjoint pair is a Drazin inverse.

In general, while not every map in a dagger category has a Moore-Penrose inverse, the Drazin inverse of the map over its adjoint (if it exists) always has a Moore-Penrose inverse.

In a dagger category $\mathbb{X}$, if $\xymatrixcolsep{1.5pc}\xymatrix{(f,f^\dagger): A<EMAIL_ADDRESS>& B<EMAIL_ADDRESS>}$ is Drazin then $f^{\frac{D}{f^\dagger}}$ has a Moore-Penrose inverse, where $\left(f^{\frac{D}{f^\dagger}} \right)^\circ = f f^{\frac{D}{f^\dagger}} f$.

If $(f,f^\dagger)$ is Drazin, by Lemma <ref>.(<ref>), we have that $((f^{\frac{D}{f^\dagger}})^\dagger, f^{\frac{D}{f^\dagger}})$ is its Drazin inverse. In other words, using Lemma <ref>, we get that $(f^{\frac{D}{f^\dagger}}, (f^{\frac{D}{f^\dagger}})^\dagger)$ is the Drazin inverse of the adjoint opposing pair $(f^\dagger, f)$. Then by Proposition <ref>, $f^{\frac{D}{f^\dagger}}$ has a Moore-Penrose inverse given by $\left(f^{\frac{D}{f^\dagger}} \right)^\circ = f f^{\frac{D}{f^\dagger}} f$.

§.§ When Moore-Penrose is Drazin

Turning our attention back to Drazin inverses for endomorphisms, we conclude with the natural question of when the Drazin inverse and Moore-Penrose inverse coincide. For complex square matrices, this happens precisely when the matrix is an EP matrix (which means that it commutes with its Moore-Penrose inverse) <cit.>. Here we show that the same holds true in an arbitrary dagger category.

In a dagger category $\mathbb{X}$, an endomorphism $x: A \to A$ is said to be Moore-Penrose-Drazin if $x$ is Drazin such that its Drazin inverse is also a Moore-Penrose inverse, $x^D = x^\circ$.

In a dagger category $\mathbb{X}$, for an endomorphism $x: A \to A$, the following are equivalent:

* $x$ is Moore-Penrose-Drazin;

* $x$ is Drazin with $\mathsf{ind}(x) \leq 1$ and the induced idempotent $e_x$ is a $\dagger$-idempotent;

* $x$ has a Moore-Penrose inverse and $x x^\circ = x^\circ x$.

($(i) \Rightarrow (ii)$) By definition $x$ is Drazin. Now using [DV.3] and [MP.1], we compute that:

\[ x^{2} x^D = x x x^D = x x^D x = x x^\circ x = x  \]

So $\mathsf{ind}(x) \leq 1$. Now using [MP.3], we compute that:

\[ e_x^\dagger = (xx^D)^\dagger = (x x^\circ)^\dagger = x x^\circ  = e_x \]

So $e_x^\dagger$ is a $\dagger$-idempotent.

($(ii) \Rightarrow (iii)$) We show that $x^D$ satisfies the four Moore-Penrose inverse axioms. However, note that since $\mathsf{ind}(x) \leq 1$, $x^D$ is a group inverse of $x$. Then [MP.1] and [MP.2] are precisely the same as [G.1] and [G.2] respectively. Now since $e_x^\dagger$ is a $\dagger$-idempotent, meaning $e_x^\dagger = e_x$, which gives us precisely that $(xx^D)^\dagger = x x^D$ and $(x^D x)^\dagger = x^D x$, which are [MP.3] and [MP.4]. So $x^\circ = x^D$ is a Moore-Penrose inverse of $x$, and by [D.3], we also have that $x x^\circ = x^\circ x$.

($(iii) \Rightarrow (i)$) We show that setting $x^D = x^\circ$ satisfies the three group inverse axioms. However, note that [G.1] and [G.2] are the same as [MP.1] and [MP.2]. While [G.3] is precisely the extra assumption $x x^\circ = x^\circ x$. So $x^\circ$ is a group inverse of $x$, which implies that $x$ is Drazin, whose Drazin inverse is its Moore-Penrose inverse. Thus $x$ is Moore-Penrose-Drazin as desired.

In particular, for any map with a Moore-Penrose inverse, we always get two Moore-Penrose-Drazin endomorphisms.

In a dagger category $\mathbb{X}$, if $f: A \to B$ has a Moore-Penrose inverse, then $ff^\dagger$ and $f^\dagger f$ are Moore-Penrose-Drazin.

By Theorem <ref>, $(f,f^\dagger)$ is Drazin with $\mathsf{ind}(f,f^\dagger) \leq 1$, which tells us that $ff^\dagger$ and $f^\dagger f$ are Drazin with $\mathsf{ind}(ff^\dagger) \leq 1$ and $\mathsf{ind}(f^\dagger f) \leq 1$. While Lemma <ref>.(<ref>) tells us that $e_{ff^\dagger}$ and $e_{ff^\dagger}$ are $\dagger$-idempotents. So by Lemma <ref>, we conclude that $ff^\dagger$ and $f^\dagger f$ are Moore-Penrose-Drazin.

[1]
Pere Ara.
Strongly $\pi$-regular rings have stable range one.
Proceedings of the American Mathematical Society,
124(11):3293–3298, 1996.

[2]
E. P. Armendariz, J. W. Fisher, and R. L. Snider.
On injective and surjective endomorphisms of finitely generated
Communications in Algebra, 6(7):659–672, 1978.

[3]
G. Azumaya.
Strongly $\pi$-regular rings.
Journal of the Faculty of Science Hokkaido University. Ser. 1
Mathematics, 13(1):034–039, 1954.

[4]
Coecke B. and Kissinger A.
Picturing Quantum Processes: A First Course in Quantum Theory
and Diagrammatic Reasoning.
Cambridge University Press, 2017.

[5]
M. Barr.
The Chu Construction.
Theory and Applications of Categories, 2:17–35, 1996.

[6]
C. Bu.
Linear maps preserving Drazin inverses of matrices over fields.
Linear algebra and its applications, 396:159–173, 2005.

[7]
Heunen C. and J. Vicary.
Categories for Quantum Theory.
Oxford University Press, 2019.

[8]
S. Campbell and C. Meyer.
Generalized inverses of linear transformations.
SIAM, 2009.

[9]
S. L. Campbell, C. D. Jr. Meyer, and N. J. Rose.
Applications of the Drazin inverse to linear systems of differential
equations with singular constant coefficients.
SIAM Journal on Applied Mathematics, 31(3):411–425, 1976.

[10]
H. Chen and M. S. Abdolyousefi.
On Hirano inverses in rings.
Turkish Journal of Mathematics, 43(4):2049–2057, 2019.

[11]
R. E. Cline.
An application of representations for the generalized inverse of
a matrix.
Mathematics Research Center, United States Army, University of
Wisconsin, 1965.

[12]
J.R.B. Cockett and J-S. P. Lemay.
Moore-Penrose Dagger Categories.
In Proceedings of the Twentieth International Conference
on Quantum Physics and Logic, Paris, France, 17-21st July 2023,
volume 384 of Electronic Proceedings in Theoretical Computer Science,
pages 171–186. Open Publishing Association, 2023.

[13]
M. F. Dischinger.
Sur les anneaux fortement $\pi$-régulier.
Comptes Rendus de Académie des Sciences, Paris, 283:571–573,

[14]
M. P. Drazin.
Pseudo-inverses in associative rings and semigroups.
The American mathematical monthly, 65(7):506–514, 1958.

[15]
M. P. Drazin.
Commuting properties of generalized inverses.
Linear and Multilinear Algebra, 61(12):1675–1681, 2013.

[16]
H. Fitting.
Die theorie der automorphismenringe abelscher gruppen und ihr
analogon bei nicht kommutativen gruppen.
Mathematische Annalen, 107:514–542, 1933.

[17]
N. Jacobson.
Basic Algebra.
Number v. 2 in Basic Algebra. W.H. Freeman, 1985.

[18]
T.Y. Lam and P. P. Nielsen.
Jacobson’s lemma for Drazin inverses.
Contemp. Math.: Ring theory and its applications, 609:185–196,

[19]
T. Leinster.
Counting nilpotent operators.
https://www.maths.ed.ac.uk/ tl/cno.pdf, 2019.

[20]
T. Leinster.
The eventual image.
arXiv preprint arXiv:2210.00302, 2022.

[21]
J. Marovt.
Orders in rings based on the core-nilpotent decomposition.
Linear and Multilinear Algebra, 66(4):803–820, 2018.

[22]
S. K. Mitra, P. Bhimasankaram, and S. B. Malik.
Matrix partial orders, shorted operators and applications,
volume 10.
World Scientific, 2010.

[23]
W. D. Munn.
Pseudo-inverses in semigroups.
In Mathematical Proceedings of the Cambridge Philosophical
Society, volume 57, pages 247–250. Cambridge University Press, 1961.

[24]
W. K. Nicholson.
Strongly clean rings and Fitting's Lemma.
Communications in algebra, 27(8):3583–3592, 1999.

[25]
R. Puystjens and M.C. Gouveia.
Drazin invertibility for matrices over an arbitrary ring.
Linear algebra and its applications, 385:105–116, 2004.

[26]
R. Puystjens and D. W. Robinson.
Generalized inverses of morphisms with kernels.
Linear Algebra and Its Applications, 96:65–86, 1987.

[27]
F. P. Romo.
Core-nilpotent decomposition and new generalized inverses of finite
potent endomorphisms.
Linear and Multilinear Algebra, 68(11):2254–2275, 2020.

[28]
P. V. Srinivasan.
Dagger Linear Logic and Categorical Quantum Mechanics.
PhD thesis, University of Calgary, 2021.

[29]
Z. Wang.
A class of drazin inverses in rings.
Filomat, 31(6):1781–1789, 2017.

[30]
Y. Wei and C. Deng.
A note on additive results for the Drazin inverse.
Linear and Multilinear Algebra, 59(12):1319–1329, 2011.

[31]
H. Yang and X. Liu.
The Drazin inverse of the sum of two matrices and its applications.
Journal of Computational and Applied Mathematics,
235(5):1412–1417, 2011.

[32]
L. Yu, T. Bu, and J. Zhou.
A note on the Drazin indices of square matrices.
The Scientific World Journal, 2014, 2014.