$i$ since we changed the row of this rook to $i'$. Note that row $i'$ is otherwise unoccupied since $P'\in M$ and $M$ is a matching. Similarly, we have solved the problem of a rook in column $j$ since we changed the column of this rook to $j'$. This column is otherwise unoccupied since $M$ is a matching and $P'\in M$. Thus, $f_1(M)$ is a matching in $B$. Moreover, $f_1(M)$ has no rooks in $T$ and so $f_1(M) \in A_{{\overline}{T}}$.

In addition, $f_1$ is injective. Define $r_1': B^+ \to B^+$ by $$r_1'(a,b):=\begin{cases}
(i',j') &\text{ if } (a,b) = (i,j)\\
(i,b) &\text{ if } a = i', b\neq j\\
(a,j) &\text{ if } a\neq i, b = j'\\
(a,b)  &\text{otherwise}
\end{cases}$$ and define $f_1': A_{{\overline}{T}} \to A_{{\overline}{S}}$ by $$f_1'(M) = \{r_1'(a,b): (a,b)\in M\}.$$ It is straightforward to check that $f_1'(f_1(M))= M$ for all $M\in A_{{\overline}{S}}$ and $f_1(f_1'(M))= M$ for all $M\in A_{{\overline}{T}}$. Thus, there is a bijection between the matchings in $B^+$ with $P'$ and no rooks in $S$ and matchings in $B^+$ with $P$ and no rooks in $T$.

*Case 2:* Suppose $M\in A_S$. That is, $M$ is a matching in $B^+$ with a rook in $P'$ and a rook 