map induced by the Busby map. Note that if $a\in \mathfrak A$ and $b\in \mathfrak A( n )$, then $ab = \pi_n(a)b\in \mathfrak A( n )$. Hence if $\{ a_n \}_{ n = 1}^{\infty} \in c_{0} ( \{ \mathfrak A( n ) \} )$ we get that $$\sigma (a) \left( \{a_n\}_{n=1}^\infty \right)= \left\{ \pi_n(a) a_n \right\}_{n=1}^\infty = (\iota \circ \ell^{\infty} ( \{ \pi_n\} ) (a)) \left( \{ a_n\}_{n=1}^\infty \right)$$ and similarly $\{ a_n\}_{n=1}^\infty \sigma (a) = \{a_n\}_{n=1}^\infty (\iota \circ \ell^{\infty} ( \{ \pi_n\} ) (a)$ and thus $\sigma = \iota \circ \ell^{\infty}(\{ \pi_n\} ) $. This implies that $\tau_{\mathfrak{A}} = \overline \iota \circ \overline \tau_{\mathfrak{A}}$ by a canonical uniqueness argument.

Let $\mathfrak{A}$ be a $C^{*}$-algebra over $\widetilde{{\ensuremath{\mathbb{N}}\xspace}}$. The $*$-homomorphism $\overline{\tau}_{\mathfrak{A}}$ in Lemma \[l:reducedBusby\], will be called the *reduced Busby map of $\mathfrak{A}$*.

To ease the notation throughout the paper, we will remove the subscript $\mathfrak{A}$ of all $*$-homomorphisms in Lemma \[l:reducedBusby\] whenever we are working with only one algebra $\mathfrak{A}$.

\[l:continuous\] Let $\mathfrak A$ be a $C^\ast$-algebra over $\widetilde {\ensuremath{\mathbb{N}}\xspace}$ and let ${  \overline \tau  \colon \mathfrak A( \infty)   \rightarrow  q_{\infty} ( \{ \mathfrak A( n ) \} )  }$ be the reduced Busby map. Then $\mathfrak A$ is continuous 