A( \infty ))$ are $\widetilde {\ensuremath{\mathbb{N}}\xspace}$-equivariant.

For continuity we check that if $U_1 \supsetneq U_2 \supsetneq \dots$ is a strictly decreasing sequence of open subsets of $\widetilde {\ensuremath{\mathbb{N}}\xspace}$, and we let $U$ denote the interior of $\bigcap_{n=1}^\infty U_n$, then $\bigcap_{n=1}^\infty \mathfrak E(U_n) = \mathfrak E(U)$. Observe that $U\subseteq {\ensuremath{\mathbb{N}}\xspace}$ since the sequence $\{ U_n \}_{n=1}^\infty$ is strictly decreasing, and thus $\mathfrak{A}(U) \subseteq \mathfrak{A}({\ensuremath{\mathbb{N}}\xspace})$. Since $\mathfrak A$ is continuous we have that $$\bigcap_{n=1}^\infty \mathfrak E(U_n) = \mathfrak{E}\cap \bigcap_{n=1}^\infty (\mathfrak A(U_n)\oplus  i_\infty(\mathfrak B)(U_{n}) ) = \mathfrak E \cap \left(\mathfrak A(U) \oplus \bigcap_{n=1}^\infty i_\infty(\mathfrak B)(U_{n}) \right).$$ Since $\pi_{\infty} ( \mathfrak{A}(U)) = 0$ and $\iota$ is injective it follows that $$\bigcap_{n=1}^\infty \mathfrak E(U_n) = \mathfrak E \cap (\mathfrak A(U) \oplus 0) = \mathfrak E\cap (\mathfrak A(U) \oplus i_\infty(\mathfrak B)(U) ) = \mathfrak E(U).$$

\[l:reducedBusby\] Let $\mathfrak A$ be a $C^\ast$-algebra over $\widetilde {\ensuremath{\mathbb{N}}\xspace}$, and consider the commutative diagram with exact rows $$\xymatrix{
0 \ar[r] & \mathfrak A({\ensuremath{\mathbb{N}}\xspace}) \ar[r] \ar[d]^\cong & \mathfrak A \ar[r] \ar[d]^{ \ell^{\infty} ( \{ \pi_n \} ) } & \mathfrak A( \infty ) \ar[r] \ar[d]^{\overline \tau_{\mathfrak{A}}} & 0 \\
0 \ar[r] & c_{0} ( \{ \mathfrak A( n ) \} ) \ar@{=}[d] \ar[r] & \ell^\infty( \{ \mathfrak A( n ) \} 