if and only if for any non-zero $a\in \mathfrak A(  \infty )$ and every lift $\{ a_n \}_{n=1}^\infty \in \ell^{\infty}( \{ \mathfrak A( n ) \} )$ of $\overline \tau(a)$, there is an $\epsilon > 0$ such that $\| a_n \| < \epsilon$ for only finitely many $n\in {\ensuremath{\mathbb{N}}\xspace}$.

Suppose that $a\in \mathfrak A( \infty ) \setminus \{ 0 \}$, $\{ a_n \}_{n=1}^\infty \in \ell^\infty ( \{ \mathfrak A( n ) \} )$ lifts $\overline \tau(a)$ such that $\| a_n\| < 1/m$ for infinitely many $n\in {\ensuremath{\mathbb{N}}\xspace}$ for every $m\in {\ensuremath{\mathbb{N}}\xspace}$. Then we may pick an infinite subsequence $\{ a_{n_k} \}_{k=1}^\infty$ such that $\lim_{k\to \infty}\| a_{n_k} \| =0$. Let $F=\{ n_1 , n_2,\dots \}$ and define $$b_n = \left\{ \begin{array}{ll} a_n & \text{ if } n\notin F \\ 0 & \text{ if } n\in F. \end{array} \right.$$ Then $\{ b_n \}_{n=1}^\infty$ is a lift of $\overline \tau(a)$. Set $U_k = \widetilde {\ensuremath{\mathbb{N}}\xspace}\setminus \{ n_1,\dots n_k\}$. Then $\bigcap_{k=1}^\infty U_k = \widetilde {\ensuremath{\mathbb{N}}\xspace}\setminus F$ and since $F$ is infinite, the interior of this set, say $U$, is a subset of ${\ensuremath{\mathbb{N}}\xspace}$. Identify $\mathfrak A$ with the pullback $\mathfrak A( \infty ) \oplus_{\overline \tau,\rho} \ell^{\infty} (\{ \mathfrak A( n ) \} )$. Then 