\lim_{m \to \infty} a_m^k.$$ By (A), $Z_k$ is complete, hence $a^k \colon X \to Z_k$ is well defined.

It follows from the definition that $p_k \circ a^{k+1}_m = a^k_m$. Passing to the limit we have $$p_k \circ a^{k+1} = \lim_{m \to \infty} p_k \circ a^{k+1}_m = \lim_{m \to\infty} a^k_m = a^k.$$ Therefore, for each $x$ the sequence $(a^k(x))_k$ is a thread in $Z$ and we can define a map $g \colon X \to Z$ by the formula $$(g(x))_k = a^k(x).$$

Observe that by (ii) and by the definition of $a^k_m$ we have $$a^k_m |_A = \pi^m_k \circ f_m |_A = \pi^m_k \circ \pi_m \circ g_0  |_A = \pi_k \circ g_0 |_A.$$ Therefore $a^k\mid_A = \pi_k \circ g_0$ for each $k$, hence $g |_A = g_0 |_A$.

\[thm:ane condition\] Let $$Z=\lim_{\longleftarrow}\left( Z_1\xleftarrow{p_1}Z_2\xleftarrow{p_2}\cdots \right).$$ Let ${{\mathcal{F}_{i}}}$ be a cover of $Z_i$. If the conditions of Definition \[def:limit\] are satisfied and $Z_1$ is an $ANE(n)$, then $Z$ is an $ANE(n)$.

We will show that $Z$ is an $ANE(n)$ directly from the definition. Let $X$ be an at most $n$-dimensional metric space and let $A\subset X$ be a closed subset. Take $g_0\colon A\to Z$. By assumption, $Z_1$ is an $ANE(n)$ so we can extend 