Map between tower of modules inducing an isomorphism Consider two towers of $R$ modules
$$E_1\subset E_2\subset \cdots \subset \bigcup_n E_n\subset \bigcap_n Z_n \subset\cdots\subset Z_2\subset Z_1$$
and
$$\overline{E_1}\subset \overline{E_2}\subset \cdots \subset \bigcup_n \overline{E_n}\subset \bigcap_n \overline{Z_n} \subset\cdots\subset \overline{Z_2}\subset \overline{Z_1}.$$
A map $f:Z_1\to\overline{Z_1}$ so that $f(Z_n)\subset \overline{Z_n}$ and $f(E_n)\subset \overline{E_n}$ induces a map
$$\tilde f:\frac{\bigcap_n Z_n}{\bigcup_n E_n}\to\frac{\bigcap_n \overline{Z_n}}{\bigcup_n \overline{E_n}}.$$
It is easy to check that if $\tilde{f_n} : \frac{Z_n}{E_n}\to \frac{\overline{Z_n}}{\overline E_n}$ is injective for all $n$ then $\tilde f$ is also injective, and one can construct counter-examples to the analog statement with surjectivity.
My question is:

If $\tilde{f_n}$ is an isomorphism for all $n$, does it follow that $\tilde f$ is an isomorphism?

I suspect so, but all my attempts of proof have failed.
 A: Here's a counterexample.  Let $\overline{Z_1}$ be freely generated by elements $e_n$ for each $n$ and one more element $x$.  Let $\overline{E_n}$ be the submodule generated by $e_i$ for all $i\leq n$ and let $\overline{Z_n}=\overline{Z_1}$ for all $n$.  Let $E_n=0$ for all $n$ and let $Z_n$ be the submodule of $\overline{Z_n}$ generated by $x+e_i$ for all $i\geq n$.  Let $f:Z_1\to\overline{Z_1}$ be the inclusion map.
Note that for each $n$, $\overline{Z_n}/\overline{E_n}$ is freely generated by $x$ and the $e_i$ for $i>n$, while $Z_n/E_n$ is freely generated by $x+e_n$ and $e_i-e_n$ for $i>n$.  Since $e_n\in\overline{E_n}$, $\tilde{f}_n$ maps $x+e_n$ to $x$ and $(x+e_i)-(x+e_n)=e_i-e_n$ to $e_i$, so it is an isomorphism.
However, $\bigcap Z_n/\bigcup E_n=0/0=0$ while $\bigcap\overline{Z_n}/\bigcup\overline{E_n}$ is freely generated by $x$.  Thus $\tilde{f}$ is not an isomorphism.
(This is the universal example, in the following sense.  If you have $x\in\bigcap\overline{Z_n}$ and each $\tilde{f}_n$ is surjective, then for each $n$ there must be $e_n\in\overline{E_n}$ and $z_n\in Z_n$ such that $f(z_n)=x+e_n$.  This example is then the free diagram of the shape you describe on such elements $x$, $e_n$, and $z_n$.)
