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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.

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    It is not clear to me what the chains look like. If you have a chain $E_1\subset E_2\subset\ldots\subset E_n$ then $\bigcup_n E_n=E_n$, so why write $\bigcup_n E_n$? The same for $\bigcap_n Z_n$. – 2017-01-12
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    @Servaes The chains are infinite, for every $n$ you have a module $E_n$ and $Z_n$. – 2017-01-12
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    I clearly haven't had my coffee yet ;) thank you for the clarification. – 2017-01-12

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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 $z_n$.)

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    I had thought that I needed this statement to be true to see that a morphism of spectral sequences induces an isomorphism of the modules on the $\infty$ page if it is an isomorphism on any other page. This is a bit unsettling. – 2017-01-13
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    In the spectral sequence situation, what you need to know is not just that you have an isomorphism in a single cell on every page but that you have an isomorphism in _every_ cell on every page. This together with the map commuting with the differentials tells you that you have isomorphisms not just on the quotient $Z_n/E_n$, but also on $Z_n$ and $E_n$ themselves (since they are just kernels and images of certain maps, and you have isomorphisms between both the domains and codomains of those maps which commute with them). – 2017-01-13
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    That has really helped me, the original statement I had wanted to show is clear now. Thanks a lot! – 2017-01-14