Suppose that $G$ is an abelian topological group with a descending filtration $(G_n)_n$ of subgroups.
Suppose that $0 \to G' \xrightarrow{f} G \xrightarrow{g} G'' \to 0$ is an exact sequence of abelian groups.
Then $(f^{-1}(G_n))_n$ and $(g(G_n))_n$ are descending filtrations of subgroups in $G'$ and $G''$ respectively.
The topologies induced by these filtrations are the subspace and quotient topologies respectively.
Viewing $0\to G'\xrightarrow{f} G \xrightarrow{g} G'' \to 0$ as an exact sequence of topological groups, is it true that $$G/f(G')\cong G''$$ as topological groups?
Any help gratefully received!
Edit in response to Moishe Cohen's comment below.
Question 1. How do the $(f^{-1}(G_n))_n$ and $(g(G_n))_n$ make $G'$ and $G''$ into topological groups?
We define topologies on $G'$ and $G''$ by declaring the descending filtrations of subgroups mentioned above to be neighbourhood bases for $0$ in $G'$ and $G''$ respectively. By homogeneity, this in turn defines a unique topology on $G'$ and $G''$ which is compatible with the respective group structures.
Question 2. How do we show that the induced group isomorphism $G/f(G')\xrightarrow{\bar{g}} G''$ is continuous?
To show that a group homomorphism of topological groups, say $\phi:H\to H',$ is continuous it is sufficient (by homogenity and the fact that $\phi(0)=0$) to show that, for any neighbourhood $N$ of $0$ in $H',$ the preimage $\phi^{-1}(N)$ is a neighbourhood of $0$ in $H.$ We apply this in our case...
Note that the quotient topology on $G/f(G')$ is exactly the topology determined by the filtration:
$$(G_1+f(G'))/f(G')\supseteq (G_2+f(G'))/f(G') \supseteq \ldots$$
Now let $N$ be any neighbourhood of $0$ in $G''.$
Then $N$ contains some $g(G_n)$ and so $\bar{g}^{-1}(g(G_n))\subseteq \bar{g}^{-1}(N)$ and $0 \in \bar{g}^{-1}(g(G_n)).$
We clearly have that $(G_n+f(G'))/f(G')\subseteq \bar{g}^{-1}(g(G_n)).$
This establishes continuity of $\bar{g}.$
Completion of the proof: $\bar{g}$ is an open map.
Let $N$ be any neighbourhood of $0$ in $G/f(G').$
Then $N$ contains some $(G_n+f(G'))/f(G')$ and so
$$\bar{g}(N)\supseteq \bar{g}((G_n+f(G'))/f(G'))\supseteq g(G_n).$$
Hence we are done, as before.
Post mortem.
The proof boils down to the following identity:
$$g(G_n)=\bar{g}\big((G_n+f(G'))/f(G')\big).$$
A good choice for a counter-example are normed vector spaces, since they also are abelian topological groups.
Let $\lVert\cdot\rVert_1 $ and $\lVert\cdot\rVert_2 $ be two inequivalent norms on a vector space $X$. Then the identity map $$ \text{id} \colon (X, \lVert\cdot\rVert_1+ \lVert\cdot\rVert_2) \to (X, \lVert\cdot\rVert_1) $$ can not be an isomorphism, otherwise $\lVert\cdot\rVert_1 $ and $\lVert\cdot\rVert_2 $ would be equivalent