If $A$, $B$, and $C$ are topological groups, and $f: A \to B$ and $g: B \to C$ are two continuous group homomorphisms, what does it usually mean for
$1 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \to 1$
to be exact?
I would expect that one at least expects exactness as abstract groups (as opposed to, say $g(B)$ just being dense in $C$).
Does one also expect $f$ to be a homeomorphism of $A$ onto $f(A)$? And if so, does one further expect $C$ to be homeomorphic to $B/f(A)$ with the quotient topology>?
Would one's expectations change if $A$, $B$, and $C$ were all abelian topological groups? (And what if they were all topological $G$-modules for some topological group $G$ -- that is, if each was an abelian topological group and additionally we had continuous group action maps $G \times A \to A$ and same for $B$ and $C$, and $f$ and $g$ were $G$-equivariant?)
This seems to just be a question of semantics. I definitely see at least some authors require $A$, $C$ to have the subspace, quotient topology. But then something I read somewhere else seems to suggest that the condition should only be on the topology of $A$ only, or maybe no additional topological condition at all --- I can't quite tell...
So, what do you expect when you read or hear that $1 \to A \to B \to C \to 1$ is an exact sequence of topological groups? Is there a consensus?
UPDATE: The question has been reduced to understanding what it means for a sequence to be exact in a category that is not abelian but has kernels and cokernels (and a zero object). There are some ideas in Matthew Daws's answer and the following comments below, but a reference would be great.