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.
Edit: Added some references; thanks to t.b. for many of these.
Suppose I take as my category topological (Hausdorff) groups and continuous homomorphisms. This has a "zero object"-- just the trivial group $\{1\}$. Then the (group theoretic) kernel of a continuous homomorphism will be a closed subgroup, and hence a topological group in its own right. It's easy to see that this shows that this category has "kernels" in the abstract sense.
Forming cokernels is a little trickier: given $f:G\rightarrow H$ let $Q$ be the closed normal subgroup of $H$ generated by $f(G)$. Then $H/Q$ is a Hausdorff topological group as well (See Hewitt+Ross, Thm. 5.26). The cokernel of $f$ is then the quotient homomorphism $q:H\rightarrow H/Q$. Indeed, clearly $qf=0$; if $q'f=0$ then $f(G) \subseteq \ker q'$ and so as $\ker q'$ is a closed normal subgroup, $Q\subseteq \ker q'$.
Similarly, $f:G\rightarrow H$ is "monic" if and only if $f$ is (set-theoretic) injective. $f$ is "epi" if it has dense range (Edit: the "only if" claim fails for locally compact groups, see an example of Reid; but it's true for locally compact groups in the category of Hausdorff groups, see a paper of Nummela, "On epimorphisms of topological groups" (1978); for merely Hausdorff topological groups, "only if" is false, see paper of Uspenskij).
I then believe that we have an abstract, category-theoretic definition of "exact sequence", namely $$ 0 \rightarrow G \xrightarrow{f} H \xrightarrow{g} K \rightarrow 0 $$ with $f$ monic, $g$ epi, $f$ is the kernel of $g$ and $g$ is the cokernel of $f$. Translating, this means precisely that $f$ is a homeomorphism onto its range, which is the usual kernel of $g$, and $K$ is isomorphic to $H/f(G)$, with $g$ the quotient map. This definition is copied from Mac Lane, "Categories for the working mathematician", Chapter VIII, Section 3. Now, that section is all about abelian categories, but the definition works in more generality.
Is this the "correct" definition? I absolutely wouldn't want to say that. But if you wanted something to start with, this seems reasonable. If you want something slightly different for an application? Then what's wrong with just spelling out, very clearly, what you mean by "exact", and maybe saying something to the effect that this is not an entirely standard defintion?