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Let $1 to H_1 to G to H_2 to 1$ be a short exact sequence of abstract groups.

Question: If $H_1$, $H_2$ have fixed topologies, can we endow $G$ with a topology such that the sequence above becomes a short exact sequence of topological groups? If yes, is this topology then uniquely determined?

Motivation: I want to understand the topology of the Weil group $W_K$ which is a dense subgroup of the absolute Galois group $G_K$ (which has this weird profinite topology I find hard to get). However, we have a short exact sequence of abstract groups

$$1 to I_K to W_K to langle operatorname{Frob}_k rangle to 1$$
where $I_K$ denotes the inertia subgroup of $G_K$ and $operatorname{Frob}_k : x mapsto x^{|k|}$ is the Frobenius element of the absolute Galois group $G_k$ of the residue field $k$ of $K$.

If I understand it correctly, this sequence is not a short exact sequence of topological groups if we give $W_K$ the subspace topology of $G_K$. However, if we give $I_K$ and $langle operatorname{Frob}_k rangle$ the subspace topologies of $G_K$ and $G_k$ respectively, we can give $W_K$ a unique topology such that this sequence becomes a short exact sequence of topological groups.

I heard that this causes $W_K$ to have a finer topology than the usual subspace topology, $I_K$ to be open and the maximal compact subgroup of $W_K$ etc. but I do not really understand that.

Could you please explain this to me? Thank you!

Not saying anything about the Weil group at this point. Trying to settle the first question instead with the following construction that smacks of cheating.

Let $i:H_1to G$ be the monomorphism, and $p:Gto H_2$ be the epimorphism in this short exact sequence. Exactness means that $N:=i(H_1)=ker(p)unlhd G$. If I got it right $H_1$ and $H_2$ are topological groups, and you want to give a topology to $G$ so that $G$ becomes a topological group, and the mappings $i,p$ are continuous.

Unless I made a silly mistake this can be achieved by giving $G$ the topology defined by letting the cosets of $N$ to form a basis. In other words, the open sets of $G$ are the arbitrary unions of cosets of $N$. This is trivially a topology $tau$. Furthermore:

1. 
   $p:(G,tau)to H_2$ is continuous for much the same reason that any mapping from a discrete space to any topological space is continuous. The preimage of any subset of $H_2$ is a union of cosets of $N$.


2. 
   $i:H_1to(G,tau)$ is continuous for much the same reason that any map to a space with a trivial topology is continuous. The preimage of any open subset $intau$ is either all of $H_1$ or the empty set.


3. 
   $(G,tau)$ is a topological group. For if $x,yin G$ are arbitrary, then the product map $m:Gtimes Gto G, m(x,y)=xy$ satisfies $m(xN,yN)subseteq xyN$. Any open set containing $xy$ contains the coset $xyN$, so continuity of $m$ follows. Continuity of the inverse mapping is verified in the same way.

Seems to me that you really wanted to ask a slightly different question. I'm afraid I cannot suggest one. Anwyway, it seems morally certain to me that the topology $G$ needs to have to satisfy items 1 to 3 is not unique at all.