# Topological HNN extensions

First, let me recall what an abstract HNN extension is. Let $G$ be an abstract group, $A, B < G$ be subgroups of $G$ and $\phi : A \to B$ be an isomorphisms. Then there is a group $H$ and an element $h \in H$ such that $G < H$ and for any $a \in A$ one has $\phi(a) = hah^{-1}$. Such an extension is, certainly, not unique, but there is a canonical way to construct one. The group $H$ obtained in this canonical way is called HNN extension of $G$ by $\phi$. More information can be found in a nice wikipedia article.

My question is: for what classes of topological groups HNN extension is possible?

More formally, let $G$ be a topological group, $A, B < G$ and $\phi : A \to B$ is an isomorphism in the category of topological groups. Is there a topological group $H$ that contains $G$ and such that $A$ and $B$ are conjugated inside $H$? If yes, what properties $H$ inherits from $G$, e.g., if $G$ is metrizable can $H$ be chosen metrizable?

Let me also mention that classical HNN construction uses free products and amalgamation of abstract groups. It is known that amalgamation of two Hausdorff topological groups over a closed subgroup may not be Hausdorff (though amalgamation, that appears in HNN extension is quite special, I don't know how bad it is).

The simplest case when $A$ is generated by one element already seems interesting.

-
Maybe its useful to allow non injective maps G->H as maps above. Maybe then one gets always a group with the desired universal property. Furthermore in the example of Hausdorff groups´,I generally think it is better to think of coequalizers in that category, instead of worrying, that a quotient (in another category, here top. groups) is ot again in that category. Then one has to get back to that category (usually one takes the left adjoint to the forgetful functor); so the question is then again: has the desired category coproducts and coequalizers and how do the look like. – HenrikRüping Feb 27 '10 at 12:09
Thanks. I agree with both suggestions. Though I am mostly interested in the case when G->H is indeed injective (and would be happy to know if this is always possible for, say, metrizable groups). To take a left adjoint is a good idea. But the main example, that I have in mind, is a category of metrizable groups, and then one, probably, may just to pass to a subgroup. – Konstantin Slutsky Feb 27 '10 at 17:48