Constructing Topological Groups

Question

In general, is there a way to construct topological groups? That is, given two topological groups $X$ and $Y,$ can I construct a topological group $Z$ using $X$ and $Y$ in said construction? I have been trying various constructions but it seems that the group structure imposes great limitations.

Answer 1

The product $X\times Y$ is a topological group containing both $X$ and $Y$ as subgroups (as $X\times\{1\}$ and $\{1\}\times Y$).

More generally, if $B$ is a topological group and $f:X\to B$ and $g:Y\to B$ are continuous group homomorphisms, then we can construct the fibre product

$$X_f\times_g Y = \{(x,y)\in X\times Y: f(x)=g(y)\}$$ which is a closed subgroup of $X\times Y$. When $B$ is the trivial group $\{1\}$, this is equal to $X\times Y$.

Edit

To address your question about fibre bundles: if $g$ is a fibre bundle, then its fibre is isomorphic to $\ker(g)$ and the projection onto the first factor $X_f\times_g Y\to X$ is a fibre bundle with fibre $\ker(g)$. Likewise if $f$ is a fibre bundle, then $X_f\times_g Y\to Y$ is a fibre bundle with fibre $\ker(f)$.

Note that if $X$ and $B$ are Lie groups, then every smooth surjective homomorphism $X\to B$ is indeed a fibre bundle with fibre isomorphic to the kernel.

Another generalization of the product construction is semidirect products. Suppose we have a homomorphism $\phi:X\to \operatorname{Aut}(Y)$ such that the map $$X\times Y\to Y: (x,y)\mapsto \phi_x(y)$$ is continuous (where $\phi_x := \phi(x)$). Then the semidirect product $X\times_\phi Y$ is the set $X\times Y$ equipped with the "twisted" product $$(x,y)\cdot (x',y') = (xx',y\phi_x(y')).$$ This is a topological group which again contains $X$ and $Y$ as subgroups, and when $\phi$ is the trivial homomorphism it is the usual product $X\times Y$.