Mapping Tori and Topological Groups

Question

This post states that it is possible to define a topological group structure on a covering space of a topological group provided certain conditions are satisfied. I am wondering if it is possible (and if so, when does it happen) that the mapping torus of a homeomorphism of a given surface satisfies these conditions and is in fact a topological group?

Answer 1

The crucial fact here is that the fundamental group of a topological group must be abelian.

As you point out in the comment, $\pi_{1}(M_{f})$ contains $\pi_{1}(X)$ as a subgroup and hence if $\pi_{1}(X)$ is not abelian, $\pi_{1}(M_{f})$ is neither and so $M_{f}$ cannot be a topological group.

The only closed surfaces with abelian fundamental group are the sphere and torus. If $X = T$ is a torus, $\pi_{1}(X)$ is non-trivial and an additional requirement for $\pi_{1}(M_{f})$ to be abelian is that the homomorphism $\pi_{1}(f): \pi_{1}(T) \rightarrow \pi_{1}(T)$ is the identity. Since $T$ is an Eilenberg-Maclane space, if we know action of a map on the fundamental group, we know it up to homotopy and so $f$ is homotopic to the identity. In this case $M_{f} \simeq T \times S^{1}$ and so it is homotopy equivalent to a product of three circles. Hence, it is at least homotopy equivalent to a topological group.

If $X = S^{2}$ is the sphere, then since $f$ is a homeomorphism it is either homotopic to the identity or to the reflection by a hyperplane. If $f$ is homotopic to the identity, then $M_{f} \simeq S^{2} \times S^{1}$. I strongly doubt $M_{f}$ can be a topological group because if it was then $S^{2} \times S^{1}$ would inherit a structure of an H-space which I find rather unlikely, but I cannot find a formal argument.

However, if $f$ is a homeomorphism homotopic to the reflection by a hyperplane, then it reverses orientation and $M_{f}$ is not orientable as a 3-manifold. Hence, it cannot be a topological group as all topological groups are orientable.