2-manifold with involution without fixed points is a boundary of a 3-manifold.

Question

I am preparing for geometry/topology test and I can not handle the following problem from previous years:

Let $F$ is closed orientable 2-manifold with involution without fixed points. Prove that $F$ is a boundary of a 3-manifold.

I have only several vague ideas about it:

1. $F$ should be a sphere win $n$ handles (known theorem).

2. I feel that I saw more general problem earlier (but I am not sure if it is true or that I correctly remember the statement, sorry if it is completely false). So closed $n-$ manifold $M$ with involution without fixed points is a boundary of some $n+1$-manifold $M'$.

I am interested in solving the original problem more than in the general statement (it is true), but if there is no complications in solving more general case, please let me know.

Thanks a lot for your help!

Answer 1

This does indeed hold in a more general setting: consider $F$ a closed $n$-manifold together with a fixpoint-free smooth involution $f$. Now proceed as follows:

1. Understand that the quotient map: $F\to F/f$ is a covering map.
2. Understand that the mapping cylinder $M_{F\to F/f}$ is an $(n+1)$-dimensional manifold with boundary.
3. Verify that $F=\partial M_{F\to F/f}$.

To illustrate and more intuitively understand what is going on, you can try this with the circle and look at the Mobius band, the mapping cylinder of the 2-covering.