Fiber bundle map is proper if the model fiber is compact

algebraic-topologydifferential-geometryfiber-bundlesgeneral-topologymanifolds

This is one direction of problem 10-19 (c) from John Lee's Introduction to Smooth Manifolds.

Suppose $\pi: E \to M$ is a fiber bundle with fiber $F$.

Show that $\pi$ is a proper map if $F$ is compact.

Here, $M$, $E,F$ are topological spaces with no further assumption and $\pi:E \to F$ is a surjective continuous map with the property that for each $x\in M$, there exist a neighborhood $U$ of $x$ in $M$ and a homeomorphism $\Phi: \pi^{-1}(U)\to U\times F$, called a local trivialization of $E$ over $U$, such that we have $\pi_1 \circ \Phi = \pi$, where $\pi_1$ is the projection onto $U$.

I have seen a solution to this problem Let $\pi: E \rightarrow M$ be a fiber bundle with fiber $F$. Then $\pi$ is a proper map $\iff F$ is compact.
here that uses the manifold assumptions on $E$ and $M$, namely Hausdorff and local compactness, but I cannot figure out a way to prove this in the general case without these assumptions on the topological spaces.

Suppose that $F$ is compact and $A \subset M$ is compact. Then we need to show that $\pi^{-1}(A)$ is compact. Clearly we would need to use the local trivialization property, so that for each $p \in A$, we can find a cover $U_p$ such that $\Phi: \pi^{-1}(U_p) \to U_p \times F$ is a homeomorphism and $p \in U_p$. And cover $A$ with finitely many $U_i$'s. But I cannot think of a way to progress from here without resorting to local compactness.

How can we prove this? I would greatly appreciate some help.

Best Answer

Here is a proof using nets. Recall that a space $X$ is compact if and only if every net in $X$ contains a convergent subnet (see e.g. this question). Let $K\subset M$ be a compact, $C:= \pi^{-1}(K)$. Let $x_\bullet$ be a net in $C$. Then, by compactness of $K$, the net $\pi(x_\bullet)$ contains a subnet converging to a point $k\in K$. This subnet has the form $\pi(y_\bullet)$, where $y_\bullet$ is a subnet in $x_\bullet$. Since $\pi(y_\bullet)\to k$, WLOG we can assume that $\pi(y_\bullet)$ is contained in a neighborhood $U$ of $k$ (in $M$) over which the bundle is trivial. By trivializing the bundle over $U$, we obtain that $y_\bullet$ is contained in $U\times F$. Let $\pi': U\times F\to F$ be the projection to the second factor. Since $F$ is compact, the net $y_\bullet$ contains a further subnet $z_\bullet$ such that $\pi'(z_\bullet)$ converges to some $f\in F$. Since $\pi(z_\bullet)$ still converges to $k\in K$, we obtain that $z_\bullet$ converges to $(k,f)\in C=\pi^{-1}(K)$. Since $z_\bullet$ is a subnet in $x_\bullet$ (a subnet of a subnet is again a subnet), we obtain that the net $x_\bullet$ contains a subnet converging to a point in $C$. Since $x_\bullet$ was arbitrary, $C$ is compact.

Lastly, I am not sure that the definition of properness in terms of compacts is useful once you work with non-Hausdorff spaces. Algebraic geometers (working with Zariski topology), tend to use a different notion, the one of universally closed maps.

Related Question