Fiber bundle is compact if base and fiber are

Question

I want to show that the total space $E$ is compact if the fiber $F$ and the base space $B$ are compact. Let $\pi$ denote the fiber projection. Since every point in $B$ has an open neighborhood $U$ whose preimage $\pi^{-1}(U)$ is homeomorphic to $U\times F$ via $\phi_U=(\pi,\beta_U)$, there are finitely many such sets covering $B$. Let's call this collection $\mathcal U$. Then $\mathcal S:=\{\phi_U^{-1}(V\times W)\mid U\in\mathcal U,V\text{ open}\subseteq U,W\text{ open}\subseteq F\}$ is a subbase for the topology on $E$.

By Alexander's subbase lemma we can restrict ourselves to considering a cover $\mathcal C\subseteq\mathcal S$ of $E$. (Mariano Suárez-Alvarez recommends this in his answer here) For each $b\in B$ the fiber $\pi^{-1}(b)$ is compact and thus contained in the union ${\bigcup_{i=1}^{n_b}\phi^{-1}_{U_i^b}(V_i^b\times W_i^b)}$ of sets in $\mathcal C$. Define $V^b:=\bigcap_{i=1}^{n_b}V_i^b$. By compactness $B=\bigcup_{k=1}^l V^{b_k}$. I think that $E=\bigcup_{k=1}^l\bigcup_{i=1}^{n_{b_k}}\phi^{-1}_{U_i^{b_k}}(V_i^{b_k}\times W_i^{b_k})$, similar to the proof of compactness of the product of two compact spaces.

Unlike the trivial bundle case, however, where the homeo's are just identities, here the $\phi_{U}$'s could be pretty ugly, especially where they are overlapping, and we don't have $y\in\bigcup_{i=1}^{n_{b_k}}\phi^{-1}_{U_i^{b_k}}(V_i^{b_k}\times W_i^{b_k})\ $ whenever $\ \pi(y)\in V^{b_k}$

So I try to get to a contradiction by assuming that for each $k=1,\dots,m$ such that $\pi(y)\in V^{b_k}$ we have $\forall i=1,\dots,n_k:\ y\notin\phi^{-1}_{U_i^{b_k}}(V_i^{b_k}\times W_i^{b_k})$ which then implies $\beta_{U^k_i}(y)\notin W_i^{b_k}$. If there were a $k$ such that $U^k_1,\dots,U^k_{n_k}$ were all the same then I would be finished. But maybe the $b_k$'s should have been chosen in certain way.

Maybe someone remembers how the proof worked.

Answer 1

We use the following characterization of compactness: every infinite net has a convergent subnet.

Let $\{x_{\lambda}\}_{\Lambda}$ be an infinite net in $E$, $\Lambda$ the directed set of the net. As $B$ is compact, there is an infinite subnet $\Lambda_1 \subset \Lambda$ for which $\{\pi(x_{\lambda})\}_{\Lambda_1}$ converges to a point $b \in B$. Fix a local trivialization $\phi_U : \pi^{-1}(U) \rightarrow U \times F$ for $b \in U$. Pare off the net $\Lambda_1$ so that $\pi(x_{\lambda}) \in U$ for all $\lambda$.

Now let $\pi_2 : U \times F \rightarrow F$ be the continuous projection onto the second factor. So, $\{\pi_2 \circ \phi_U(x_{\lambda})\}$ is an infinite net in $F$, hence possesses yet another convergent subnet $\Lambda_2$. Then $\{\phi_U(x_{\lambda})\}_{\Lambda_2}$ is convergent in $U \times F$ by definition of the product topology, and the image of this convergent net under the homeomorphism $\phi_U^{-1}$ is convergent. Thus $\{x_{\lambda}\}_{\Lambda_2}$ is a convergent subnet.

Answer 2

I came across (while I was actually searching for something else) another nice proof which makes use of several facts which are also useful on their own. I'm going to post it here:

Lemma 1: If $F$ is compact, then the projection $p:F\times Y\to Y$ is closed.
Proof: Let $C$ be closed and $y\notin p(C)$. Then $\{y\}\times F$ is covered my finitely many open product sets, each of which does not intersect $C$. The intersection of their projections is an open neighborhood of $y$ not intersecting $p(C)$. Thus $p(C)$ is closed.

Lemma 2: Let $\{A_i\}$ be a family of subsets of a topological space $Y$ such that $\{\mathrm{int}A_i\}$ covers $Y$. A subset $B$ is closed if $B\cap A_i$ is closed in $A_i$ for every $i$.
Proof: Let $B\subseteq Y$ such that $B\cap A_i$ is closed in $A_i$ for every $i$. Let $D=Y-B$. Since $D\cap A_i$ is open in $A_i$, thus $D\cap\mathrm{int}A_i$ is open in $Y$. Their union equals $D$, hence $B$ is closed.

Lemma 3: Let $f:X\to Y$ be a function, where $Y$ is covered as in lemma 2. Then $f$ is closed if the restrictions $f_i:f^{-1}(A_i)\to A_i$ are closed for all $i$.
Proof: Let $C$ be closed in $X$. Then $f(C)\cap A_i=f_i(C\cap f^{-1}(A_i))$ which is closed in $A_i$ by hypothesis. By lemma 2 $f(C)$ is closed.

Definition: A continuous surjection is called perfect if it is closed and the preimages of points are compact.

Definition: A continuous function is called proper if the preimages of compact sets are compact.

Lemma 4 (without proof): A perfect map is proper.

Proposition: Let $F\rightarrow E\xrightarrow{\pi}B $ be a fiber bundle with compact fiber $F$. Then $\pi$ is a proper map.
Proof: $B$ is covered by trivialized open sets $\{U_i\}$. Since $F$ is compact and $\pi_i:\pi^{-1}(U_i)\to U_i$ is equal to the projection $U_i\times F\to U_i$ up to homeomorphism, it is a closed map by Lemma 1. By lemma 3 the entire map $\pi$ is closed. Since fibers are compact, $\pi$ is also a perfect mapping, hence it is proper by Lemma 4.

Answer 3

I decided to answer this question because it seems that there is a plethora of proofs, none of which is the one I think of first!

Note: I assume that compact means "compact and Hausdorff", which is the way I was taught at Warsaw University. This implies normality.

Claim: There is a finite cover $U_{i}$ of $B$ such that $\pi$ is trivial over some open neighbourhood $V_{i}$ of the closure of $U_{i}$, ie. $\bar{U_{i}} \subseteq V_{i}$.

Proof of the claim: By compactness it's enough to find a neigbourhood with that property of every point in $B$, so let $b \in B$. Let $V_{b}$ be an open neighbourhood of $b$. By normality, there are open sets $\{ b \} \subseteq U_{b}, (B \setminus V_{b}) \subseteq W_{b}, U_{b} \cap W_{b} = \emptyset$, because $\{ b \}, (B \setminus V_{B})$ are closed and disjoint. Now $\bar{U_{b}} \subseteq V_{b}$, since $\bar{U_{b}} \subseteq (B \setminus W_{b}) \subseteq (B \setminus (B \setminus V_{b})) \subseteq V_{b}$, so $U_{b}$ is the required neighbourhood.

This is a basic property of (locally) compact spaces which makes precise the statement that in such spaces "neighbourhoods of points can be made arbitrarily small". The theorem you're after follows easily.

Let $U_{i}$ be a cover of $B$ as above. Since $\bar{U_{i}} \subseteq V_{i}$, the subsets $\pi^{-1}(\bar{U_{i}}) \simeq \bar{U_{i}} \times F$ are compact and their union is $E$ by construction. So $E$ is the union of a finite number of compact subsets and thus is compact itself.