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Let $f:X\to X$ be a homeomorphism. Consider the mapping torus $T_f=X\times I/(x,0)\sim (f(x),1)$ for $x\in X$. Then $X\to T_f\to S^1$ is a fiber bundle. I can see this.

Every fiber bundle over $S^1$ is a mapping torus.

The above statement was said as a side note in the class. I tried to prove it but I'm quite new in this subject. I can't see any reference of the above statement on Google. Could you help?

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    $\begingroup$ You need to suppose that $f$ is a homeomorphism in order to get a fiber bundle. To prove the claim in question, I suggest using that fiber bundles over paracompact base spaces are (Hurewicz) fibrations. The intuition should be that the homeomorphism of the fiber is the monodromy obtained by traversing the base $S^1$ once. $\endgroup$
    – Thorgott
    Commented Dec 3, 2021 at 15:06
  • $\begingroup$ @Thorgott I've never heard monodromy before. Wikipedia says that it's a map between groups not topological spaces. Could you explain in more detail? $\endgroup$ Commented Dec 4, 2021 at 8:56
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    $\begingroup$ @love_sodam In this context you're just supposed to think about transporting $x \in X$ around $S^1$ through all of the fibers to arrive at some $x' \in X$ in the original fiber: the induced map $x \mapsto x'$ is the monodromy map. $\endgroup$ Commented Dec 4, 2021 at 9:01
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    $\begingroup$ @KeeleyHoek Now it makes sense and intuitive. To make that definition in rigor, maybe I need to consider section of the given fiber bundle (am I right?). How can I define the map so that it really a map $x\mapsto x'$? $\endgroup$ Commented Dec 4, 2021 at 9:30

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Despite what was said in comments, I don't think that homotopic properties such as the homotopy extension property are going to be enough for us here, because we need to use that we have a fiber bundle (and not merely a fibration) in an essential way. Instead we'll use the fact that fiber bundles over a contractible base are trivial, proved perhaps on this site and e.g. this is Corollary 1.8 on p21 of Hatcher's VBKT (note that this is stated for vector bundles, but Hatcher remarks that the same proof for fiber bundles is fine).

If you're comfortable with the definition of a mapping torus and the basic construction below then I think you'd be safe saying that your question is a "trivial corollary" of this result about fiber bundles, but we're going to check everything very carefully.


To begin, let's think of $S^1$ as a quotient of $I = [0, 1]$ with $p : I \to S^1$ the quotient map and let $\pi : E \to S^1$ be a fiber bundle. Let $s_0 := p(0) \in S^1$ be a basepoint. We can then build the pullback bundle $p^* \pi : p^* E \to I$, and remember the associated bundle map $\widetilde{p} : p^* E \to E$ for later. Since $p^* \pi$ is a fiber bundle over a contractible base it is trivial, so let $\phi(t, x) : p^* E \to I \times F$ be a bundle isomorphism with $F := \pi^{-1}(s_0)$ a standard fiber. Then we can build the composite $$ f : \pi^{-1}(s_0) \xrightarrow{(\widetilde{p}\rvert_{(p^*\pi)^{-1}(0)})^{-1}} (p^* \pi)^{-1}(0) \xrightarrow{\phi^{-1}} \{0\} \times F \to \{1\}\times F \xrightarrow{\phi} (p^* \pi)^{-1}(1) \xrightarrow{\widetilde{p}} \pi^{-1}(s_0). $$ Every map in this composite is a homeomorphism, so the entire composite is as well.

(The intuition is that this map: fixes an element $x$ of the fiber $F$ above $s_0 \in S^1$, chops the $S^1$ to get a fiber bundle over $I$, then "untwists" the bundle, pushes $x \in F$ from above $0 \in I$ in $I \times F$ to now sit above $1 \in I$---which essentially does nothing---then retwists the bundle and glues it back together. We return to a point in the fiber above $s_0$, but we need not have gotten $x$ back. Allowing $x \in F$ to vary in this procedure yields a homeomorphism $F \to F$.)

With the map $f : F \to F$ in hand, it remains to show that $\pi : E \to S^1$ is isomorphic to $M_f$. But this isn't so bad: by definition $M_f$ is a quotient of $I \times F$, so upon composing with the quotient map $q : I \times F \to M_f$ we have a bundle map $p^* E \xrightarrow{\phi} I \times F \xrightarrow{q} M_f$ which is a fiberwise isomorphism. This composite descends to a map $E \to M_f$ now over the same bases, hence an isomorphsim of fiber bundles, exactly when for all $y_1, y_2 \in p^* E$ such that $\widetilde{p}(y_1) = \widetilde{p}(y_2)$ we have $(q \circ \phi)(y_1) = (q \circ \phi)(y_2)$. (Don't forget the map $\widetilde{p} : p^* E \to E$ from the beginning, and if you like the condition we are checking is just the universal property of quotients.)

So, if we write $y_1 = (t_1, z_1) \in I \times E$ and $y_2 = (t_2, z_2)$ similarly then $\widetilde{p}(y_1) = \widetilde{p}(y_2)$ just says $z_1 = z_2$ and $p(t_1) = p(t_2)$. Obviously if $t_1 = t_2$ there is nothing to check, and the only way that $t_1 \not = t_2$ is if one is $0$ and the other is $1$. WLOG we can assume $t_1 = 0$ and $t_2 = 1$, in which case the elements $\phi(y_1) = \phi(0, z_1) = (0, x_1)$ and $\phi(y_2) = \phi(1, z_1) = (1, x_2)$ of $I \times F$ become equal in the quotient defining $M_f$ exactly when $f(x_1) = x_2$. But following the successive composites defining $f$ we trace out $$ x_1 \mapsto (0, x_1) \mapsto \phi^{-1}(0, x_1) = y_1 \mapsto y_2 \mapsto \phi(y_2) = (1, x_2) \mapsto x_2, $$ exactly as desired. This completes the proof.

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  • $\begingroup$ While monodromy is the right picture intuitively speaking (as you also explain), my initial comment was not fully thought out, so I agree with your first paragraph. On second thought, I arrived at a solution pretty much agreeing with yours, but by the time I wanted to post a clarifying comment, you'd already beaten me to it with an entire elaborate answer (+1). $\endgroup$
    – Thorgott
    Commented Dec 4, 2021 at 13:20

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