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This is an example/exercise from section 13 in Differential Forms in Algebraic Topology by Bott & Tu. I've been posting a few questions to get through this part of the book. It was pretty tough getting through the previous 12 sections, but this section seems particularly difficult for me.

Let $X = S^1 \vee S^2$ and let $\tilde{X}$ be the universal covering of X. (Note that $H^*(X)$ is finite whereas $H^*(\tilde{X})$ is not). Define a fiber bundle over $S^1$ with fiber $\tilde{X}$ by setting $$E=\tilde{X}\times I / (x,0)\sim(s(x),1)$$ where $s$ is the deck transformation that shifts everything up one unit. The projection $\pi: E \rightarrow S^1$ is given by $\pi(\tilde{x},t)=t $. Note that the fundamental group of the base $\pi_1(S^1)$ acts on $H_2(\text{fiber})$ by shifting each sphere one up.

I am then asked to find the homotopy type of the space $E$.

This chapter deals how the fundamental group of a manifold with a good cover $\mathfrak{U}$ is isomorphic to that of $N(\mathfrak{U})$, the nerve of the cover. I'm guessing this should be relevant? (I assume that there is a direct way of computing the fundamental group via Van Kampen etc. which I am less interested in...)

Also, the previous two examples finds the cohomology of the total space using Mayer - Vietoris principle and double complex. Here's a link : A simple example calculating Čech comohomology Would finding $H^*(E)$ be useful in finding $\pi_{1}(E)$? I can't seem to relate these two.

Thanks, always!

shaine
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    I think $E$ is homotopy equivalent to $X$. – Tyrone Jul 07 '20 at 17:30
  • Some explanations would be helpful! – shaine Jul 08 '20 at 05:39
  • The projection onto the first factor induces a map $p:E\rightarrow X$. Uncoil the circle and run it backwards from $1$ to $0$ in $E$ to get a section $s$. I'll let you write down the homotopy $sp\simeq id_E$. You can compute the homotopy groups of $E$ using the homotopy sequence eq. (17.4). It's not difficult to see dircetly from this that $p$ is a weak equivalence, and so in this case a homotopy equivalence. I don't know how this fits in with the chapter so I left it as a hint. – Tyrone Jul 08 '20 at 14:53

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Since you have the fibration $\tilde{X}\hookrightarrow E \to S^1$, you can use the fibration long exact sequence. If you have not encountered it yet, you may try to prove it using the homotopy lifting property of fibrations.

For $k\ge 2$ one gets, $$\underbrace{\pi_{k+1}S^1}_{0} \to \pi_{k} \tilde X \to \pi_k E \to \underbrace{\pi_k S^1}_{0}$$ and hence, the inclusion map $\tilde{X}\to E$ induces isomorphism $\pi_k\tilde X \cong \pi_k E$.

And for $k=1$ we have, $$\underbrace{\pi_{2}S^1}_{0} \to \pi_{1} \tilde X \to \pi_1 E \to \underbrace{\pi_1 S^1}_{\mathbb{Z}}\to \pi_0\tilde X$$ Since $\tilde X$ is an universal cover, it is simply connected and so $\pi_1\tilde X = 0 = \pi_0 \tilde X$. Then we get, $\pi_1 E \cong \mathbb{Z}$.

Thus we have, $$\pi_k E= \cases{ \pi_k \tilde X, \text{if $k\ge 2$}\\ \mathbb{Z}, \text{if $k=1$}}$$

Note: The above argument works for any mapping torus. To compute the fundamental group directly, you can use the vanKampen's theorem, as explained here

Hope this helps. Cheers!

ChesterX
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