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I'm trying Hatcher (Algebraic Topology) exercise 4.2.16.

Show that closed surfaces with infinite fundamental group are $K(\pi,1)$'s by showing that their universal covers are contractible, via the Hurewicz theorem and results of section 3.3.

So far, my solution goes as follows:

Let $X$ be the closed surface in question. I want to show that $\pi_n(X)=0 \text{ } \forall n \neq 1$.

Let us consider the universal cover of $X$, $\tilde{X}$, so $\tilde{X}$ is simply connected. Therefore, by Hurewicz, we have $\tilde{H}_n(\tilde{X})=0 \text{ } \forall n \leqslant 1$, and $H_2(\tilde{X})=\pi_2(\tilde{X})$.

Since $X$ is a surface, $H_n(X)=0 \text{ for all } n >2$. We also have $H_n(\tilde{X})=0 \text{ } \forall n >2$.

$\color{blue}{\text{If } \pi_2(X)=\pi_2(\tilde{X})=0 \text{, then we have } H_n(\tilde{X})=0 \text{ for all } n}$.

We have a map $f \colon \tilde{X} \to *$, inducing isomorphism $f_* \colon H_n(\tilde{X}) \to H_n(*)$, so by Whitehead's theorem (second formulation), we have $f$ is a homotopy equivalence. Hence $\tilde{X}$ is contractible, so $\pi_n(\tilde{X})=0 \text{ for all } n$, and since $\pi_n(\tilde{X})\cong \pi_n(X) \text{ for all } n\neq 1$, then we have $\pi_n(X)=0 \text{ } \forall n \neq 1$.

As you can see, I have a hole in the middle of my argument, most important of which is, does $\pi_2(X)=0$? I feel like this must depend on $X$ being closed and $\pi_1(X)$ infinite, otherwise the above would hold for any surface. I thought about using Poincare duality, but this seems to give me $H_2(\tilde{X}) \cong H^0(\tilde{X})\cong \mathbb{Z} \neq 0$ since it seems Poincare duality doesn't use reduced homology.

Help would be appreciated, thanks.

Ali
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  • Note: I just edited the question after realising a covering space is locally homeomorphic to the space, so if the space is n-dimensional, the covering space is also at most n-dimensional. – Ali Feb 22 '16 at 12:02

1 Answers1

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For your first question, yes, $H_n(\tilde{X}) = 0$ for $n > 2$ because $\tilde{x}$ is also a surface. (This is true whether or not $X$ is closed or $\pi_1(X)$ is infinite.)

The question for $n=2$ is more delicate, and relies on the fact that $\pi_1(X)$ is infinite. Namely, the fact that $\pi_1(X)$ is infinite implies $\tilde{X}$ is non-compact. Idea of proof: Pick a point $p\in X$. Since $\pi_1(X,p)$ is infinite, the covering has infinitely many sheets, so $\pi^{-1}(p)\subseteq \tilde{X}$ is infinite. Moreover, the covering property implies $\pi^{-}(p)$ is discrete, so $\tilde{X}$ admits an infinite discrete subset. This tells you $\tilde{X}$ is non-compact.

Now, we also know that for non-compact surfaces, that $H_2$ is trivial. This follows, for examples, from Poicare duality with compact supports. (Poincare duality, in its typical presentation, only applies to closed orientable manifolds, but there are many generalizations.)

Jason DeVito - on hiatus
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  • "Now, we also know that for non-compact surfaces, that $H_2$ is trivial." You need to assume connectivity for this, which is fine. – Cheerful Parsnip Feb 22 '16 at 12:03
  • Ah, thanks! Thank you for your proof of X being non-compact. Homology of a non-compact surface is a new thing for me but I've found sections of Hatcher called "Cohomology with Compact Supports" and "Duality for Noncompact Manifolds" so perhaps with added reading I will be able to flesh out your last paragraph. – Ali Feb 22 '16 at 12:19
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    @Grumpy: You're right. I always seem to forget that manifolds can be disconnected. But if you allow disconnected manifolds, the OP's theorem is false: For example, $S^2 \coprod T^2$ is closed with inifinite $\pi_1$, but the universal cover is $S^2\coprod \mathbb{R}^2$, which has $\pi_2$ (and many other $\pi_k$ nontrivial, if the base point is in the $S^2$ piece... – Jason DeVito - on hiatus Feb 22 '16 at 14:57
  • @JasonDeVito: good point. – Cheerful Parsnip Feb 22 '16 at 23:10
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    $H_n=0$ for non-compact $n$-manifolds is proven in Hatcher (before even fully developing the P.D. machinery). – PVAL-inactive Feb 22 '16 at 23:32