2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Ali Feb 22 '16 at 12:02
2
$\begingroup$

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.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ "Now, we also know that for non-compact surfaces, that $H_2$ is trivial." You need to assume connectivity for this, which is fine. $\endgroup$ – Cheerful Parsnip Feb 22 '16 at 12:03
  • $\begingroup$ 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. $\endgroup$ – Ali Feb 22 '16 at 12:19
  • 1
    $\begingroup$ @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... $\endgroup$ – Jason DeVito Feb 22 '16 at 14:57
  • $\begingroup$ @JasonDeVito: good point. $\endgroup$ – Cheerful Parsnip Feb 22 '16 at 23:10
  • $\begingroup$ $H_n=0$ for non-compact $n$-manifolds is proven in Hatcher (before even fully developing the P.D. machinery). $\endgroup$ – PVAL-inactive Feb 22 '16 at 23:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.