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I'm doing past papers for a first course in algebraic topology.

The question is:

Let $M$ be a 3-dimensional, closed, connected, non-orientable manifold. Show that $M$ has infinite fundamental group.

Is there any way of answering this question without simply quoting a classification theorem for 3-manifolds with finite fundamental group?

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What definition are you using for orientability? – dfeuer Jun 15 '13 at 18:34
@dfeuer Good question. I guess I can assume that top homology vanishes. – Earthliŋ Jun 15 '13 at 23:07
@Earthliŋ, do you know a reference for such classification theorem? – Sigur Jun 8 '15 at 19:42
up vote 13 down vote accepted

$\def\QQ{\mathbb Q}$If $\pi_1(M)$ is finite, $H_1(M;\QQ)=0$. If $M$ is non-orientable, $H_3(M;\QQ)=0$. So $\chi(M)=h_0(M;\QQ)-h_1(M;\QQ)+h_2(M;\QQ)-h_3(M;\QQ)=1+h_2(M;\QQ)>0$.

But by Poincaré duality, any odd-dimensional manifold has zero Euler characteristic.

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A little remark: the Euler characteristic of any closed odd-dimensional manifold is 0, but in this case it doesn't follow by Poincaré duality, since $M$ is supposed to be non-orientable. – Dario Jun 23 '14 at 7:09
@Dario just use $\mathbb Z_2$ coefficients. – Grumpy Parsnip Nov 7 '14 at 19:21
@GrumpyParsnip Sure! I didn't think to that. Thanks! – Dario Nov 10 '14 at 22:20
@Dario Or use the two-fold orientable cover $\tilde M\to M$. $\chi(\tilde M)=0$, and $\chi(\tilde M)=2\chi(M)$, hence $\chi(M)=0$. – mathreader Apr 21 '15 at 23:15

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