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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

1 Answer 1

up vote 10 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 at 7:09
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@Dario just use $\mathbb Z_2$ coefficients. –  Grumpy Parsnip Nov 7 at 19:21
    
@GrumpyParsnip Sure! I didn't think to that. Thanks! –  Dario Nov 10 at 22:20

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