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This question looks like pretty easy and missing a crucial point to solve it but I am not sure. Here it is:

Let $M$ be an n dimensional compact manifold such that $H_i(M;\mathbb{Z}_2) = H_i(S^n;\mathbb{Z}_2)$ for all $i$. Then $M$ is orientable.

What I am wondering is since, from the given equation $H_i(M;\mathbb{Z}_2)$ is $\mathbb{Z}_2$ for $i=n$ and any manifold is already $\mathbb{Z}_2$ orientable, without M being orientable, how can we conclude that M is orientable? Do we also mean $\mathbb{Z}$-orientable?

quefaire
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    Recall that $H^1(M; \mathbb{Z}_2) = \operatorname{Hom}(\pi_1(M), \mathbb{Z}_2)$. Now, if $M$ was non-orientable, I claim there would be a non-trivial map $\pi_1(M) \to \mathbb{Z}_2$. Hint: consider the orientable double cover of $M$. – Michael Albanese May 19 '18 at 23:22
  • @MichaelAlbanese The hypothesis is about $H_$, how do you turn it into $H^$ here? – Arnaud Mortier May 19 '18 at 23:30
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    @ArnaudMortier: over a field (in this case $\mathbb Z_2$), homology and cohomology are isomorphic. – Cheerful Parsnip May 19 '18 at 23:38
  • @MichaelAlbanese How did you get $\pi_1(M)$ in $\operatorname{Hom}(\pi_1(M), \mathbb{Z}_2)$? Shouldn't it be $\operatorname{Hom}(H_1(M;\mathbb{Z}_2); \mathbb{Z}_2$? – quefaire May 20 '18 at 10:18
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    It should be more or less - from UCT you get that $H^1(M;\mathbb{Z}_2)\cong Hom(H_1(M;\mathbb{Z}),\mathbb{Z}_2)$. But $H_1(M;\mathbb{Z})$ is an abelianization of $\pi_1(M)$ - so every map from $\pi_1(M)$ to an abelian group factors through the abelianization. – Igor Sikora May 20 '18 at 13:15

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$M$ is orientable if and only if its first Stiefel Whitney class is not zero.

The first Stiefel-Whitney class is zero if and only if the bundle is orientable

Tsemo Aristide
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