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Anybody could help me with this exercise, please? If $M$ is a compact, connected, orientable and smooth $n$-manifold:

1) Show that there is a one-to-one correspondence between orientations of $M$ and orientations of the vector space of its de Rham cohomology, under which the cohomology class of a smooth orientation form is an oriented basis for $H_{dR}(M)$.

2) Now suppose $M$ and $N$ are smooth $n$-manifolds with given orientations. Show that a diffeomorphism $F\colon M \rightarrow N$ is orientation preserving if and only if the pullback between their rham cohomologies is orientation preserving.

Gibbs
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MGF01
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  • What do you know / have you tried already? Helping is hard without knowing where to start. – T'x Dec 10 '17 at 18:01
  • Actually, I am lost, I don't know how to start. – MGF01 Dec 10 '17 at 18:04
  • Do you know the fact that an orientation on $M$ is equivalent to a nowhere vanishing $n$-form on $M$? – T'x Dec 10 '17 at 18:07
  • Yes, I do, but how can I relate this fact with the cohomology? – MGF01 Dec 10 '17 at 18:10
  • $H^n(M)$ consists of equivalence classes of $n$-forms. – T'x Dec 10 '17 at 18:16
  • We know that an orientation on M is given by a nonvanishing n-form, and the de Rham cohomology is given by the quotient space between closed forms and exacts ones, so we would like to construct a correspondence between nonvanishing n-forms and equivalence classes in that quotient space. But, if w=dn i.e. w is exact, how can I relate this with a nonvanishing form? – MGF01 Dec 10 '17 at 18:28
  • Remember that you have exactly two orientations on $M$. Moreover your first assignemt somehow suggests that you already know that in your case $H^n(M)$ is one dimensional. – T'x Dec 10 '17 at 18:42
  • Maybe what I am going to say is stupid but: if M is orientable, I take the forms whose jacobian is positive, i.e. I choose one orientation and then send those forms to exact forms, since if the cohomology is 1D, all closed forms are exact. – MGF01 Dec 10 '17 at 18:57
  • I think this does not really work out. If all closed forms would be exact the $n$-th cohomology would be 0. – T'x Dec 11 '17 at 09:26
  • Another hint: think about how different orientations on $M$ are related (considered as forms). Then think about how two different orientations on the vectorspace $H^n$ are related. – T'x Dec 11 '17 at 09:29

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