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Q: Let $M$ be a smooth compact manifold, and suppose there is a smooth map $F:M \rightarrow S^{1}$ whose derivative is non-zero at every point. Prove that the de Rham cohomology space $H^{1}(M)$ is non-zero.

So here is my trouble with this question. I have the $1$-form $d\theta$ on $S^{1}$ and I would like to say that $F^{*}(d\theta)$ is a non-exact form. If $M$ was $1$-dimensional then I could conclude the map has a non-zero degree, hence the $\int_{M}F^{*}(d\theta)=2\pi k$ is non zero. However the problem does not say that $M$ is $1$-dimensional. However, by the way it says "derivative" instead of "Jacobian" I'm inclined to think $M$ is $1$-dimensional and the problem is just sloppy, but then $M$ would have to be either the circle or an interval.

So, my question is, is there a way to solve this replacing "derivative" with "Jacobian" in the problem for an arbitrary dimensional manifold $M$? Are there any general tools other than degree for dealing with pullbacks when little is known about one or either of the manifolds in question?

Many thanks!

Edit: So above instead of "Jacobian" I should have said $F$ is a smooth submersion, or something along those lines.

TheManWhoNeverSleeps
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  • The fact that they commute, means that $dF^{}(d\theta)=F^{}(d \circ d \theta)=0$ so the form is closed. I need to try to show that it is not exact. What you wrote would imply that in fact it is exact, but I think the problem is that $\theta$ is not well defined since it's not single valued. I could be wrong though :p. Edit: Sorry I am new to this commenting system, did I accidentally delete someone's comment? There was one above this a few minutes ago... – TheManWhoNeverSleeps Dec 09 '14 at 00:10
  • Well, but $M$ is arbitrary and all we know is that $F$ is smooth with non-zero derivative. Even though you gave an example where it does hold for a specific manifold and a specific map, that is not enough, I think. – TheManWhoNeverSleeps Dec 09 '14 at 00:22
  • Ah, alright sorry. So you were just saying, there are cases where it can happen then? Any thoughts on whether it is true in general? – TheManWhoNeverSleeps Dec 09 '14 at 00:27
  • Here is the best I could come up with. A submersion $g:M \rightarrow N$ with $M,N$ compact and $N$ connected is surjective. (see http://math.stackexchange.com/questions/470164/a-submersion-f-mathcalx-to-mathcaly-must-be-surjective). When $g$ is surjective, the induced map on cohomology groups $g^*:H_1(N) \rightarrow H_1(M)$ is injective. Therefore, the generator of $H_1(N)$ is a nonzero element in $H_1(M)$. Sorry I can't fill in the details. – ndruiven Dec 09 '14 at 03:01
  • No worries, thanks. But how do we know that if $g$ is surjective then $g^{*}$ is injective? – TheManWhoNeverSleeps Dec 09 '14 at 03:08
  • That was the detail I couldn't figure out. – ndruiven Dec 09 '14 at 03:09
  • Ah I see, I will ponder that then, thanks for the help :) – TheManWhoNeverSleeps Dec 09 '14 at 03:18
  • because it is not true I guess. Just look at a non trivial covering $R^n \to M$. – Daniel Valenzuela Dec 09 '14 at 13:35
  • One way to show the statement, is to find a loop $S^1 \to M$ which is a homotopy section of $F$. – Daniel Valenzuela Dec 09 '14 at 13:52

1 Answers1

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$\newcommand{\dd}{\mathrm{d}}$Step 1. Note that $\omega := F^{\ast}(\dd \theta)$ is closed.

Step 2. Note that $\omega$ is nonvanishing.
Explicitly, we have $$\omega(x)(X) = \dd \theta(F(x))(D_x F(X)).$$ Since $DF$ is nonzero at every point, it is onto. As $\dd \theta$ is nonvanishing, we get the result.

Step 3. Conclude that $\omega$ cannot be exact.
Indeed, if $\omega = \dd f$ for some $f \in C^{\infty}(M)$, then consider a point of global extremum (here is where we use compactness). At that point, we have $\dd f = 0$. (We are assuming that $M$ is boundaryless.)

Aryaman Maithani
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