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Let $M, N$ be smooth manifolds, and let $f: M \rightarrow N$ be a surjective submersion, i.e. a surjective smooth map such that the differential $f_{*}$ is also surjective.

I have shown that for all $k \geq 0$, the pullback map of $k$-forms $$ f^{*} : \Omega^{k}(N) \rightarrow \Omega^{k} (M)$$ is injective.

However, the problem now asks me whether the induced map on de Rham cohomology $$ H_{dR}^k (N) \rightarrow H_{dR}^k (M) : [\omega] \mapsto [f^{*} \omega] $$ is also injective?

I was trying to prove this. I took $[\omega] \in H_{dR}^{k} (N)$ and assumed $[f^{*} \omega] = [0]$. This means $f^{*} \omega \sim 0$ or $$ f^{*} \omega = d \tau $$ for some $(k-1)$ form $\tau$ on $M$.

I want to conclude from this somehow that $[\omega ] = [0]$ or $\omega = d \sigma$ for some $(k-1)$ form $\sigma$ on $N$. But I'm not sure if the statement is even true. I tried to find a counter example, but couldn't.

Any help is appreciated!

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2 Answers 2

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Consider the convering of the torus $T^2$ by $\mathbb{R}^2$, $H^2_{DR}(T^2)\neq 0$ and $H_{DR}^2(\mathbb{R}^2)=0$ since it is contractible so the induced map on cohomology is not injective.

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  • $\begingroup$ Thanks. Do you take $N = T^{2}$ in this case? How do you see the map $f: \mathbb{R}^2 \rightarrow T^2$ is a submersion? $\endgroup$
    – Kamil
    Jan 3, 2019 at 10:15
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    $\begingroup$ @Kamil: (Smooth) covering maps are always local diffeomorphisms, hence, in particular, submersions. $\endgroup$ Jan 3, 2019 at 19:31
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This is actually true as long as the fibers of the submersion are $(k-1)$-connected, ie. the homotopy groups vanish up to and including degree $(k-1)\,.$ It's a fact that follows from theorem 1.9.4 in Bernstein's equivariant sheaves and functors, though there is another way of seeing it which uses the submersion groupoid $M\times_f M\rightrightarrows N$ (I don't mind explaining).

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