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!


2 Answers 2


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @Kamil: (Smooth) covering maps are always local diffeomorphisms, hence, in particular, submersions. $\endgroup$ Jan 3, 2019 at 19:31

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.