# Whether the induced map in de Rham cohomology is injective

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 $$[f^{*} \omega] = $$. 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 $$[\omega ] = $$ 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!

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.
• 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? Jan 3, 2019 at 10:15
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).