$f:N\to M$ with degree $\neq 0\Rightarrow f^*:H^k(M)\to H^k(N)$ injective for all $k$

Let $M^n,N^n$ be compact, orientable, smooth manifolds. If $f:N\to M$ is a nonzero degree smooth map, prove that $f^*:H^k(M)\to H^k(N)$ is injective for all $k=0,...,n$.

For $k=n$ and $\omega\in\Omega^n(M)$ with $f^*[\omega]=[0]$, we have: $$0=\int_Nf^*[\omega]=\deg(f)\int_M[\omega]$$

And since $\deg(f)\neq 0$, we get $\int_M[\omega]=0\Rightarrow [\omega]=[0]$.

My initial idea for $k<n-1$ was to take embbeded submanifolds $S^k\subset M^n, T^k\subset N^n$ of dimension $k$ so that if $\omega\in\Omega^k(M)$ is such that $f^*[\omega]=[0]$, then $$0=\int_Tf^*[\omega]=\deg(f)\int_S[\omega]$$

then I could use the same argument above.

But this implicitly assumes that $f|_S$ maps $S$ to $T$, which is not necessarily true, so it doesn't really work, now I'm stuck.

Any ideas?

• Do you know of Poincare duality? – Jason DeVito Nov 25 '17 at 19:50
• It says $H^k(M)\simeq H_{n-k}(M)$, right? But I can't see how it can help me – rmdmc89 Nov 25 '17 at 20:24
• How is that isomorphism defined? – Mariano Suárez-Álvarez Nov 25 '17 at 20:34
• I'd say that it is impossible to prove what you want without understanding the map that gives duality. I suggest you read up on that if it seems too complicated and then come back to this problem. – Mariano Suárez-Álvarez Nov 25 '17 at 22:07
• The part of Poincare duality I was referring to is the perfect pairing $H^{n-k}\times H^k \rightarrow H^n$... – Jason DeVito Nov 25 '17 at 23:01

I had a hard time finding good material about the Poncaré Duality, but now I get it. Since $M^n$ is compact and orientable, every form has compact support, so the map bellow is an isomorphism:
In particular, $\ker(\mathcal{PD})=0$, so if $\int_M[\omega\wedge\eta]=0$ for all $[\eta]$, then $[\omega]=[0]$.
Back to the problem, take $[\omega]\in H^k(M)$ with $f^*[\omega]=[0]$. For all $[\eta]\in H^{n-k}(M)$, we get:
$$\deg(f)\int_M[\omega\wedge\eta]=\int_M f^*[\omega]\wedge f^*[\eta]=\int_M [0]\wedge f^*[\eta]=0$$
Since $\deg(f)\neq 0$, we must have $\int_M[\omega\wedge\eta]=0$ for all $[\eta]$, so $[\omega]=[0]$ and $f^*$ is injective.