2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Do you know of Poincare duality? $\endgroup$ – Jason DeVito Nov 25 '17 at 19:50
  • 1
    $\begingroup$ It says $H^k(M)\simeq H_{n-k}(M)$, right? But I can't see how it can help me $\endgroup$ – rmdmc89 Nov 25 '17 at 20:24
  • 1
    $\begingroup$ How is that isomorphism defined? $\endgroup$ – Mariano Suárez-Álvarez Nov 25 '17 at 20:34
  • 2
    $\begingroup$ 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. $\endgroup$ – Mariano Suárez-Álvarez Nov 25 '17 at 22:07
  • 2
    $\begingroup$ The part of Poincare duality I was referring to is the perfect pairing $H^{n-k}\times H^k \rightarrow H^n$... $\endgroup$ – Jason DeVito Nov 25 '17 at 23:01
3
$\begingroup$

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:

\begin{align*} \mathcal{PD}:H^k(M)&\to (H^{n-k}(M))^*\\ [\omega]&\mapsto \left([\eta]\mapsto\int_M [w\wedge\eta]\right) \end{align*}

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.

$\endgroup$

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.