5

I am trying to show that when you're given a continuous map $f:M\rightarrow N$ between compact orientable $n$-dimensional manifolds and $f^*:H^n(N)\rightarrow H^n(M)$ is an isomorphism, then $f^*:H^k(N)\rightarrow H^k(M)$ is injective for every $k$.

I'm unsure how to proceed in this. Hatcher doesn't have much about induced maps on cohomologies. Obviously this is true when $k=n$ and $k=0$, but for all the other cases I'm stumped. Any help is appreciated!

Qiaochu Yuan
  • 419,620
TinaBelcher
  • 701
  • 3
  • 13

2 Answers2

6

Here's a proof assuming $M, N$ are closed, connected, and oriented $n$-manifolds. Below the coefficients for cohomology can be any commutative ring $R$.

Proposition: Let $f : M \to N$ be a map of degree $k$. Let $f^{\ast} : H^{\bullet}(N) \to H^{\bullet}(M)$ be the usual pullback map on cohomology and let $f_{!} : H^{\bullet}(M) \to H^{\bullet}(N)$ be the pushforward map on cohomology induced by Poincaré duality. Then

$$f_{!} f^{\ast} : H^{\bullet}(N) \to H^{\bullet}(N)$$

is multiplication by $k$.

Corollary: If $k$ is invertible in $R$ (equivalently, if $f$ induces an isomorphism on top (co)homology over $R$), then $f^{\ast}$ has a left inverse and hence is injective.

Proof. Let $[M] \in H_n(M), [N] \in H_n(N)$ denote the fundamental classes of $M$ and $N$. By definition, the pushforward $f_{!}$ satisfies

$$[N] \cap f_!(\alpha) = f_{\ast}([M] \cap \alpha)$$

where $\alpha \in H^{\bullet}(M)$, $f_{\ast} : H_{\bullet}(M) \to H_{\bullet}(N)$ denotes the usual pushforward map on homology, and $\cap$ denotes the cap product. It follows that

$$[N] \cap f_! f^{\ast}(\beta) = f_{\ast}([M] \cap f^{\ast}(\beta))$$

where $\beta \in H^{\bullet}(N)$. Now, by the naturality of the cap product,

$$f_{\ast}([M] \cap f^{\ast}(\beta)) = f_{\ast}([M]) \cap \beta$$

and since by hypothesis $f_{\ast}([M]) = k [N]$, we have

$$f_{\ast}([M] \cap f^{\ast}(\beta)) = k [N] \cap \beta$$

from which it follows that $f_! f^{\ast}(\beta) = k \beta$ as desired. $\Box$

Community
  • 1
Qiaochu Yuan
  • 419,620
  • +1 Nice! Now I only hope that the asker of the question knows about Poincare duality. – Daniel Valenzuela Mar 30 '15 at 05:59
  • 1
    Thank you! Yes, I am familiar with Poincare duality, though I will have to read a bit more through Hatcher to have a complete understanding of it. A big thank you to both of you! – TinaBelcher Mar 30 '15 at 06:04
  • @Tina: for what it's worth, there's an intuitive picture of what's going on here which I've left completely unsaid. Roughly you should imagine that a map of degree $k$ looks like a $k$-fold covering map, and that the pullback and pushforward maps on cohomology look like taking the preimage and the image respectively of oriented submanifolds (possibly after perturbing them suitably). When you pull back a submanifold along a $k$-fold covering map it gets $k$ times bigger, and then when you push it forward again you've got $k$ copies of what you had before. – Qiaochu Yuan Mar 30 '15 at 06:08
  • This picture is easiest to make precise for the nicest maps of degree $k$ between tori (namely the ones given by integer matrices of determinant $k$), where you can be really explicit about what pullback and pushforward of oriented submanifolds representing classes in cohomology (all of which can themselves be taken to be tori) look like. For $2$-dimensional tori you can draw some nice pictures involving $k$ copies of a parallelogram sitting inside another one. – Qiaochu Yuan Mar 30 '15 at 06:10
  • Great, thank you! It's very intuitive to think of the map that way. I think there might be a typo in your solution - when you talk about the usual pushforward map on homology, it should be $f_*:H_{\bullet}(M)\rightarrow H_{\bullet}(N)$ should it not? – TinaBelcher Mar 30 '15 at 06:17
  • @Tina: oops. Yes, that's right. – Qiaochu Yuan Mar 30 '15 at 06:18
2

$f:M \to N$ induces $H^nN \to H^nM$ isomorphism. Here we go:

Let $0 \neq\gamma \in H^kN$. By universal coefficient theorem it comes either from dualizing $H_k$ or from the torsion of $H_{k-1}$, say the former. Let $\tilde \gamma$ be the Kronecker dual of the Poincaré dual. Then we have $\langle\gamma \smile \tilde \gamma,[M]\rangle = <\tilde \gamma,\gamma \cap [M]> =1$. Now we use naturality of cupping, i.e. $0 \neq f^*(\gamma \smile \tilde \gamma) = f^* \gamma \smile f^* \tilde \gamma$, hence $\gamma \not \mapsto 0$ and we have injectivity for such gammas.

In the other case that $\gamma$ is torsion you have to be more careful, since such elements don't cup non-trivially into the top cohomology... with $\mathbb Z$ coefficients! So assume $\gamma$ has order $p$. Then it appears in $H^{k-1}(N;\mathbb Z_p)$ as Kronecker dual (i.e. arising from $H_{k-1}$) and you may proceed with caution as above.

Daniel Valenzuela
  • 6,305
  • 12
  • 20
  • 1
    Yeah, I noticed this nice argument for the torsion-free part but I was hoping for a more uniform argument for everything at once. Do you know if the following is true: if $f$ has degree $d$, then the composite $f_{\ast} f^{\ast}$ is multiplication by $d$, where $f_{\ast}$ is the pushforward $H^{\bullet}(M) \to H^{\bullet}(M)$? – Qiaochu Yuan Mar 30 '15 at 05:16
  • 1
    You mean $H^M \to H^N$? Since it is the composition $H^nN \to H^nM \to H_0M \to H_0N \to H^nN$ that should be certainly true (because all maps beside the degree map are iso's, which is due to the fact that we consider $H_0). – Daniel Valenzuela Mar 30 '15 at 05:24
  • It could have order $p^{8323}$, though. – Mariano Suárez-Álvarez Mar 30 '15 at 05:52
  • I don't think I said $p$ to be prime. – Daniel Valenzuela Mar 30 '15 at 05:55