TITLE: When the Induced Homomorphism on the $n$-th Cohomology is an Isomorphism
QUESTION [5 upvotes]: 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!

REPLY [6 votes]: 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. $\Box$