When the Induced Homomorphism on the $n$-th Cohomology is an Isomorphism 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!
 A: 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. $\Box$
A: $f:M \to N$ 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.
