closed, connected, orientable manifold with nonzero induced homology maps $S^n \rightarrow M$. Compute $H_k(M;\mathbb{Q})$

Question

Let $M$ be a closed, connected, orientable $n$-dimensional manifold, and suppose there is a map $f:S^n\rightarrow M$ such that the induced homomorphism $f_*:H_n(S^n;\mathbb{Z})\rightarrow H_n(M;\mathbb{Z})$ is non-trivial. Compute $H_k(M;\mathbb{Q})$ for all $k$. Hint: Use the naturality of the cap product.

I know $\smallfrown: H^k(M) \times H_n(M) \rightarrow H_{n-k}(M)$, but I don't understand how to get naturality involved.

I think of the Universal Coefficent Theorem for changing the coefficients of the homology groups of $M$ to $\mathbb{Q}$, but I don't know the integral homology groups, other than $H_0(M;\mathbb{Z})\cong H_n(M;\mathbb{Z})\cong \mathbb{Z}$.

I'd appreciate any help!

Answer 1

Correspondingly the map on $H^n(-;\Bbb Q)$ is nonzero. Now the point is that the cup product pairing is nondegenerate. If $\alpha$ is a nonzero class in $H^k(M;\Bbb Q)$, there is a class $\beta$ of complementary degree so that $\alpha \smile \beta = [M]$ (this uses that $\Bbb Q$ is a field). But then $f^* \alpha \smile f^*\beta = k[S^n]$ for some nonzero $k$. Hence $f^*\alpha \neq 0$, which is preposterous because you know the cohomology of a sphere. Therefore $H^k(M;\Bbb Q) = 0$ for all $0<k<n$.