$f_*: H_*(S^n, \mathbb{Z}_p) \to H_*(M, \mathbb{Z}_p)$ an isomorphism if $p$ is a prime that does not divide $q$.

Question

Let $M$ be a compact connected $n$-manifold (without boundary), where $n \ge 2$. Suppose $M$ is oriented with fundamental class $z$. Let $f: S^n \to M$ be a map such that $f_*(i_n) = qz$ where $i_n \in H_n(S^n, \mathbb{Z})$ is the fundamental class and $q \neq 0$. Is $f_*: H_*(S^n, \mathbb{Z}_p) \to H_*(M, \mathbb{Z}_p)$ an isomorphism if $p$ is a prime that does not divide $q$?

Answer 1

Yes, this is true. Let us first consider $H_n$. The natural map $H_n(X,\mathbb{Z})\otimes \mathbb{Z}_p\to H_n(X,\mathbb{Z}_p)$ is an isomorphism for $X=S^n$ and $X=M$. We thus have the following commutative diagram, where the vertical maps are isomorphisms: $$\require{AMScd} \begin{CD} H_n(S^n,\mathbb{Z})\otimes\mathbb{Z}_p @>{f_*\otimes 1}>> H_n(M,\mathbb{Z})\otimes \mathbb{Z}_p\\ @V{}VV @V{}VV \\ H_n(S^n,\mathbb{Z}_p) @>{f_*}>> H_n(M,\mathbb{Z}_p) \end{CD}$$

By assumption, the top map is multiplication by $q$, when we identify $H_n(S^n,\mathbb{Z})$ with $\mathbb{Z}$ using $i_n$ and $H_n(M,\mathbb{Z})$ with $\mathbb{Z}$ using $z$. Since $q$ is a unit in $\mathbb{Z}_p$, this means the top map is an isomorphism. Thus the bottom map is also an isomorphism.

Now for $0<*<n$, we can show $f_*:H_*(S^n,\mathbb{Z})\to H_*(M,\mathbb{Z}_p)$ is an isomorphism by showing that $H_*(M,\mathbb{Z}_p)$ is trivial. To show this, it suffices to show $H^*(M,\mathbb{Z}_p)$ is trivial. Suppose $a\in H^*(M,\mathbb{Z}_p)$ is nonzero. Then by Poincaré duality, there exists $b\in H^{n-*}(M,\mathbb{Z}_p)$ such that $ab\neq 0$. But $f^*(a)=0$ since $H^*(S^n,\mathbb{Z}_p)$ is trivial, so $f^*(ab)=0$. This is a contradiction, since $f$ induces an isomorphism on $H_n$ and hence also on $H^n$.