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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 $q \neq 0$. How do I see that multiplication by $q$ annihilates $H_i(M, \mathbb{Z})$ if $1 \le i \le n - 1$?

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  • $\begingroup$ This question and this question are closely related; in particular, my answer is adapted from the argument at the end of the accepted answer to the latter question. Note that the first argument using only Poincare duality over fields is not quite strong enough, since that only give that $H_i(M,\mathbb{Z})$ is annihilated by some power of $q$ in this case. $\endgroup$
    – Eric Wofsey
    Nov 25, 2015 at 7:47

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[This argument is stolen from the end of this answer (which handles the case $q=1$).]

Let $1\leq i\leq n-1$ and $\alpha\in H_i(M,\mathbb{Z})$. By Poincare duality, there exists $\beta\in H^{n-i}(M,\mathbb{Z})$ such that $\alpha=z\cap\beta$. Since $H^{n-i}(S^n,\mathbb{Z})=0$, $f^*(\beta)=0$. Thus $0=f_*(i_n\cap f^*(\beta))=f_*(i_n)\cap\beta=qz\cap\beta=q\alpha$.

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  • $\begingroup$ "$0 = f_*(i_n \cap f^*(\beta))$"... what does this mean? $\endgroup$
    – user235289
    Nov 26, 2015 at 23:08
  • $\begingroup$ $f^*(\beta)$ is a cohomology class on $S^n$ and $i_n$ is a homology class on $S^n$, so we can take their cap product to get another homology class on $S^n$. Since $f^*(\beta)=0$, the cap product is $0$. The following equality uses the fact that cap products are natural in the sense that that $f_*(x\cap f^*(y))=f_*(x)\cap y$. $\endgroup$
    – Eric Wofsey
    Nov 26, 2015 at 23:11

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