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I'm looking through some old algebraic topology problems to study for an exam, and I came across the following: Let $M$ be a compact, orientable $n$-manifold, and let $f:S^n\to M$ be a map of degree $d(f)$. For $0<i<n$, show that for every $u\in H_i(M;\mathbb Z)$, $d(f)u=0$.

I know that the degree of a map $f$ is the number $d\in \mathbb Z$ such that $f_*:H_n(S^n)\to H_n(M)$ takes $[S^n]$ to $d\cdot [M]$, where $[X]$ represents to fundamental class of $X$.

However, I don't understand how to find the effect of the degree of the map on the lower homology groups. In fact, it seems to me that this result cannot be true: maybe my reasoning is flawed, but if $M$ is a compact, orientable manifold, then $H_{n-1}(M;\mathbb Z)$ is torsion-free, so for $u\in H_{n-1}(M;\mathbb Z)$, if it is true that $d(f)u=0$ then $u$ would be torsion, a contradiction.

Any help would be appreciated.

user161142
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1 Answers1

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Any $u\in H_i(M)$ is of the form $u=|M|\frown\alpha$ for a unique $\alpha\in H^{n-i}(M)$ (Poincaré duality, $\frown$ is the cap product). We thus have $d(f)u=(f_*|S^n|)\frown\alpha=f_*(|S^n|\frown f^*\alpha)=0$, as $f^*\alpha=0\in H^{n-i}(S^n)=0$.

user8268
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  • That makes perfect sense -- thanks! Do you happen to have any insight on why my reasoning was incorrect? That is, why does this not violate the property that $H_{n-1}(M;\mathbb Z)$ is torsion-free, if $u\in H_{n-1}(M)$? – user161142 Jul 02 '14 at 19:50
  • @user161142 Your reasoning and the result imply that if $f$ with $d(f)\neq0$ exists then $H_{n-1}(M)=0$. Maybe it's not so bad :) – user8268 Jul 02 '14 at 20:57