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If $X$ is a connected $n$-dimensional manifold and $Y \subseteq X$ a proper subspace. Prove that $H_n(Y;\mathbb{Z}) \to H_n(X;\mathbb{Z})$ is the zero map.

My idea was to look at the long exact sequence in homology and use exactness properties to deduce that $H_n(Y;\mathbb{Z}) \to H_n(X;\mathbb{Z})$ is in fact the zero map. I'm pretty sure that $H_{n+1}(X, Y;\mathbb{Z}) =0 $ which would imply that $H_n(Y;\mathbb{Z}) \to H_n(X;\mathbb{Z})$ has trivial kernel.

If I can show that $H_n(X;\mathbb{Z})\to H_n(X, Y ;\mathbb{Z})$ has trivial kernel then I would be done by exactness, but I don't see any reason why this should hold which leads me to believe that there might be some other approach to attack this problem that I'm not seeing.

Perturbative
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1 Answers1

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The case $n = 0$ is trivial. So let us assume $n \ge 1$.

It suffices to consider the case $Y(x) = X \setminus \{x\}$ with $x \in X$ because each inclusion $Y \hookrightarrow X$ factors through some $Y(x)$. The space $Y(x)$ is a non-compact manifold. For $n \ge 2$ it is connected, thus $H_n(Y(x)) = 0$ and the result follows. See Proposition 3.29 in Hatcher. For $n = 1$ we need a special argument.

Case 1. $Y(x)$ is connected. Then again $H_1(Y(x)) = 0$.

Case 2. $Y(x)$ is not connected. This is possible only when $X \approx \mathbb R$ (see The only 1-manifolds are $\mathbb R$ and $S^1$). Then $H_1(X) = 0$.

Paul Frost
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    Very nice - I was worried about, e.g., taking $Y\subseteq S^3$ to be the sphere version of the Hawaiian earring, so that $H_3(Y)$ is large. I like how using $Y(x)$ just skirts all these issues. – Jason DeVito - on hiatus Aug 31 '21 at 13:54