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Let $X$ be a space whose homology groups are finitely generated. In order to avoid trivial cases, suppose that $X$ is not a singleton. Must there exist a point $p \in X$ such that $X\setminus\{p\}$ does not have the same homology groups as $X$? This seems implausible, but I have been unable to think of a counterexample. It seems to hold for most of the "usual" spaces one considers, eg. spheres, tori, $\mathbb R^n$, and various products and wedge sums thereof. Any ideas?

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    Perhaps you want to include the condition that the isomorphism of homology groups is actually induced by inclusion of $X\setminus p$ into $X$.2012-02-09
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    Are you assuming $X$ is Hausdorff?2012-02-09
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    Yeah, lets impose the Hausdorff condition.2012-02-09
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    My algebraic topology is extremely rusty, but what about an infinite dimensional space such as $\ell^2$? Intuitively, it seems like finite dimensional objects such as simplices should not be able to detect a hole, because there are always plenty of directions to nudge them in to avoid it.2012-02-09
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    I think @NateEldredge is onto something. The infinite dimensional sphere $S^\infty=\cup_{i=1}^\infty S^n$ (including each finite-dimensional sphere in the next as a hemisphere) is contractible by Whitehead's theorem. So I think puncturing an infinite dimensional space will be homotopy-equivalent to $S^\infty$. So it will remain contractible.2012-02-09
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    hemisphere->equator in my last comment.2012-02-09

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A simple example, though perhaps not what you had in mind, is to let $X$ be any set with the coarse topology, and with at least two points. Removing a point from $X$ will also inherit the coarse topology, and the homology groups will be the same as that of a single point in both cases.

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    I think asking for a Hausdorff example is a fascinating problem.2012-02-09
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    Unless $X$ has one element or is empty!2012-02-09
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    @Qiaochu: Indeed.2012-02-09
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    Thanks, guys. Typos fixed.2012-02-09
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    This is a very nice example :D2018-05-10
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If $M$ is a manifold, then let $p\in M$ be the point you want to remove, and $A$ a Euclidean neighborhood of that point. Then we have by excision $$ H_\ast(M)\cong H_\ast(M,A)\cong H_\ast(M-\lbrace p\rbrace,A-\lbrace p\rbrace).$$

Since $A-\lbrace p\rbrace$ is homotopic to $S^{n-1}$, the LES on relative homology shows $H_{n}(M)\cong H_{n}(M-\lbrace p\rbrace)\oplus\mathbb{Z}$.

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    No ..it's not true in general ...you need orientability of M. E.g take RP^2 whose 2nd homology vanish so there is no Z ..2016-08-18