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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 $X$. – Grumpy Parsnip Feb 9 '12 at 1:04
Are you assuming $X$ is Hausdorff? – Grumpy Parsnip Feb 9 '12 at 1:05
Yeah, lets impose the Hausdorff condition. – user15464 Feb 9 '12 at 1:29
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. – Nate Eldredge Feb 9 '12 at 2:56
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. – Grumpy Parsnip Feb 9 '12 at 12:02
up vote 1 down vote accepted

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. – Grumpy Parsnip Feb 9 '12 at 1:08
Unless $X$ has one element or is empty! – Qiaochu Yuan Feb 9 '12 at 1:23
@Qiaochu: Indeed. – Grumpy Parsnip Feb 9 '12 at 2:05
Thanks, guys. Typos fixed. – Grumpy Parsnip Feb 9 '12 at 11:58

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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