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?
Removing a single point alters homology
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algebraic-topology
homology-cohomology
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0hemisphere->equator in my last comment. – 2012-02-09
2 Answers
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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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0This is a very nice example :D – 2018-05-10
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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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0No ..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