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
6
$\begingroup$
algebraic-topology
homology-cohomology
-
1Perhaps 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
-
1Are you assuming $X$ is Hausdorff? – 2012-02-09
-
0Yeah, lets impose the Hausdorff condition. – 2012-02-09
-
2My 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
-
4I 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
-
0hemisphere->equator in my last comment. – 2012-02-09
2 Answers
2
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.
-
0I think asking for a Hausdorff example is a fascinating problem. – 2012-02-09
-
0Unless $X$ has one element or is empty! – 2012-02-09
-
0@Qiaochu: Indeed. – 2012-02-09
-
0Thanks, guys. Typos fixed. – 2012-02-09
-
0This is a very nice example :D – 2018-05-10
2
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 $H_{n}(M)\cong H_{n}(M-\lbrace p\rbrace)\oplus\mathbb{Z}$.
-
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