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What is an example of a compact Hausdorff space $X$ and a contractible subspace $A\subset X$ such that $H^2(X)\ncong H^2(X/A)$?

Note that $A\subset X$ must not be a cofibration. I was thinking that $X=[0,1]^2$ and $A$ the comb space might work but I am not sure.

user448378
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1 Answers1

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Let $X=S^2$ and let $A$ be the complement of a point in $X$. Then $X/A$ has only two points, and is in fact contractible. So $H^2(X)\cong\mathbb{Z}$ but $H^2(X/A)$ is trivial.

Here's an example for which $A$ is closed. Start with a sphere $S^2\subset\mathbb{R}^3$, and then replace a neighborhood of a point by circular waves that oscillate faster and faster as you get near the point, accumulating at an entire line segment (like a 2-dimensional version of the topologist's sine curve). Let $A$ be the line segment where the waves accumulate and let $X$ be the union of the modified sphere and $A$. There is no path in $X$ from a point of $A$ to a point of the sphere, so $X$ is weak homotopy equivalent to the disjoint union of the punctured sphere and $A$; in particular, $H^2(X)$ is trivial. But $X/A$ is an ordinary sphere, so $H^2(X/A)\cong \mathbb{Z}$.

Eric Wofsey
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  • Thanks for the answer. So you start with $S^2$, remove a neighborhood of a point and replace it by circular waves that oscillate faster and faster near the point. Why does it accumulate at a line segment? What exactly is $A$? – user448378 May 22 '17 at 05:57
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    Let's pretend that instead of $S^2$, we had $\mathbb{R}^2$ sitting inside $\mathbb{R}^3$. What I'm doing is taking points of the form $(r\cos\theta,r\sin\theta,\sin(1/r))$ for $r>0$. As $r$ approaches $0$, the set of such points accumulates at the entire vertical line segment ${(0,0)}\times[-1,1]$. That vertical line segment is $A$ (except we are doing all this in a little patch of $S^2$ sitting inside $\mathbb{R}^3$, rather than on $\mathbb{R}^2$ sitting inside $\mathbb{R}^3$). – Eric Wofsey May 22 '17 at 06:03
  • Thank you, makes sense. The space $X$ isn't compact, but is it paracompact? – user448378 May 22 '17 at 06:15
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    The space $X$ is compact...why do you think it isn't? Also, any metric space (in particular, any subspace of $\mathbb{R}^n$) is paracompact. – Eric Wofsey May 22 '17 at 06:18
  • Right, I just noticed my mistake, I agree now on compactness. Another thing that's not yet entirely clear to me is why $X/A$ is an ordinary sphere. Could you make that a bit more explicit? And by the way, $X$ has no cohomology at all, right? – user448378 May 22 '17 at 06:21
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    There is a map $X\to S^2$ which "flattens" all the waves and sends $A$ to a point. This induces a continuous bijection $X/A\to S^2$, which is a homeomorphism since $X/A$ is compact and $S^2$ is Hausdorff. – Eric Wofsey May 22 '17 at 06:23
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    $X$ has the weak homotopy type of $S^0$ (since as far as paths are concerned, it's a disjoint union of two contractible pieces). So it has no cohomology in positive degrees, but $H^0(X)$ has rank $2$. – Eric Wofsey May 22 '17 at 06:24
  • Thanks. I agree that there are no paths from point of $A$ to points of the sphere, but why does this imply that $X$ is weakly equivalent to $S^0$? – user448378 May 22 '17 at 09:28
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    Consider the obvious continuous bijection $f:X\setminus A\coprod A\to X$. Any map from a path-connected space to $X$ factors through $f$ uniquely. It follows that $f$ is a weak equivalence. But $X\setminus A\coprod A\cong \mathbb{R}^2\coprod [0,1]$ is homotopy equivalent to $S^0$. – Eric Wofsey May 22 '17 at 15:51