# Can contractible subspace be ignored/collapsed when computing $\pi_n$ or $H_n$?

Can contractible subspace be ignored/collapsed when computing $\pi_n$ or $H_n$?

Motivation: I took this for granted for a long time, as I thought collapsing the contractible subspace does not change the homotopy type. Now it seems that this is only true for a CW pair...

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Please make the body of the message self-contained; think of the title as the index entry for the post. A book wouldn't begin the theorem saying "As in the index..." –  Arturo Magidin Feb 12 '11 at 21:55
Sorry for that, will pay attention to it next time. –  Soarer Feb 13 '11 at 5:20

Yes, you need a hypothesis like being a CW pair. A more general condition you could impose is that the inclusion of the contractible subspace is a cofibration.

For an example of why this is necessary, consider $X = S^1$ and $A = X\setminus\{\ast\}$. Then collapsing A gives you a two-point space which is contractible. There are several ways to see the contractibility of this space, one is that this space is the non-Hausdorff cone over a point.

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That is a nice and simple example :) –  Mariano Suárez-Alvarez Feb 12 '11 at 20:47
Can you shed some light on the contractibility of $X/A$ you mentioned? I can manually construct a contraction, but I don't have intuition about why it's true. –  Soarer Feb 13 '11 at 5:31
The argument I gave in terms of non-Hausdorff cone fits into the general framework of finite topological spaces. You can look at the notes on finite spaces on J.P. May's homepage. Here is another argument: a map f from X to the two-point space we constructed is the same thing as an open subset of X. (Take the preimage of the open point.) Now any such map is homotopic to the map given by the empty subset of X. Indeed, take the homotopy such that you take the map f for all elements in $[0,1)$ and the empty set for $[1]$. Then this is an open subset of $X \times [0,1]$ so... –  Dan Petersen Feb 13 '11 at 10:00
...it defines a continuous map to the two-point space. Hence the hom-sets of maps into this space (working in the category of topological spaces and homotopy classes of maps) are singletons, and it follows then from the Yoneda lemma that the space is homotopy equivalent to a point. –  Dan Petersen Feb 13 '11 at 10:05

You are right. An interesting example of this kind of behavior consists of taking two copies of the Hawaiian Earring space and connecting their basepoints by a line segment. Contracting the middle segment gives you the standard Hawaiian Earrings. However this contraction is not a homotopy equivalence! The fundamental groups are different!

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Why are fundamental groups different? I'd really like to understand this bit. –  Max Feb 15 '11 at 18:23
Well, if we let $G$ denote the fundamental group of the Hawaiian Earrings, the fundamental group of two copies joined by an edge is the free product $G*G$. So one has to show that $G*G\not\cong G$. –  Grumpy Parsnip Feb 15 '11 at 18:56
Right. I got that far too. And how would one do that? –  Max Feb 15 '11 at 19:04
The details are tricky, but it's not too hard to show that the obvious map (contracting the central edge) does not induce a surjection on $\pi_1$. Take a path that alternates between the two copies infinitely many times. This is clearly not in the image of the contraction map. –  Grumpy Parsnip Feb 15 '11 at 20:35
@Max: I heard this fact from Greg Conner. He has a series of three papers with Jim Cannon on these types of questions, but I failed to find this exact computation in their papers. :( If you look at Figure 2 of "On the fundamental groups of one-dimensional spaces" they give a shocking example of two cones on the Hawaiian earrings which become non-contractible when identified at their "bad" points. –  Grumpy Parsnip Feb 16 '11 at 11:42

Let me note a general fact: if the inclusion $A \hookrightarrow X$ (for $A$ a closed subspace) is a cofibration, and $A$ is contractible, then the map $X \to X/A$ is a homotopy equivalence. See Corollary 5.13 in chapter 1 of Whitehead's "Elements of homotopy theory."

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