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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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    Sorry for that, will pay attention to it next time.2011-02-13

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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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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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    @Jim. Thanks for the reference. This came up in my teaching and I did a quick literature search, and did not find anything. I would imagine experts (Conner, Cannon, maybe Eda) would know this, but it would be nice to see it written/explained.2011-02-16
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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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    ...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.2011-02-13