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Let $X$ be a contractible topological space. Let $A\subseteq X$ be a contractible subspace. Is the quotient space $X/A$ necessarily contractible?

It is not hard to show that this is true if, for example, the pair $(X,A)$ has the homotopy extension property (see, e.g. Proposition 0.17 of Hatcher, Algebraic Topology).

Some friends and I amused ourselves by trying to answer this question without assuming that $(X,A)$ has the HEP, but we were unsuccessful. Any insight would be appreciated.

Mr. Frog
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1 Answers1

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Cannon and Conner have a counterexample in their paper "On the Fundamental Group of One-Dimensional Spaces." See Example 2.0.2. This is a doubled cone on the Hawaiian Earring space, which they show is not contractible. However, if you expand the point of contact between the two cones to a line segment, then it is contractible.

  • Could you elaborate on why the space is contractible when the cones are joined by a line segment? Simple-connectedivity is clear from Van Kampen, but I don't see why the space is contractible. – Mr. Frog Sep 01 '16 at 16:57
  • @Mr.Frog: I am simply quoting their paper, which on closer inspection actually cites their paper "The combinatorial structure of the Hawaiian Earring group" for proofs. I actually haven't thought about how the proof goes myself! – Cheerful Parsnip Sep 01 '16 at 17:27
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    @Mr.Frog: I think this is how the contraction goes. First homotop each copy of the Hawaiian earrings up to their cone point, dragging a bit of the ends of the line segment up with them, but fixing the middle ofthe line segment. Now everything has been homotoped into a homeomorphic copy of an interval connecting the two cone points, which can then be easily contracted to its center. – Cheerful Parsnip Sep 01 '16 at 17:38
  • @GrumpyParsnip: I suppose that the logic behind this is that a line segment allows for stretching, but a single point does not? – Transcendental Sep 02 '16 at 23:59
  • @Transcendental: yes that's intuitively what's happening. – Cheerful Parsnip Sep 03 '16 at 01:28