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It is well known that, if $A \subset X$ is a reasonable contractible subspace, then the quotient map $X \to X/A$ is a homotopy equivalence ("reasonable" means that the pair $(X,A)$ has the homotopy extension property, e.g. if it is a CW pair). For example, it's proposition 0.17 in Hatcher's celebrated Algebraic Topology.

It seems to me that this result should be a consequence of the following result, which seems true but for which I've been unable to find a proof or a reference.

Proposition? If $X$ is Hausdorff and $\sim$ is an equivalence relation whose classes $C$ are contractible and such that every pair $(X, C)$ has the homotopy extension property, then the quotient map $X \to X/\sim$ is a homotopy equivalence.

The classical result would of course be a direct corollary of this one.

So, is this proposition true?

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  • $\begingroup$ When you say classes contractible you also assume connected? What about the quotient maps from spheres to projective spaces? $\endgroup$ – Sigur Nov 15 '13 at 15:50
  • $\begingroup$ Yes: for me, contractibility implies connectedness. $\endgroup$ – PseudoNeo Nov 15 '13 at 16:24
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    $\begingroup$ Contractible means connected for everyone. It's defined as "homotopy equivalent to a point." $\endgroup$ – Cheerful Parsnip Nov 18 '13 at 14:24
  • $\begingroup$ Great question, by the way. $\endgroup$ – Cheerful Parsnip Nov 18 '13 at 14:55
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I think there's something wrong with your statement that $(X,A)$ has the homotopy extension property if $X$ is Hausdorff and $A$ is closed. Take the Hawaiian earring space $E$ and form a new space $X$ by joining two copies of $E$ by an edge from the wedge point to the wedge point. Then contracting the central edge to get a new space $E\vee E$ is not a homotopy equivalence. For example, it is not surjective on the fundamental group. There are loops in $E\vee E$ which travel back and forth over each copy of $E$ infinitely many times. However there are no such paths in $X$ since they would have to travel over the central edge infinitely many times.

Hatcher assumes $(X,A)$ are a CW pair, which is a far stronger condition.

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  • $\begingroup$ I think you can get away with $A$ have a contractible neighbourhood. $\endgroup$ – Dan Rust Nov 19 '13 at 2:00
  • $\begingroup$ Oh, drat. You're right, of course. I wanted to avoid technicalities, but I can't. I will modify the question. Thank you very much. $\endgroup$ – PseudoNeo Nov 19 '13 at 13:32
  • $\begingroup$ @PseudoNeo: Okay, thanks for the revised version. $\endgroup$ – Cheerful Parsnip Nov 19 '13 at 13:54
  • $\begingroup$ @PseudoNeo: Thanks for the bonus. I guess it's good not to let it evaporate. The revised question is quite interesting, but I haven't had the time to think about it. $\endgroup$ – Cheerful Parsnip Nov 26 '13 at 16:42
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The conjectured proposition is false. Even though it is not the main subject of the paper, a counter-example is given by the "Example 2.11" in the paper On fundamental groups of quotient spaces by Calcut, Gompf and McCarthy. They give a more detailed description of this example in section 3.7 of their paper Quotient maps with connected fibers and the fundamental group. For the sake of completeness, I give the highlights of this counter-example below.

Let $H = \{(x,y,z) \in \mathbb{R}^3 \, | \, z \ge 0\}$. It is an Hausdorff contractible space. It admits an equivalence relation $\sim$ such that all classes are properly smoothly embedded copies of $[0,1]$ or of $\mathbb{R}$ which are closed in $H$; As such, given any class $C$, $C$ is contractible and the pair $(H, C)$ is a good pair, so it has the homotopy extension property. However, the quotient $H/\sim$ is homeomorphic to $S^2$ which is not contractible, hence not homotopy equivalent to $H$.

In fact, if $\sigma : S^3 \setminus \{(0,0,0,1)\} \to \mathbb{R}^3$ denotes the stereographic projection, then $\sim$ is the image by $\sigma$ of the relation induced on $\sigma^{-1}(H)$ by the Hopf fibration.

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