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It is well-known that for a pair of spaces $(X,A)$ satisfying the homotopy extension property, with $A$ contractible, that the quotient $X \to X/A$ is a homotopy equivalence. Does the converse hold as well? If the quotient map $X \to X/A$ is a homotopy equivalence, is $A$ contractible?

  • There is probably a whitehead-type argument showing this is true for CW pairs, so I imagine that a counterexample would have to be some non-contractible, but still maybe weakly contractible subset of $X$. – Dan Rust Feb 11 '18 at 01:41
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    Your "well-known" statement is false. If you let $X$ be two copies of the Hawaiian earring space joined by a line segment $A$ connecting the basepoints, then the natural projection $X\to X/A$ is not a homotopy equivalence. In fact, the fundamental groups are distinct. – Cheerful Parsnip Feb 11 '18 at 02:48
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    Alternatively, take $A$ to be the complement of a point in $X=S^1$, then $X/A$ is a 2 point space ${x,y}$ with topology ${\emptyset,{x},{x,y}}$. It is perhaps not obvious, but this has trivial $\pi_1$. – Cheerful Parsnip Feb 11 '18 at 02:54
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    Eh, sorry, I forgot an adjective! The pair (X,A) must also satisfy the homotopy extension property! –  Feb 11 '18 at 05:09

2 Answers2

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  1. Here is an example for the updated question. Let $M$ be the (triangulated) Poincare homology sphere and let $A$ be obtained by removing from $M$ the interior of a 3-dimensional simplex. Then $A$ is acyclic (has homology of a point) but is not simply-connected since $\pi_1(M)\cong \pi_1(A)$. The cone $X=CA$ is obviously contractible. The pair $(X,A)$ satisfies HEP since $A$ is a subcomplex in $X$. The quotient $X/A$ is homotopy-equivalent to $X$ with the cone over $A$ attached; in other words, $X/A$ is homotopy-equivalent to the suspension $SA$ of $A$. But $SA$ is simply-connected (by the Seifert-Van Kampen theorem) and is easily seen to be acyclic (by the Mayer-Vietoris theorem). Hence, by the Hurewicz theorem, $SA$ is weakly contractible. Since $SA$ is a cell-complex, it is also contractible (by Whitehead's theorem). Thus, the projection map $X\to X/A$ is a continuous map of two contractible spaces, hence, a homotopy-equivalence. But $A$ is not contractible.

  2. This was my answer to the original question where the HEP was not assumed:

Consider $X=E^2$ and $A\subset E^2$, the double comb space. Then $A$ is known to be nonontractible, but at the same time, $A$ is cell-like (shape-equivalent to a point). Hence, by Moore's theorem (see the discussion and references here), $X/A$ is homeomorphic to $X$. The quotient map $f: X\to X/A\cong X$ is continuous. Then $f$ is a homotopy-equivalence (as any continuous map between contractible CW-complexes).

Moishe Kohan
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This is incorrect, see paragraph below.

Not necessarily. Let $X$ be the wedge sum of infinitely many circles and $A$ one of the circles, then $X/A$ is again the wedge sum of infinitely many circles.

As discussed in the comments, while $X$ and $X/A$ are homotopy equivalent, the projection map is not a homotopy equivalence. To see this, note that $\pi_1(X) = \pi_1(\bigvee_IS^1) = \bigoplus_I\pi_1(S^1) = \bigoplus_I \mathbb{Z}$ while $\pi_1(X/A) = \pi_1(\bigvee_{I\setminus\{i_0\}}S^1) = \bigoplus_{I\setminus\{i_0\}}\pi_1(S^1) = \bigoplus_{I\setminus\{i_0\}}\mathbb{Z}$, and the induced map is the natural projection map $\bigoplus_I \mathbb{Z} \to \bigoplus_{I\setminus\{i_0\}}\mathbb{Z}$ which has kernel $\mathbb{Z}$.

Michael Albanese
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