A pair of spaces $(X,A)$, with $A$ non-contractible, such that the quotient $X \to X/A$ is a homotopy equivalence. 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?
 A: *

*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.


*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).
A: 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 $\mathbb{Z}$.
