Does homotopy equivalence to a subspace imply (weak) deformation retract? Here is a nearly same question, but with a slight difference in interpretation. We use the definition of wikipedia, which is different from that of Hatcher or Munkres (where deformation retract refers to strong deformation retract on wikipedia).
My question is :

Suppose $X$ is a path-connected topological space, and $A$ is its subspace. If $X$ and $A$ are homotopic equivalent, is it true that $A$ is a (weak) deformation retract of $X$?

Both Hatcher and Munkres have provided counterexamples for the case of strong deformation retract, but not the one above. Please help.

This community wiki solution is intended to clear the question from the unanswered queue.
Noel Lundström has sketched a counterexample in a comment. Let $X = \bigvee_{i=1}^\infty S^1_i$ be the countably infinite wedge of copies of the circle $S^1$. This is a $CW$-complex with one $0$-cell and countably infinitely many $1$-cells. The space $A = \bigvee_{i=2}^\infty S^1_i$ is a subcomplex of $X$ which is homeomorphic to $X$; it is also a retract of $X$. However, the inclusion $i : A \to X$ is not a homotopy equivalence, thus $A$ is not a (weak) deformation retract of $X$.
To prove this, assume $i$ has a homotopy inverse $f : X \to A$. Then $i \circ f \simeq id_X$. Let $i_1$ be the inclusion of $S^1_1$ into $X$ and $r_1 : X \to S^1_1$ be the retraction which maps $A$ to the wedge point. We get $r_1 \circ f \circ i_1 \simeq r_1 \circ id_X \circ i_1 = id$. But $r_1 \circ f \circ i_1$ is constant, thus the identity on $S^1_1$ must be inessential. This is a contradiction.
