Homotopy equivalence in terms of strong deformation retract Visualizing homotopy equivalence maps are not so easy. I thought before that $f:X\to Y$ and $g:Y\to X$ are homotopy equivalence iff one can deform $X$ continuously to $Y$. But this is wrong in general. So I tried the following:
Q1:

$f:X\to Y$ and $g:Y\to X$ are homotopy equivalence iff one can deform $X$ and $Y$ continuously to a third space $Z$.

or I think equivalently

$f:X\to Y$ and $g:Y\to X$ are homotopy equivalence iff there is a $A\subset X$ such that $A$ be a strong deformation retract of $X$ and $f(A)$ be a strong deformation retract of $Y$

Q2: What about this one?

$f:X\to Y$ and $g:Y\to X$ are homotopy equivalence iff there is a $A\subset X$ such that $A$ be a strong deformation retract of $X$ and $f(A)$ be a strong deformation retract of $Y$ (added after Paul's answer) AND there is a $B\subset Y$ such that $B$ be a strong deformation retract of $Y$ and $g(B)$ be a strong deformation retract of $X$.

Are the above statements true? any proof or counterexample?
 A: It is wrong. You ask whether the following two conditions are equivalent for two maps $f:X\to Y$ and $g:Y\to X$:

*

*$f$ and $g$ are homotopy equivalences.


*There is a $A\subset X$ such that $A$ be a strong deformation retract of $X$ and $f(A)$ be a strong deformation retract of $Y$.
First note that $g$ does not play any role in 2.
Now let $f : S^1 \to *$ be the constant map, where $*$ is a one-point space, and $g : * \to S^1$ be any map.  $f$ is no homotopy equivalence. Now take $A = S^1$. Then you see that 2. is satisfied.
Update for Q2:
It is wrong. For $n \le 0$ let $C_n \subset \mathbb R^2$ be the circle with radius $1/3$ and center $(n,0)$, for $n > 0$ let $C_n = \{(n,0)\}$. Define
$$X = Y = \bigcup_{n \in \mathbb Z} C_n ,$$
$$f : X \to Y, f(z) = \begin{cases} z + (1,0) & z \in C_n, n  \ne 0 \\ (1,0) & z \in C_0 \end{cases}$$
and $g = f$. This map translates $C_n$ to $C_{n+1}$ if $n \ne 0$ and collapses the circle $C_0$ to the point $C_1$. $f$ is not a homotopy equivalence. Let $A = X$. Then $f(A) = X$ and your condition on $f$ is satisfied. Since $g = f$, also the condition on $g$ is satisfied.
