Does Homotopy Equivalence Lead to a Homeomorphism? I've read online that "Intuitively, two spaces X and Y are homotopy equivalent if they can be transformed into one another (i.e., made homeomorphic) by bending, shrinking and expanding operations", for example on Wikipedia. Am I to assume then that if I have two spaces $X,Y$ and two continuous functions $f:X\to Y,$ $g:Y\to X$ such that $f\circ g$ and $g\circ f$ are homotopic to the identity functions, then $f\circ g(Y)$ and $g\circ f(X)$ are homeomorphic?
If not, then what's with the "intuition"? I've tried to figure it out and also searched online but haven't found nothing.
Edit: To be more clear, my question roughly is if $X,Y$ are homotopy equivalent, are there necessarily subsets $X'\subset X$ and $Y'\subset Y$ such that $X'$ is homotopy equivalent  to $X$ with inclusion, $Y'$ is homotopy equivalent to $Y$ with inclusion, and $X',Y'$ are homeomorphic?
But I'll accept an answer that is "homotopy equivalent" to this one.
 A: Consider two 1-complexes: 


*

*$X$ is a rose with two petals: a 1-complex with one vertex $v$ and with two edges $E_1,E_2$ each with both endpoints attached to $v$

*$Y$ is the "theta" graph, literally the letter $\theta$: a 1-complex with two vertices $v,w$ and with three edges $E_1,E_2,E_3$ each with one endpoint at $v$ and the other at $w$.


The graphs $X,Y$ are homotopy equivalent to each other. They are not homeomorphic to each other. And neither of them is homotopy equivalent to any proper subset.
A: Homotopy is a much looser equivalence than homeomorphism.  As an example, the circle $X = S^1$ and the punctured plane $Y = \Bbb{R}^2 \setminus \{0\}$ are homotopy equivalent, but they are not homeomorphic.
To see the homotopy equivalence explicitly, let $f: X \hookrightarrow Y$ be the natural inclusion, and let $g: Y \to X$ be the radial projection, defined by
$$
g(y_1, y_2) = \Bigl( \frac{y_1}{r}, \frac{y_2}{r} \Bigr), \qquad \text{where } r = \sqrt{y_1^2 + y_2^2}.
$$
In one direction, $g \circ f = \operatorname{Id}_X$ on the nose (so the constant homotopy suffices).
Now, consider $f \circ g: Y \to Y$.  Now, the map $H: Y \times [0, 1] \to Y$, defined by
$$
H(y, t) = \biggl( \frac{y_1}{r + (1-r)t}, \frac{y_2}{r + (1-r)t} \biggr), \qquad \text{where } r = \sqrt{y_1^2 + y_2^2}
$$
is a homotopy satisfying $f \circ g = H( \;\cdot\;, 0)$ and $H( \;\cdot\;, 1) = \operatorname{Id}_Y$.
(This is an example of deformation retraction.)
