According to wikipedia, two spaces $X$ and $Y$ are intuitively homotopy equivalent, if they can be transformed into one another by bending, shrinking and expanding operations. I wondered why this is an accurate intuition of the notion.
Let $I:=[0,1]$. I know that a deformation retraction of a space $X$ onto a subspace $A$ is a family of maps $(f_t)_{t \in I}$, where $f_t:X \to X$ for all $t$ with the properties that $f_0=\mathrm{id}_X$, $f_1(X)=A$ and $f_t|_A=\mathrm{id}_A$, such that the function $F:X \times I \to X, (x,t)\mapsto f_t(x)$ is a map. Intuitively that means that one continuously shrinks the space $X$ to the space $A$, where the points of $A$ are fixed/never moved. To my mind, this represents the shrinking operations, so I would say that, intuitively, that a subspace $A$ is a deformation retract of $X$ if one can continuously shrink $X$ to $A$ without moving the points of $A$.
If this is the case, we have a resulting retraction $r:X \to A$, that is, a map $r:X \to A$ such that $r|_A=\mathrm{id}_A$ and with $r(X)=A$, namely $r:=f_1|_A.$ If $\iota$ denotes the inclusion of $A$ into $X$, we have $r \circ \iota =\mathrm{id}_A$ and $\iota \circ r \simeq \mathrm{id}_X$ since $\iota \circ r=f_1.$ In particular, this is a homotopy equivalence:
Let $X$ and $Y$ be spaces, then they are homotopy equivalent, if there are maps $f:X \to Y$ and $g:Y \to X$ such that $g \circ f \simeq \mathrm{id}_X$ and $f \circ g \simeq \mathrm{id}_Y$.
Thus, intuitively, the notion of a shrinking process as above gives a homotopy equivalence. However, this doesn't justify the other "implication", that is, why spaces are homotopy equivalent if they are obtained in the way wikipedia claims.
I have read this post with a similar question and one of the answers mentions that every homotopy equivalence is the composition of deformation retractions. Why is that the case? Perhaps this would help me gain more intuition, but as of now, I don't see why the claim made by wikipedia is correct.
