A deformation retraction in the weak sense of a space $X$ to a subspace $A$ is a homotopy $f_t:X \to X$ such that $f_0=1_X, f_1(X) \subset A$ and $f_t(A)\subset A$ for all $t$. Show that if $X$ deformation retracts to $A$ in this weak sense, then the inclusion $\iota:A \to X$ is a homotopy equivalence.
I've seen the proof for this problem which follows just by taking $g:X \to A$ as $g=f_1$ for the homotopy inverse for $\iota$. I'm trying to get sorta "visual" understanding of what is tried to communicate here.
If I take $X=S^1 \times [0,1]$ i.e. the cylinder in $\Bbb R^3$, then we know that $S^1 \times [0,1]$ deformation retracts to $S^1$ so in this case I would have that $\iota :S^1 \to S^1 \times [0,1]$ is a homotopy equivalence, but I have a slight concern about this.
Wikipedia states that
Intuitively, two spaces $X$ and $Y$ are homotopy equivalent if they can be transformed into one another by bending, shrinking and expanding operations.
However I don't think I can use $\iota$ to "get back to" $S^1 \times [0,1]$ in any way so how can these two spaces be homotopy equivalent?