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Consider $$T=\left\{ \left( x, \sin \tfrac{1}{x} \right ) : x \in [-1, 0)\cup(0,1] \right\} \cup \{(0,0)\}\subset \mathbb{R}^2$$ with the subspace topology.

Denote $p=(-1, \sin -1), q=(1, \sin 1)\in T$.

A TSC-homotopy between continuous maps $f, g:X\to Y$ is a continuous map $H:X\times T\to Y$ such that $H(x, p)=f, H(x, q)=g$.

Consider the following equivalence relation on topological spaces. $X$ and $Y$ are equivalent if there are continuous maps $f:X\to Y, g:Y\to X$ and homotopies $f\circ g\sim \mathrm{id}_Y, g\circ f\sim \mathrm{id}_X$.

What can be said about this equivalence relation?

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    $\begingroup$ This kind of homotopy is probably neither symmetric nor transitive, so it is not an equivalence relation. This makes it unlikely that there is any useful answer to your question. $\endgroup$
    – Neil Strickland
    Commented Apr 21, 2021 at 12:18
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    $\begingroup$ After your latest edit your relation on maps is now reflexive and symmetric, but still not transitive. Because of that, the resulting relation on spaces is probably not transitive either. $\endgroup$
    – Neil Strickland
    Commented Apr 21, 2021 at 16:22

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