Consider two topological spaces $X$ and $Y$. 


The notion of homotopy equivalence between $X$ and $Y$ is defined as a pair of continuous maps $f:X\to Y$ and $g:Y\to X$ such that $f\circ g$ and $g\circ f$ are homotopic to the identities. 


But what if we weaken the definition a little bit, by considering the following "weaker" homotopy equivalence between $X$ and $Y$ : it is defined as a pair of pairs of continuous maps, $(f,g)$ and $(f',g')$, where $f,f':X\to Y$, and $g,g':Y\to X$ such that $g\circ f$ is homotopic to $Id\_X$ and $f'\circ g'$ is homotopic to $Id\_Y$. 


I'm asking this question because I don't see why intuitively we would require $f'$ and $g'$ to be actually equal to $f$ and $g$.


Would this definition give an interesting notion of homotopy equivalence, or would this just coincide with the usual definition ? Or would this be simply too weak to do anything with it ?



On a weaker version of homotopy equivalence between topological spaces
Saying that there are maps $f \colon X \to Y$ and $g \colon Y \to X$ such that $g f$ is homotopic to $\mathrm{id}\_X$ means that $X$ is a homotopy retract of $Y$. (By the way, we say that maps are *homotopic* not *homotopy equivalent*.) In your situation you have two spaces that are homotopy retracts of each other and it turns out that they need not be homotopy equivalent.


Let $T^k$ be the $k$-torus (the product of $k$ copies of $S^1$). Set $X = \bigsqcup\_{k \ge 0} T^k$ and $Y = \bigsqcup\_{k \ge 1} T^k$. Let $f \colon X \to Y$ map each $T^k$ onto $T^k \times \{ 1 \} \subset T^{k + 1}$, this has a (strict) retraction that is given by projections $T^{k + 1} \to T^k$ (on the first $k$ coordinates). Let $g' \colon Y \to X$ map each $T^k$ identically onto $T^k$, it also has a (strict) retraction that just maps $T^0$ anywhere. However, $X$ and $Y$ are not homotopy equivalent since $X$ has a contractible component, but $Y$ does not.


