 Solution

If $fg\simeq 1_Y$ and $hf\simeq 1_X$, composing homotopies givesThus $gf\simeq hf\simeq 1_X$, while $fg\simeq 1_Y$ already holds, so $g$ is a homotopy inverse of $f$.

More generally let $a:Y\to Y$ be a homotopy inverse of $fg$ and $b:X\to X$ one of $hf$. Then $g^{\prime }=ga$ satisfies $f{g}^{\prime }\simeq 1_Y$, and $h^{\prime }=bh$ satisfies $h^{\prime }f\simeq 1_X$. The first argument applies. Hence the separate one-sided hypotheses imply that $f$ is a homotopy equivalence, as expressed by separate homotopy inverses combine into a homotopy equivalence.