Homotopy dimension of homotopy domination

Question

The homotopy dimension of a space $X$ is the smallest covering dimension of any space homotopy equivalent to $X$.

Assume that for topological spaces $X$ and $Y$, there exist maps $f:X\to Y$ and $g:Y\to X$ so that $g\circ f\simeq 1_X$. If $X\not \simeq Y$, then is it true that homotopy dimension of $X$ is strictly smaller than homotopy dimension of $Y$?

Answer 1

No. Take $X=S^n$ and $Y=S^n\vee S^n$, with $f$ the inclusion into one of the wedge summands. Then it is clear that $X\not\simeq Y$, and that the homotopy dimensions of $X$ and $Y$ are equal.