# Homotopy dimension of homotopy domination

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 $$Y$$?

## 1 Answer

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.