A question about homotopy dimension

Question

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

There's a simple fact that if $X$ is homotopy dominated by $Y$ (denoted by $X\leqslant Y$), i.e. there exist maps $f:X\to Y$ and $g:Y\to X$ so that $g\circ f\simeq 1_X$., then $hd(X)\leq hd(Y)$ and if $X\not \simeq Y$, then it's not $hd(X)<hd(Y)$, for example, take $X=S^n$ and $Y=S^n\vee S^n$, with $f$ the inclusion into one of the wedge summands.

My question is that:
Is there any result which tells us with conditions on $Y$ we have "$X\leq Y$ and $X\not \simeq Y$ implies $hd(X)<hd(Y)$"?

Answer 1

Spheres have this property: If $Y=S^n$ and $X\le Y$ then since the identity homomorphism on $\pi_i(X)$ factors through $\pi_i(S^n)$ for all $i$, we have that $\pi_i(X)=0$ for $i<n$ and $\pi_n(X)=0$ or $\mathbb{Z}$ (these being the only retracts of the infinite cyclic group). Furthermore, the identity map on $H_i(X)$ factors through $H_i(S^n)=0$ for all $i>n$. From the Hurewicz Theorem, we conclude that either $X$ is weakly contractible or the map $f:X\to Y$ is a homology isomorphism.

Now I will assume that everything has the homotopy type of a CW complex. Then either $X$ is contractible or $X\simeq Y$.

The same argument applies when $Y$ is a Moore space $M(G,n)$ with $G$ cyclic, or an Eilenberg-Mac Lane space $K(G,1)$ with $G$ a group without non-trivial retracts.