Let $f:X\longrightarrow Y$ be a map between CW-complexes $X$ and $Y$. By the Whitehead Theorems, if one of the conditions: 


1- (homotopy version) $\pi\_n (f):\pi\_n (X)\longrightarrow \pi\_n (Y)$ is an isomorphism for all $n\geq 1$, 


or


2- (homology version) $\pi\_1 (f):\pi\_1 (X)\longrightarrow \pi\_1 (Y)$ and $H\_n (\tilde{f}):H\_n (\tilde{X})\longrightarrow H\_n (\tilde{Y})$ are isomorphisms for all $n\geq 2$,


hold, then there is a map $g:Y\longrightarrow X$ such that $g\circ f\simeq id\_X$ and $f\circ g\simeq id\_Y$. 


Question: Is there any weaker condition (with respect to above conditions) under which there is a map $g:Y\longrightarrow X$ such that we have only $g\circ f\simeq id\_X$? 



A weak version of the Whitehead Theorems
Suppose you have a map $f\colon X\to Y$ of finite CW complexes such that $K(p,n)\_\*(f)$ is injective for all primes $p$ and integers $n\geq 0$ (where $K(p,n)$ is Morava $K$-theory). Then the Nilpotence Theorem of Hopkins, Devinatz and Smith implies that the map $\Sigma^kf^{(m)}\colon \Sigma^kX^{(m)}\to\Sigma^kY^{(m)}$ has a left inverse $g$ for sufficiently large $k$ and $m$. This may not seem very satisfactory but I think that it is the best that you can reasonably hope to do. The question of whether a map has a left inverse is just intrinsically much more subtle than the question of whether it is an equivalence. 


One can also say that in the category of finite spectra, a map $f\colon X\to Y$ has a left inverse iff $\pi\_\*^S(f)\colon \pi\_\*^S(X)\to\pi\_\*^S(Y)$ is injective, provided that Freyd's Generating Hypothesis is correct. But the Generating Hypothesis has been any open question for fifty years now, which is another indication that the problem is intrinsically hard.


