Does weak equivalence determine the homotopy classes of maps to a CW complex?

Question

We work in the category of pointed CGWH topological spaces. Say $X$ is some arbitrary space and $f: \Gamma X \to X$ is a CW approximation of $X$, so in particular $f$ is a weak equivalence. For any CW-complex $Y$, is it true that the pullback $$f^\ast: [X, Y] \to [\Gamma X, Y]$$ induces an isomorphism? Here, $[A, B]$ denotes the set of based homotopy classes of maps $A \to B$. This is certainly true if $X$ has the homotopy type of a CW complex, but I'm not sure in general. I'm aware that if the positions of $X, Y$ were reversed, and $f^\ast$ were replaced with $f_\ast$, then this is true, but I'm not sure about the pullback.

Answer 1

This is false. Let $X$ be the Warsaw circle, i.e. $X$ is the the topologist's sine curve in the plane together with its limit points and a choice of arc connecting the origin to 1 not intersecting the rest of the construction. Then $X$ is weakly contractible, so $\Gamma X$ is contractible. Taking now $Y = S^1$, we of course have $[\Gamma X, Y] = 0$, but $[X, Y] \neq 0$: This answer shows there is a non-nullhomotopic map from $X$ to $S^1$.

Answer 2

Consider any space $X$ that is totally path-disconnected but not discrete. Then taking $\Gamma X$ to be $X$ with the discrete topology, the identity function $f:\Gamma X\to X$ is a CW approximation. If $f^*:[X,\Gamma X]\to[\Gamma X,\Gamma X]$ were surjective, then there would exist $g:X\to\Gamma X$ such that $gf\simeq 1_{\Gamma X}$. Such a $g$ could only be the identity function, but this is not continuous as a map $X\to\Gamma X$ since $X$ is not discrete.