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Let $f \colon X \to Y$ be a weak homotopy equivalence between topological spaces. If I am not mistaken, then one can rephrase this by stating that the induced map $[K,X] \to [K,Y]$ between homotopy classes of maps is a bijection for every finite simplicial complex $K$, right?

Now in a paper that I was reading the authors proved the same bijection, but for any compact Hausdorff space $K$, and I was wondering: How much more general than a weak homotopy equivalence is this?

(Note that in the paper the spaces $X$ and $Y$ are point-set topologically not well-behaved; especially, they are not CW-complexes. So one can not just conclude that they are homotopy equivalent to each other.)

One can phrase my question a bit more concretely in the following way: If the induced map $[K,X] \to [K,Y]$ is a bijection for every finite simplicial complex, and $X$ and $Y$ are CW-complexes (or more generally, have the homotopy type of CW-complexes), then $X$ and $Y$ are actually homotopy equivalent to each other. So if we can now prove that $[K,X] \to [K,Y]$ is a bijection for every compact Hausdorff space $K$, what kind of topological spaces can $X$ and $Y$ be in order to conclude that they are already homotopy equivalent?

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    $\begingroup$ It's not true that $f:X\rightarrow Y$ need be a weak homotopy equivalence if it induces bijections $f_*:[K,X]\rightarrow [K,Y]$ for each finite CW complex $K$. It's only true if either $(a)$ some nilpotency conditions are placed on $X,Y$, or $(b)$ $f_*$ is a bijection for each CW complex $K$. $\endgroup$
    – Tyrone
    Commented May 3 at 15:32
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    $\begingroup$ And it is not true that a weak homotopy equivalence induces a bijection on homotopy classes from a compact Hausdorff space. Counterexample: The Warsaw circle is compact Hausdorff , not contractible but weakly contractible. $\endgroup$
    – Johannes Ebert
    Commented May 3 at 17:06
  • $\begingroup$ @Tyrone: I am confused. If I take $K$ to be the $n$-sphere, then $[K,X]$ is, up to basepoint stuff, $\pi_n(X)$. So what goes wrong here with $f$ not being a weak homotopy equivalence, is it really the basepoint issue? $\endgroup$
    – AlexE
    Commented May 4 at 6:23
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    $\begingroup$ There is an action of $\pi_1Y$ on the pointed homotopy sets. It turns out that this is quite important. Assuming $X,Y$ are path-connected, a map $f:X\rightarrow Y$ inducing a bijection $[S^1,X]\xrightarrow{\cong}[S^1,Y]$ induces a bijection between conjugacy classes in $\pi_1X$ and $\pi_1Y$. But it might not induce an isomorphism between the $\pi_1$'s. This is essentially the problem: $f:X\rightarrow Y$ is a weak equivalence if and only if $(i)$ $f_*:[S^n,X]\rightarrow[S^n,Y]$ is bijective for all $n\geq0$, and $(ii)$ $f_*:\pi_1(X,x)\rightarrow \pi_1(Y,f(x))$ is surjective for all $x\in X$. $\endgroup$
    – Tyrone
    Commented May 5 at 10:20
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    $\begingroup$ This is the obvious answer, but since it hasn't been made explicit yet: If $f_{\ast}\colon[K,X]\rightarrow[K,Y]$ is a bijection for all ch(compact hausdorff)-spaces $K$ and $X,Y$ are ch-spaces, then $f$ is a homotopy equivalence (Yoneda lemma in the homotopy category of ch-spaces) and of course this holds equally well if $X,Y$ only have the homotopy type of ch-spaces. If a space has both the homotopy type of a ch-space and of a CW-complex, its (co-)homology is f.g., so this is a statement in a different direction. $\endgroup$
    – Thorgott
    Commented May 5 at 21:51

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