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Are there any nontrivial spaces $Y$ so that for all weak homotopy equivalences $A\to B$, the induced map $[B, Y]\to [A,Y]$ is bijective?

This would be a property of the homotopy type of $Y$, and if $Y$ is homotopy equivalent to a space with has some kind of local structure under which very close maps (probably of compact spaces like $S^k$) are necessarily homotopic, then it probably won't have this property.

My idea is to use the following construction: let $L^+ = \{ {1\over n} \mid n \geq 1\}$ and let $L = \{ 0\} \cup L^+$. Then this hypothetical local property of $Y$ would ensure that the restriction $L \times X \to L^+\times X$ would induce an injection on homotopy sets. But $L\times X$ is weakly equivalent to $\coprod_{0}^\infty X$, and in the latter space we can have maps which are $f$ on $X\times {n}$ for $n > 0$ and $g$ on $X\times 0$, where $f\not \simeq g$.

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If the space $Y$ is also a $T_1$ space, then its path components are weakly contractible, by choosing $A=S^n$, $n>0$, and $B$ a suitable space with finitely many points. Each of these path components is then contractible, by choosing $A$ to be the path-component and $B$ to be a point. For any two path components there is a sequence in one converging to a point in the other, by choosing $B=${0,1, 1/2, 1/3,...} and $A$ discrete. Thus $Y$ is either connected or empty. Can one take this further and see that then $Y$ is either or contractible or empty? –  Tom Goodwillie Nov 23 '10 at 14:22
    
Great! So in particular, singular cohomology is not represented by any $T_1$-space in the homotopy category of spaces. –  Lennart Meier Feb 5 at 20:20

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Here is an interesting test case: let $B$ be the Stone-Cech compactification of a set $S$, let $A$ be the underlying set of $B$ with the discrete topology, and let $f$ be the identity map. Then $B$ is totally disconnected so every map from a simplex to $B$ is constant, and it follows that $f$ is a weak equivalence, so we must have $[B,Y]=[A,Y]=\text{Map}(A,\pi_0(Y))$. Note that $B$ is always compact and that if $S$ is large enough we can choose a surjective map $A\to Y$; it follows that there is a compact subset of $Y$ that meets every path component. I think it should be possible to extract a lot more from this line of argument, but I do not see it at the moment.

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