Assume two $CW$ complexes $X,Y$ give two functors $h\_X=[-,X], h\_Y=[-,Y]$ on the homotopy category of $CW$ complexes whose restrictions to the full subcategory of finite $CW$ complexes are naturally equivalent. Does this imply that the two functors are naturally equivalent (i.e. that X, Y are homotopy equivalent)? 



Are Brown representable functors determined by restriction to finite complexes?
The answer depends on which category of CW-complexes you have in mind. It is true in the category of based connected CW-complexes simply by the Whitehead's Theorem.


**EDIT:** The concern raised by Matthias in the comment above is addressed in Brown's paper (Theorem 2.8 of *Abstract Homotopy Theory*). This is only proven in the based case, but even if it works in the unbased case too, then the following still applies.


However, dropping any of these assumptions makes the answer negative. In fact, Heller proves in Corollary 2.3 of *On the representability of homotopy functors* that there is no set of CW-complexes such that functors represented by them detect weak equivalences.
(This is closely related to the failure of Brown's representability in the unbased case, discussed [here](https://mathoverflow.net/questions/104866/brown-representability-for-non-connected-spaces/105049) based on another paper by Freyd and Heller.)


In particular, finite CW-complexes do not suffice since a map inducing bijections on homotopy classes of maps out of CW-complexes might not be a $\pi\_1$-isomorphism. In general, it will only induce a bijection on conjugacy classes of $\pi\_1$s. On the other hand, [this answer](https://mathoverflow.net/a/55428/12547) shows that this works for maps between CW-complexes with finitely generated fundamental groups.


