I'm currently reading Edgar H. Brown's paper, Abstract Homotopy Theory, which proves a categorical version of the Brown Representability Theorem. I've come across a statement that I don't really how to verify (if it is in fact true).
First the notation and definitions. Let $\mathscr{C}$ be the homotopy category of pointed CW complexes. Let $\mathscr{C}_0$ be (as I understand it) a countable (though small is apparently sufficient) full subcategory of the subcategory of finite pointed CW complexes that contains a representative of every isomorphism class of finite pointed CW complexes.
Now we define $\newcommand\oln\overline\oln{\mathscr{C}_0}$ to be the set of all objects $Y\in\mathscr{C}$ such that
for all homotopy classes of maps $f:Y\to Y'$, if we have that for all $X\in\mathscr{C}_0$, $f_* : [X,Y]\to [X,Y']$ is an isomorphism, then we have that $f$ is an isomorphism. In other words, $\oln{\mathscr{C}_0}$ consists of the objects $Y$ for which isomorphisms out of $Y$ can be detected by restricting the Yoneda embedding to finite CW complexes.
Brown claims that the set $\oln{\mathscr{C}_0}$ is the set of all connected CW complexes. I can see why the Whitehead theorem implies that the CW complexes are a subset of $\oln{\mathscr{C}_0}$, but I can't find any example of a space not in $\oln{\mathscr{C}_0}$, although it appears such a space must exist, due to a claim that there is no small full subcategory of the homotopy category of spaces that reflects isomorphisms in this manner appearing in this arxiv paper. It doesn't seem to really matter for the purposes of Brown's paper, but I do want to find some resolution to this question for my own personal curiosity.
To that end, Brown appears to be claiming that for any space $Y$ with at least 2 connected components, one can construct a space $Y'$ and a map $f:Y\to Y'$ which induces bijections $f_*:[X,Y]\to [X,Y']$ for all finite CW complexes $X$. If this is in fact true, I'd appreciate any reference to a proof of this fact, or even just a proof of this for one space, like $S^0$. Alternatively, a disproof of the claim would also resolve my question. Perhaps this is an easy question, but my first thoughts seem to have failed.
Actually, since I know that not all spaces can be in $\oln{\mathscr{C}_0}$, I know the following must be wrong, but I can't figure out where. I can't see why we can't apply the Whitehead theorem in the disconnected case by taking $X=S^0$ and $X=S^n_+$ (unpointed $S^n$ plus a disjoint base point). The first forces $f$ to induce a bijection on $\pi_0$, and the others appear to force $f$ to induce bijections on homotopy classes of maps from $S^n$ to the paired up components of $Y$ and $Y'$, which seems like it would imply that the induced map of homotopy groups for each component is an isomorphism. I would also appreciate if someone could explain where I'm going wrong.
Best Answer
Let $f:A\to B$ be a map of non-pointed CW complexes which looks like an isomorphism to the finite complexes in the homotopy category. (For example as in the paper you link we might take $A=B$ the classifying space of the group of permutations of a countable infinite set with all but finitely many points fixed, and $f$ the map induced by the “suspension” homomorphism, which shifts the action of a permutation up by one.) Then $f_+:A_+\to B_+$ has the same properties in the pointed category. Indeed the maps $X\to A_+$ in the pointed homotopy category are just the maps from subsets of the connected components of $X\setminus X_0$ to $A$ in the unpointed category ($X_0$ the component containing the basepoint).
The fact that unpointed homotopy rears its head in the pointed but not connected category away from the basepoint also explains the error in your last paragraph. Maps from $S^n_+$ in the pointed homotopy category are free homotopy classes of maps from $S^n$, which is not what $\pi_n$ is. In case $\pi_1$ is being annoying, this difference ruins everything.