The following statement is the version of Brown's representability theorem I learned:

Let $\mathbf{CW}$ be the category of based, connected CW complexes together with based homotopy classes of based maps, i.e. $\mathrm{Hom}(X,Y)=[X,Y]_0$. Furthermore, let $\mathscr{F}:\mathbf{CW}^{\operatorname{op}}\to \mathbf{Set}$ be a contravariant functor satisfying the wedge and the Mayer–Vietoris axiom. Then $\mathscr{F}$ is representable.

The master examples were the following:

  1. For each $n\ge 0$ and each abelian group $G$, the singular cohomology functor $H^n(-;G)$ is representable, the classifying space is the Eilenberg–MacLane space $K(G,n)$, so we get $H^n(-;G)\cong [-,K(G,n)]_0$.

  2. For each topological group $G$, the functor $\mathrm{Prin}_G$ assigning to each connected and based CW complex the set of isomorphism classes of principal $G$-bundles (and each morphism the pullback) is representable, the classifying space is given by $BG$, so we get $\mathrm{Prin}_G\cong [-,BG]_0$.

The question is: Is it correct to consider based homotopy classes of based maps? In many articles, I just read something like $H^n(-;G)\cong [-,K(G,n)]$. Apparently, there is a difference between $[X,Y]$ and $[X,Y]_0$ and since we are working in a category with based spaces, it seems natural to use the morphisms from above …

  • $\begingroup$ The based-category is so prevalent that this is often assumed in the notation. What would the identity element be without it? $\endgroup$ – Randall Jun 29 '18 at 16:41
  • $\begingroup$ You also could consider free homotopy classes of based maps and have an identity element, but this seems strange … $\endgroup$ – FKranhold Jun 29 '18 at 16:45

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