Let $\mathcal{C}_0$ be the homotopy category of path connected pointed spaces with nondegenerate base points. A part of the definition of the homotopy functor is that whenever $j: X \to Z$ is an equalizer for the pair of maps $f_0,f_1: A \to X$, and $u\in H(X)$ is such that $H(f_0)u=H(f_1)u$, then $\exists v \in H(Z)$ such that $j^*v=u$. (See Spanier, Algebraic Topology p.407). How does one prove, as Spanier claims, that this condition is met for the singular cohomology functor with coefficients in a group $G$, i.e. $H^n(X;G)$ ?
Edit : This seems to be a puzzling problem to me. Needless to say, we need to do this without using the Brown representability theorem. After lot of thinking I am still not able to see how to proceed. Nor am I able to show how the condition is met for the $Hom(\pi_1(-),G)$ where $G$ is a fixed abelian group (example 5 on page 407).
Lemma 7.1 (p. 406 on Google Books) gives an explicit construction of the "equalizer" (homotopy colimit) of $f_0,f_1\colon A\to X$. You can check it can be covered by two spaces each homotopy equivalent to $X$ and with intersection equal to $A$.
For $H^n(X;G)$, use the exactness of the Mayer-Vietoris sequence applied to this cover.
For $\pi_1(X,x_0)\to G$ use the Seifert–van Kampen theorem applied to this cover, plus the formal fact that representable functors are continuous: $\mathrm{hom}_{Gp}(-,G)$ takes pushouts in $\mathrm{Gp}$ to pullbacks in $\mathrm{Set}$.