When proving that colimits are universal (stable under pullback), why is it sufficient to prove it for coproducts and coequalizers?

Question

I am trying to understand Borceux's proof that colimits are universal in Set. He opens by saying that it is sufficient to prove this for coproducts and coequalizers. I saw this answer, but I am missing steps.

Suppose we have a functor $F: \mathcal{D}\to\mathcal{C}$ with $\mathcal{C}$ cocomplete. Let $L$ be the colimit of $F$ and let $f: K\to L$ be a morphism that we would like to pull back along. Let $G$ be the pulled-back functor. We'd like to argue that $K$ is the colimit of $G$.

We can construct coproducts $M_1, M_2$ with morphisms $\alpha, \beta: M_2\to M_1$, $\gamma: M_1\to L$ such that $\gamma=\mathrm{Coker}(\alpha,\beta)$. I see that we can first pullback the coequalizer diagram along $f$, and then pullback each of $M_1$ and $M_2$'s coproduct diagrams along the resulting morphisms. This would show that $K$ is the coequalizer of some pair of coproducts. But why would this show that $K$ is the coequalizer of the correct pair of coproducts for computing the colimit of $G$? The pullbacks for constructing this coequalizer and the coproducts don't involve $G$.

Answer 1

This is basically a matter of sitting down and checking all the details. Since coproducts are assumed to be universal, the situation can be summarised by the following diagram:

But $f^* \circ F$ is exactly your pulled back functor $G$.