Let $\mathcal{D}$ be a finite category and let $F:\mathcal{D}\to\mathbf{Set}$ be a functor. Since $\mathbf{Set}$ has (co)equalizers and finite (co)products, by the Existence Theorem $F$ has a limit and (dually) a colimit, and the Theorem tells us how to construct them (see for instance this answer for an explicit construction of the colimit).
Here is another neat way to find the limit. Recall that, for $A$ a set and $\Delta_A:\mathcal{D}\to\mathbf{Set}$ the constant functor, we have a bijection $\hom(A,\lim F)=[\mathcal{D},\mathbf{Set}](\Delta_A,F)$. Setting $A=\{ 1 \}=\ast$ to be a one-element set, we therefore find
$$
\begin{aligned}
\lim F &= \hom(\ast,\lim F) = [\mathcal{D},\mathbf{Set}](\Delta_\ast,F)\\
&= \Big\{ \big(f_i:\ast\to F(i)\big)_i\in \prod_{i\in\mathcal{D}} \hom\big(\ast,F(i)\big) \;\Big\vert\; f_j=F(\alpha)\circ f_i \text{ for all }\alpha:i\to j \Big\}.
\end{aligned}
$$
Using that $\hom(\ast,F(i))=F(i)$ for each $i$, we obtain the usual description of the limit in $\mathbf{Set}$!
Question: Is there a similar trick, for example using $\hom(\text{colim} F,A)=[\mathcal{D},\mathbf{Set}](F,\Delta_A)$, to find a description of the colimit?
My approach so far has been to first reduce the question to the case $\mathcal{D}=\{1,2\}$ and to find the right set $A$ to deduce that $F(1)\sqcup F(2)=\big(F(1)\times\{1\}\big)\cup\big(F(2)\times \{2\}\big)$, but I haven't found such a set yet. I suspect that dividing out by the equivalence relation will somehow follow from the condition $f_i=f_j\circ F(\alpha)$ on a cone $(f_i:F(i)\to A)_i$.
Best Answer
The power set functor $P : \mathbf{Set}^{\mathrm{op}} \to \mathbf{Set}$ is monadic, for example by using the Crude Monadicity Criterion; see Theorem 5.1.1 in Toposes, Triples and Theories by Barr-Wells, where it is proven for every elementary topos. In particular, $P$ creates limits. This means that we can construct colimits in $\mathbf{Set}$ from limits in $\mathbf{Set}$. This is also the standard way to show that every elementary topos is finitely cocomplete (Corollary 5.1.2 in loc.cit.).