Presentation of the covariant power set functor How to write the covariant power set functor (restricted to finite sets for simplicity)
$$P : \mathsf{FinSet} \to \mathsf{Set}$$
concisely as a colimt of representable functors? There is an epimorphism $$\coprod_{n \geq 0} \hom(\{1,\dotsc,n\},-) \to P,$$
mapping $f \in \hom(\{1,\dotsc,n\},X)$ to $\mathrm{im}(f) \in P(X)$. This already provides a generating set of $P$. Compare this with the contravariant power set functor, which is already a representable functor.
This question is motivated by the exercise to find all morphisms of functors $P \to P$ without too much calculations.
 A: It helps a lot that in this case $\hom(\{1,...,n\},-)\cong\hom(\{1\},-)^n$. Denote $\hom(\{1\},-)$ by $X$ and take more generally in any category with finite products and countable colimits distributing over each other the free monoid on $X$, that is, $1\sqcup X\sqcup X^2\sqcup X^3\sqcup\cdots$. Now in addition factor out by the following: identify, for any surjection $\pi:\{1,...,m\}\twoheadrightarrow\{1,...,n\}$, the summand $X^n$ with its image under $X^\pi:X^n\rightarrowtail X^m$ (note that in particular this forces quotienting $X^n$ by the action of the $n$th symmetric group, so we get the free commutative monoid on $X$ as an intermediate step).
I believe what we get is the free internal semilattice on $X$, and for $X=\hom(\{1\},-)$ one obtains the covariant powerset. This answers the question since the diagram we start from consists of representables since, as already said, $X^n$ is isomorphic to $\hom(\{1,...,n\},-)$.
As HeinrichD notes in the comment below, we probably only need quotients by $X^k\times\text{switch}\times X^l:X^k\times X^2\times X^l\to X^k\times X^2\times X^l$ and by $X^n\times\text{diagonal}:X^n\times X\to X^n\times X^2$, although I fail to organize this into a honest diagram for the moment.
In any case, seems like the resulting description of natural transformations $P\to F$ is as follows: they are in one-to-one correspondence with families
$$
\left(\xi_n\right)_{n=0,1,2,...}\in\prod_{n=0}^\infty F(n)^{\Sigma_n}
$$
satisfying
$$
F(\pi)(\xi_{n+1})=\xi_n
$$
for all $n>0$, where $\pi:\{1,...,n+1\}\to\{1,...,n\}$ is given by $1\mapsto1,...,n\mapsto n,n+1\mapsto n$.
Later:
Here is a sketch of an alternative calculation for the above $\hom$, using the fact that $P\cong i_!(1)$, where $i:\mathsf{Finepi}\hookrightarrow\mathsf{FinSet}$ is the embedding of the subcategory with the morphisms surjections only, while $i_!$ is the left adjoint to the restriction $i^*:\mathsf{Set}^{\mathsf{FinSet}}\to\mathsf{Set}^{\mathsf{Finepi}}$.
Indeed, using the left Kan extension formula,$$i_!(1)(n)=\varinjlim\left(i/n\xrightarrow{\text{(take domain)}}\mathsf{Finepi}\xrightarrow{\text{constant $1$}}\mathsf{Set}\right);$$
now every object $f:i(m)\to n$ of $i/n$ admits a morphism to the object $\operatorname{image}(f)\hookrightarrow n$, so that we may restrict from $i/n$ to its cofinal subcategory whose objects are inclusions of subsets into $n$. From this it is easy to see that indeed $P=i_!(1)$.
Using this then, $\hom(P,F)$ $=$ $\hom(i_!(1),F)$ $=$ $\hom(1,i^*(F))$ $=$ $\varprojlim(i^*(F))$, which more or less amounts to the same expression as in the first version above.
