Why isn't the covariant powerset functor representable? I put my background and a worked out a related example, but it may not be necessary to read that.
My question is: Why isn't the covariant powerset functor representable?

$\newcommand{\mc}{\mathcal}\newcommand{\ms}{\mathscr}$We can curry the Hom-set bifunctor $$\mc C(-,-) : \mc C^{\text{op}}\times\mc C \to \text{Set}$$ in two different ways to get Yoneda functors $$Y : \mc C^{\text{op}}\to[\mc C,\text{Set}]$$ and  $$Y' : \mc C^{}\to[\mc C^{\text{op}},\text{Set}].$$ The Yoneda lemma states that there is a natural isomorphism between natural transformations $\mc C(A,-) \to F$ and elements $F(A)$. A functor $F$ is representable if it is naturally isomorphic to some $\mc C(A,-)$ and when it is we say it is represented by a pair containing $A$ and the isomorphism. This definition only applies to covariant functors because $\mc C(A,-)$ is covariant. By duality we can define representability of a contravariant functor. I don't see the connection from Yoneda to representable functors, is it just motivation for the idea?

Example: The contravariant powerset functor $\ms P : \text{Set} \to \text{Set}$ takes objects to their powersets and morphisms to their inverse images. Since for a given set $X$ its powerset is bijective with the maps $X \to \{0,1\}$ (think of $f(x) = 1$ meaning $x$ is in the set, and $0$ meaning it's not) and in fact there is a natural isomorphism between $\ms P$ and $\text{Set}(-,\{0,1\})$ because given $g : X \to Y$ we have $\text{Set}(g,\{0,1\}) : \text{Set}(Y,\{0,1\}) \to \text{Set}(X,\{0,1\})$ which maps subsets of $Y$ to their inverse images.
 A: Just to make it clear, the covarinat power set functor $\mathcal{P}:\text{Set}\rightarrow \text{Set}$ is defined as

*

*$\mathcal{P}(X)$  is the power set of $X$.

*for a set map $X\rightarrow Y$ the map $\mathcal{P}(X)\rightarrow \mathcal{P}(Y)$ is given by the image; that is, $S\mapsto f(S)$.

Where as, the  contravariant power set functor $\mathcal{P}:Set\rightarrow \text{Set}$ is defined as

*

*$\mathcal{P}(X)$  is the power set of $X$.

*for a set map $X\rightarrow Y$ the map $\mathcal{P}(Y)\rightarrow \mathcal{P}(X)$ is given by the inverse image; that is, $S\mapsto f^{-1}(S)$.

Though contravariant power set functor is representable, the covariant power set functor is not representable.
Almost same idea as what Peter LeFanu Lumsdaine said in comments to  Zhen Lin 's answer.
Suppose $X$ is a set that represents the covariant functor $\mathcal{P}:\text{Set}\rightarrow \text{Set}$; then, among other things, we need to have $|\mathcal{P}(Y)|=|\text{Maps}(X,Y)|$ for all $Y\in \text{Set}$.
Let $Y$ be a singleton set. Then, $|\text{Maps}(X,Y)|=1$ (as there can be only one functor from any ($\neq \emptyset$) set to a singleton set) but $|\mathcal{P}(Y)|=2$. Thus $|\mathcal{P}(Y)|=|\text{Maps}(X,Y)|$. So, there can be no object $X$ in $\text{Set}$ that represents the covariant power set functor $\mathcal{P}$.
Thus, the covariant power set functor is not representable.
A: It's not representable because it doesn't preserve products. (Representable functors preserve all limits.) Indeed, $2 \times 3 = 6$, but $2^{2 \times 3} = 2^6 \ne 2^5 = 2^2 \times 2^3$.
