Why isn't the covariant powerset functor representable?

It's not representable because it doesn't preserve products. (Representable functors preserve all limits.) Indeed, $2 \times 3 = 6$, but $2^{2 \times 3} = 2^6 \ne 2^5 = 2^2 \times 2^3$.

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.