# 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?


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.

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Why do you write: "$\mc C(A,-)$ is contravariant" ? – magma Dec 8 '12 at 10:04

It's not representable because it doesn't preserve products. (Representable functors preserve all limits.) Indeed, $2 \times 3 = 6$, but $2^{2 \times 3} = 2^6 \ne 2^5 = 2^2 \times 2^3$.
Very nice. An even simpler argument from the same basic principle: it doesn’t preserve the terminal object — $\mathcal{P}{1} \not \cong 1$, whereas $[A,1] \cong 1$ for any $A$. – Peter LeFanu Lumsdaine Dec 7 '12 at 21:01