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.

  • $\begingroup$ Why do you write: "$\mc C(A,-)$ is contravariant" ? $\endgroup$ – 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 $2^{2 \times 3} = 2^6 \ne 2^5 = 2^2 \times 2^3$.

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    $\begingroup$ 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 $A$. $\endgroup$ – Peter LeFanu Lumsdaine Dec 7 '12 at 21:01

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