Leinster Basic Category theory gives two equivalent definitions of representable functors:

$X: \mathcal{A} \to Set$ is representable if $X \cong H^A = Hom(A,-)$

$X: \mathcal{A}^{op} \to Set$ is representable if $X \cong H_A = Hom(-,A)$

where $A \in \mathcal{A}$.

I don't understand why one definition can be derived from the other. More precisely, the text says:

We now define representability for contravariant set-valued functors. Strictly speaking, this is unnecessary, as a contravariant functor on $\mathcal{A}$ is a covariant functor on $A^{op}$, and we already know what it means for a covariant set-valued functor to be representable. But it is useful to have a direct definition.

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    $\begingroup$ These are not definitions for the same thing. We call functors of either form representable. (Actually, I think its more common to call functors of the form $$\mathrm{Hom}(A,-)$$ corepresentable.) $\endgroup$ – Brian Shin Oct 17 '19 at 12:50
  • $\begingroup$ One is covariant representable, the other contravariant. $\endgroup$ – Randall Oct 17 '19 at 12:52

Note that $\mathcal A$ is really the same thing as $(\mathcal A^{\mathrm{op}})^{\mathrm{op}}$. So, if we apply the second definition to a functor $X : \mathcal A = (\mathcal A^{\mathrm{op}})^{\mathrm{op}} \to \mathsf{Set}$, we get that we call it representable if $$ X \cong \mathrm{Hom}_{\mathcal A^{\mathrm{op}}}(-, A) \cong \mathrm{Hom}_{\mathcal A}(A, -). $$ However, I think this perspective is a lot less useful than the perspective in the comments, that these are really just two separate definitions.

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  • $\begingroup$ On the other hand, your answer addresses my concern. However, the last isomorphism does not convince me. An isomorphism between functors should be in particular a natural transformation right? But my definition of natural transformation requires both functors to have the same domain and codomain while here the first has domain $\mathcal{A}^{op}$ and the second $\mathcal{A}$ $\endgroup$ – Rodrigo Oct 25 '19 at 9:13
  • $\begingroup$ @Javier, no, they both have domain $\mathcal A$. Alternatively, you could say that the first one has domain $(\mathcal A^{\mathrm{op}})^{\mathrm{op}}$, but in the set-up that I am used to these are literally the same thing. Specifically, if $f: X \to Y$ is a morphism in $\mathcal A$ (or equivalently, in the double opposite), then $\mathrm{Hom}_{\mathcal A^{\mathrm{op}}}(f, A)$ takes an element $g \in \mathrm{Hom}_{\mathcal A^{\mathrm{op}}}(X, A)$ -- which is to say, $g \in \mathrm{Hom}_{\mathcal A}(A, X)$ -- and produces $f \circ g \in \mathrm{Hom}_{\mathcal A}(A, Y)$, that is to say (cont.) $\endgroup$ – Mees de Vries Oct 25 '19 at 9:21
  • $\begingroup$ $f \circ g \in \mathrm{Hom}_{\mathcal A^{\mathrm{op}}}(Y, A)$. Note the covariance of $\mathrm{Hom}_{\mathcal A^{\mathrm{op}}}(f, A)$! $\endgroup$ – Mees de Vries Oct 25 '19 at 9:21

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