In a small category $\cal{A}$, if a functor $F$ in $[\cal{A}^\text{op},\mathbf{Set}]$ is a retract of a representable functor (i.e., a functor isomorphic to the hom-functor), and $\cal{A}$ considered as a full subcategory of $[\cal{A}^\text{op}, \mathbf{Set}]$ under the Yoneda embedding, has split idempotents, then $F$ is isomorphic to a representable functor. Why? The assumptions mean that there are two natural transformations $u$ and $v$ with $u\circ v=\text{id}_F$ and $v\circ u:\cal{A}(-,a) \longrightarrow \cal{A}(-,a)$.
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1$\begingroup$ What have you tried? Remember that splittings induce idempotents (in your notation $v \circ u$ is idempotent). $\endgroup$– Martin BrandenburgCommented Jan 24, 2014 at 19:37
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$\begingroup$ vou is obviously an idempotent, but we would need vou=id_A(-,a) (which is not true) to conclude that F is isomorphic to A(-,a).Then u and v would be inverse isomorphisms. $\endgroup$– user122424Commented Jan 24, 2014 at 19:49
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1$\begingroup$ It isn't an isomorphism. Go use the hypothesis of idempotents splitting. $\endgroup$– Zhen LinCommented Jan 24, 2014 at 20:25
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1 Answer
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So, $v\circ u:\mathcal A(-,a)\to \mathcal A(-,a)$ is idempotent. By the Yoneda lemma, it corresponds to an arrow $a\to a$ (namely, to $\varepsilon:=(v\circ u)_a\left(1_a\right)$). Check that $\varepsilon$ is idempotent, too.
Then, by the condition that idempotents split in $\mathcal A$, there is an object $e$ with arrows $\varphi:a\to e,\,\psi:e\to a\,$ such that $\,\psi\circ\varphi=\varepsilon\,$ and $\, \varphi\circ\psi=1_e$.
Again, by the Yoneda lemma, these imply that $\mathcal A(-,e)$ is also a retract of $\mathcal A(-,a)$, in the functor category, that corresponds to the idempotent $v\circ u$. Since, such must be unique up to isomorphism in any category, we get that the object $e$ represents $F$.
