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I'm trying to prove that the free cocompletion of a small category $\mathcal{C}$ gives an equivalence of categories $$Cat_+[\widehat{\mathcal{C}},\mathcal{D}] \longrightarrow Cat[\mathcal{C},\mathcal{D}]$$ by precomposition with the Yoneda embedding $h$ (take $\widehat{\mathcal{C}}$ as the presheaf category on $\mathcal{C}$, and $Cat_+[\widehat{\mathcal{C}},\mathcal{D}]$ means the category of cocontinuous functors and natural transformations between them). For example, see https://qchu.wordpress.com/2014/04/01/the-free-cocompletion-i/. What isn't clear to me is that this functor is full. I see that given a natural transformation $$\varepsilon : Fh \Rightarrow Gh : \mathcal{C} \rightarrow \mathcal{D}$$ there is a unique way to define the components of a potential natural transformation $$\alpha : F \Rightarrow G : \widehat{\mathcal{C}} \rightarrow \mathcal{D}$$ such that $\alpha h = \varepsilon$, but why must $\alpha$ be natural?

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A teacher of mine gave me this answer.

Define for each presheaf $P$, the morphism $\alpha_P : FP \rightarrow GP$ in the unique possible way to make the naturality square of $\alpha$ commutative for all $\lambda_{C,p}$ (these $\lambda_{C,p} : [-,C] \rightarrow P$ form the colimiting cone that establishes $P$ as a colimit of representable functors). Then the naturality square of $\alpha$ commutes for any arrow $[-,C] \rightarrow P$ because this arrow is $\lambda_{C,p}$ if $p \in PC$ is obtained by the Yoneda bijection.

Now fix any $f : P \Rightarrow Q$. To see that $\alpha_Q Ff = Gf \alpha_P$, we can precompose with the colimiting cone $F\lambda_{C,p} : F[-,C] \rightarrow FP$ (F is cocontinuous), so it remains to be seen that $\alpha_Q F(f\lambda_{C,p}) = G(f\lambda_{C,p}) \varepsilon_C$, but this holds because of naturality of $\alpha$ with respect to any arrow $[-,C] \rightarrow P$.

This works for any dense subcategory.

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