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In the text, "Topology: A Categorical Approach", there is an exercise generalizing the identity $\mathsf C(\coprod X_i, Y) \cong \prod \mathsf C(X_i, Y)$:

Prove that a functor $F: \mathsf B \to \mathsf C$ has a colimit if and only if for all objects $Y \in \mathsf C$ there is a natural isomorphism: $$\mathsf C(\operatorname{colim} F, Y) \cong \lim \mathsf C(F-, Y).$$

A similar statement and its dual can also be found here: https://ncatlab.org/nlab/show/hom-functor+preserves+limits.

Suppose the colimit of $F$ exists. I can understand why these are isomorphic as sets, and this much is proven in the nlab page, but I am not quite sure how to interpret "natural isomorphism" here. I think it is meant to be understood that the isomorphism is natural in $Y$. In this case, the left hand side becomes a regular hom-functor, $\mathsf C(\operatorname{colim} F, -)$, but if this is correct, then how would one interpret the right hand side? It seems likely that for objects, we have $Y \mapsto \lim \mathsf C(F-, Y)$, but I have not been able figure out how such a functor should act on morphisms. Does anyone have any insight on this?

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Let $f : Y_1 \to Y_2$ be a morphism in $\mathsf C$. Considering the object $\lim \mathsf C(F-,Y_1)$ as a constant functor, there is a natural transformation $$\lim \mathsf C(F-,Y_1) \longrightarrow \mathsf C(F-,Y_1) \stackrel{f_*}\longrightarrow \mathsf C(F-,Y_2)$$ (the first one is precisely the cone from $\lim \mathsf C(F-,Y_1)$ to $\mathsf C(F-,Y_1)$, and the second one is the natural transformation whose components are pre-composition by $f$) and so, the universal property of $\lim \mathsf C(F-,Y_2)$ tells us that there is a unique morphism $f_\flat : \lim \mathsf C(F-,Y_1) \to \lim \mathsf C(F-,Y_2)$.

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