# Does precomposition with opposite functor preserve colimits?

For $$F: \mathcal{C} \to \mathcal{D}$$ we can define a functor $$- \circ F^{op}: Set^{\mathcal{D}^{op}} \to Set^{\mathcal{C}^{op}}$$ which sends an object $$X$$ to $$X \circ F^{op}$$ and natural transformation $$(\mu_D)_{D \in \mathcal{D}}$$ to $$(\mu_{FC})_{C \in \mathcal{C}}$$.

Does this functor preserve colimits? Certainly, if $$(X, \mu)$$ is a limiting cone for $$G: \mathcal{I} \to Set^{\mathcal{D}^{op}}$$, we have that $$(X \circ F^{op}, \mu F)$$ is a cone for $$(- \circ F^{op}) \circ G$$.

Normally, we would assume a cone for $$(- \circ F^{op}) \circ G$$, then turn it into a cone of $$G$$ and do something with the unique morphism for $$G$$ to get a unique morphism for $$(- \circ F^{op}) \circ G$$, but I do not see how to do that here.

• Hint : in a presheaf category, colimits are computed pointwise Nov 20, 2019 at 20:57

Consider where $$i \to j$$ is an arrow in $$\mathcal I$$ and the double arrows indicate that they are natural transformations. By functoriality, this is certainly a cone $$(X \circ F^{op}, \mu F^{op})$$ under $$(- \circ F^{op}) \circ G$$.

Then for each $$c \in \mathcal C$$, we have the following commutative diagram by dissembling the above cone.

Now since $$F^{op}$$ is a functor, we may write $$F^{op}c = d$$ for $$d \in \mathcal D$$. Thus we have the following commutative diagram Since $$(X, \mu)$$ is a colimit cone under $$G$$, there is a unique map $$Xd \to Zd$$ for any cone with nadir $$Z$$ under $$G$$. Assembling these maps for each $$F^{op}c = d$$, we obtain the unique morphism for $$(- \circ F^{op}) \circ G$$ as desired.