Does precomposition with opposite functor preserve colimits?

Question

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.

Answer 1

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.