Let $F: \mathcal{C} \to \mathcal{D}$ be an equivalence of categories where $\mathcal{C}$ is a complete category, i.e. any functor $F: I \to \mathcal{C}$ with $I$ small admits a limit.
I'm trying to understand a proof that claims that $\mathcal{D}$ is complete as well.
The proof uses the following result:
Suppose $F: \mathcal{C} \to \mathcal{D}, G : \mathcal{D} \to \mathcal{C}$ are functors such that $F$ is a left adjoint of $G$. Then $F$ preserves small colimits while $G$ preserves small limits.
The proof I read goes as follows:
The functor $F$ admits both a left and a right adjoint (this is proven earlier). From the above proposition the result immediately follows.
I'm not sure how exactly this follows. I tried to argue in the following way:
Let $H: I \to \mathcal{D}$ be a functor with $I$ a small category. We show $H$ admits a limit.
Consider a right adjoint $G$ for $F$. Then $G$ preserves limits. Maybe I can use that?
Thanks in advance for any help!
Let $G$ be an inverse equivalence. Let $H:I\to \mathcal{D}$ be a diagram. Then $GH : I\to \mathcal{C}$ admits a limit, $L$, and $FGH \cong H$ has limit $FL$ because $F$ admits a left adjoint, and therefore preserves limits.