Suppose we have $inc:C \rightarrow D$ a full subcategory and an adjunction $F:D\leftrightarrow C: inc$ where 


1. $C$ and $D$ are complete and cocomplete categories.
2. $F$ is a left adjoint such that $F\circ F=F$
3. $inc$ commutes with colimits.


Does $F$ commute with finite limits in general and with pullback and product in particular?



idempotent functor
In general $F$ preserves neither pullbacks nor even products. In a comment I mentioned that the "discrete graph" functor $\text{Set} \to \text{Set}^{\bullet \rightrightarrows \bullet}$ is full and faithful and has a left adjoint $\pi\_0$ which sends a graph to its set of connected components and a right adjoint which sends a graph to its set of vertices. The functor $F = \pi\_0$ preserves neither finite products nor equalizers. For products, consider the product of two copies of $0 \to 1$, which has four vertices but only one edge. For equalizers, consider the equalizer of an obvious pair of maps from $0 \to 1 \to 2$ to the "diamond" consisting of edges $S \to W \to N$ and $S \to E \to N$; the equalizing object is the discrete graph on vertices $0, 2$). 


(In an earlier version of this answer, I had mistakenly said that $\pi\_0$ preserves products, but I was conflating the case of (directed) graphs = quivers with the case of *reflexive graphs*, where the $\pi\_0$ does preserve products.) 


Similarly, let $\text{Set}^\mathbb{Z}$ denote the category of $\mathbb{Z}$-sets, i.e., sets equipped with an automorphism. The functor $\text{Set} \to \text{Set}^\mathbb{Z}$ which assigns to a set $S$ the trivial $\mathbb{Z}$-action is full and faithful, and has a right adjoint which assigns to a $\mathbb{Z}$-set $X$ its set of fixed points $X^\mathbb{Z} = \hom(1, X)$ under the action, and a left adjoint which assigns to $X$ its set of orbits = connected components $\pi\_0(X) = X \otimes\_\mathbb{Z} 1$. The functor $\pi\_0$ does not preserve products, as we easily see looking at $X = \mathbb{Z} \times \mathbb{Z}$: here $\mathbb{Z} \times \mathbb{Z}$ has infinitely many orbits under the action by the automorphism $(m, n) \mapsto (m+1, n+1)$, but $\mathbb{Z}$ has only one. 


In general, we *can* say that $F$ preserves the terminal object: if $1$ denotes the terminal in $D$, then we have a unit map $1 \to i F 1$ which in fact is an isomorphism (because any algebra for an idempotent monad such as $i F$ has this property). In that case, for any object $c$ of $C$ there is a unique map $i c \to i F 1$, but then by full faithfulness of $i$, there is a unique map $c \to F 1$. 


