Does a fully faithful functor always preserve limits and colimits? I read on this n-lab page that a fully faithful functor $F: C\to D$ reflects all limits and colimits by the universal property.
On the other hand, I think a fully faithful functor does not always preserve limits and colimits since a "testing object" in $D$ is not necessarily in the image of $F$.
Are there any easy counter examples?
 A: The point is that the cone starts out in $C$, then you test it in $D$.
Being a limit cone in $D$ is more than you need by virtue of what you mentioned about some test objects in $D$ not being visible to $F$.
Suppose you had a cone $\ell$ in $C$ whose image under $F$ is a limit cone in $D$. Let $c$ be another cone in $C$ for the same diagram, then $Fc$ will be a cone in $D$ and so by the universal property (of $F\ell$ in $D$) this admits a unique morphism $Fc\to F\ell$.
Now, since $F$ is fully faithful, this morphism arises as a unique morphism $c\to\ell$ in $C$, proving that $\ell$ has the universal property of being a limit cone in $C$ as well.
I suspect you might be confusing this with the property of preserving limits. If a fully faithful functor cannot see all the objects of $D$ (i.e., is not essentially surjective) then the fact that there are test objects in $D$ that $F$ cannot see will make it possible that a limit cone in $C$ will no longer be a limit cone in $D$.
For an explicit example, let $C=\{0\}$ be a one object category, and $D=\{0\to1\}$ the walking arrow category. Take $F:C\to D$ to be the inclusion sending $0\mapsto0$, then $F$ is fully faithful. However, $0$ is a limit cone for the empty diagram in $C$ (since it is the terminal object), but $F(0)=0$ is not a limit cone for the empty diagram in $D$.
A: The forgetful functor from abelian groups to groups is fully faithful, and does not preserve coproducts. For example, in abelian groups, $\mathbb Z\coprod \mathbb Z=\mathbb Z\times \mathbb Z$, but in groups $\mathbb Z\coprod \mathbb Z$ is the free group on two generators.
