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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?

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    $\begingroup$ What your concern points at is that limits and colimits are not necessarily preserved by a fully faithful functor. $\endgroup$
    – Tim Campion
    Commented Sep 1, 2020 at 17:59
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    $\begingroup$ I provide an explicit counterexample to a fully faithful functor preserving limits in my answer $\endgroup$
    – Zach Goldthorpe
    Commented Sep 1, 2020 at 18:38
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    $\begingroup$ In fact, I'd say that it's relatively rare for a fully faithful functor $F: \mathcal C \to \mathcal D$ to preserve both limits and colimits. After all, if things are nice enough so that adjoint functor theorems can be applied everywhere, and if $F$ preserves both limits and colimits, then $F$ admits both a left and a right adjoint, $\mathcal C$ is simultaneously a reflective and coreflective subcategory of $\mathcal D$. $\endgroup$
    – Tim Campion
    Commented Sep 1, 2020 at 18:56

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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.

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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 $D$.

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    $\begingroup$ The problem is that not all cones in $D$ are of the form $Fc$. $\endgroup$
    – Zhaoting Wei
    Commented Sep 1, 2020 at 18:04
  • $\begingroup$ When you're talking about reflecting limits, you only care about cones that come from $C$. What you know is that the image of $\ell$ is a limit cone in $D$, so fully faithfulness allows you to move the problem to $D$ (where you know by assumption that there is a factoring morphism) and then bring it back to $C$. $\endgroup$
    – Zach Goldthorpe
    Commented Sep 1, 2020 at 18:07
  • $\begingroup$ Reflecting limits means that "cones in $C$ that are limits in $D$ become limits in $C$" $\endgroup$
    – Zach Goldthorpe
    Commented Sep 1, 2020 at 18:09
  • $\begingroup$ Reflecting limits means that "cones in $C$ that are limits in $C$ become limits in $D$". $\endgroup$
    – Zhaoting Wei
    Commented Sep 1, 2020 at 18:21
  • $\begingroup$ Let $l$ be a limit cone in $C$. Although for any cone $c$ in $C$, there is a unique morphism $Fc\to F l$, this does not mean that for any cone $d$ in $D$ there is a unique morphism $d\to F l$ since $d$ is not necessarily of the form $Fc$. $\endgroup$
    – Zhaoting Wei
    Commented Sep 1, 2020 at 18:23

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