Is an exact functor to FSet faithful?

Question

I'm wondering if the following argument is correct. Let $\mathcal C$ be a category with finite limits and let $F: \mathcal C \to \textsf{FSet}$ be left exact, i.e. $F$ preserves finite limits. Then $F$ is faithful.

Proof: let $\phi, \psi : X \to Y$ be morphisms in $\mathcal C$ such that $F(\phi) = F(\psi)$. Then the equaliser of these latter two morphisms is $(F(X),\text{id})$. By exactness, the equaliser of $\phi$ and $\psi$ in $\mathcal C$ is $(X,\text{id})$, so $\phi = \psi$. $\square$

I just wonder if the exactness suffices? Surely $(X,\text{id})$ maps to the equaliser of $F(\phi)$ and $F(\psi)$ in $\textsf{FSet}$, and equalisers are unique up to unique isomorphism, but I can't mold this into an argument that says that $(X,\text{id})$ really is the equaliser.

Even if we additionally have that $F$ is conservative, I can't make it work - if we denote the equaliser in $\mathcal C$ by $E$, we have $F(E) \cong F(X)$. We'd like to deduce that then $E \cong X$, but I don't see why this isomorphism should come from a morphism in $\mathcal C$. The universal property of $E$ only induces one if $X$ equalises $\phi$ and $\psi$, which is what we're trying to prove. [Edit: Actually this works - in this case the equalising morphism $F(E) \to F(X)$ coming from $E \to X$ is an isomorphism, so if we have conservativity, also $E \to X$ is an isomorphism. So the question is: do we need conservativity?]

Any help is appreciated!

Answer 1

No, a left exact functor to the category of finite sets need not be faithful. As a trivial example, the functor $\mathcal{C} \to \mathrm{FinSet}$ sending everything to a singleton is left exact, but not faithful unless $\mathcal{C}$ is equivalent to a meet-semilattice. For a less trivial example, the functor $\mathrm{FinCat} \to \mathrm{FinSet}$ that sends a finite category to its set of objects is also left exact, but not faithful.

Of course, neither of the above examples is conservative either. An example of a faithful non-conservative left exact functor to the category of finite sets is the forgetful functor from the category of finite topological spaces.

Answer 2

Yes, conservativity is needed for this result. The problem with your original argument is that it tries to conclude from the fact that $F(\mathrm{id}_X)$ is an equalizer of $F(\phi)$ and $F(\psi)$ that $\mathrm{id}_X$ is an equalizer of $\phi$ and $\psi$. But this reasoning is backwards: you are entitled to use the converse reasoning, so that if $f:E\to X$ is an equalizer of $\phi$ and $\psi$ then $F(f)$ is an equalizer of $F(\phi)$ and $F(\psi)$. So this shows that $F(f)$ is an isomorphism, where your argument about conservativity comes in.

Note that no use is made of a particular codomain here: any conservative functor between categories with equalizers which preserves equalizers is faithful.