# Any function preserving morphism typing between two thin categories is a full functor?

Let $$F$$ is a function which maps objects and morphisms from one thin category $$\mathcal C$$ to another thin category $$\mathcal D$$. If the function preserve morphisms typing ($$F$$ associates to each morphism $$f: X \to Y$$ in $$\mathcal C$$ a morphism $$F(f): F(X)\to F(Y)$$ in $$\mathcal D$$), then

1. $$F$$ also preserve identity morphisms and morphism composition and thus the function is a functor.
2. $$F$$ is a full functor, because there is at most one morphism in each target Hom-set $$\operatorname{Hom}_D(F(X),F(Y))$$. And each of these morphisms is mapped from some source morphism.

If $$F$$ is injective on objects, then

1. $$F$$ is a faithful functor, because there is at most one morphism in each source Hom-set. There is no distinct morphisms, which could be mapped to the same target morphism.

Are this statements right?

• I added a definition to the question. Commented Oct 14, 2018 at 7:49

## 1 Answer

Your statements 1 and 3 are correct, but note that you don't even need injectivity on objects to prove that your functor is faithful.

Your statement 2 is incorrect, however. For example, you could take $$\mathcal{C}$$ and $$\mathcal{D}$$ to be discrete categories, then if $$X\neq Y$$ are objects in $$\mathcal{C}$$ such that $$F(X)=F(Y)$$, you will have a morphism in $$\mathcal{D}(F(X),F(Y))$$ but none in $$\mathcal{C}(X,Y)$$, so in that case $$F$$ is not full.

In fact if you see your thin categories as preorders $$(C,\leq_C )$$ and $$(D,\leq_D)$$, then a functor is the same thing as a preorder-preserving map, i.e. a map such that $$x\leq_C y\Rightarrow f(x)\leq_D f(y)$$and a full functor is the same thing as a pre-order-preserving and preorder-reflecting map, i.e. a map $$f$$ such that $$x\leq_C y\Leftrightarrow f(x)\leq_D f(y).$$

• Could you please clarify the following point? Does injectivity of $F$ on objects implies that $F$ is a full functor? Commented Oct 14, 2018 at 8:47
• It does in my example with discrete categories, but not in general (perhaps I should have put another example). The simplest counterexample is to take $\mathcal{C}$ and $\mathcal{D}$ with the same two objects $X$ and $Y$, and : no arrows between them in $\mathcal{C}$; one arrow $X\to Y$ in $\mathcal{D}$. Then you have an inclusion functor $\mathcal{C}\to \mathcal{D}$ which is injective (the identity in fact) on objects but not full. Commented Oct 14, 2018 at 9:03
• Thanks a lot! Everything is clear now. Commented Oct 14, 2018 at 9:40