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
- $F$ also preserve identity morphisms and morphism composition and thus the function is a functor.
- $F$ 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
- $F$ 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?
Best 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 $$x\leq_C y\Leftrightarrow f(x)\leq_D f(y).$$