Functor maps a morphism to empty morphism

Question

Let $F$ be a functor from categories $C$ to $D$. Is it allowed for the functor to map a morphism to empty map? So $f:A\rightarrow B$ in $C$ is mapped to nothing in $\operatorname{Hom}_D(F(A),F(B))$.

Answer 1

As requested in comments above, I'll expand my comment into an answer.

The definition of "functor" is as follows:

Definition Let $C$ and $D$ be categories. A functor from $C$ to $D$ consists of the following assignments:

1. For each object $X$ of $C$, an object $F(X)$ of $D$
2. For each pair $(X,Y)$ of objects in $C$, a function $F_{X,Y} : \operatorname{Hom}_C(X,Y) \to \operatorname{Hom}_D(F(X),F(Y))$

Satisfying the following conditions:

1. $F(\operatorname{id}_X) = \operatorname{id}_{F(X)}$ for all objects $X$ of $C$
2. $F_{Y,Z}(g) \circ F_{X,Y}(h) = F_{X,Z}(g \circ h)$ for all composable pairs of morphisms $(g : Y \to Z, h : X \to Y)$ in $C$

Normally the subscripts on the functions $F_{X,Y}$ are omitted, making the second condition much easier to read: $F(g) \circ F(h) = F(g \circ h)$.

In any case, by the very definition of functor, every morphism $f : X \to Y$ in $C$ must be sent to a morphism $F(f) : F(X) \to F(Y)$. In other words, it is illegal to have "$f : A \to B$ ... mapped to nothing".

However, as @Andreas Blass wisely points out in the comments, this is very different from asking if a functor may "map a morphism to empty map" – if there is a morphism $e$ in $D$ called "empty map", there is definitely a category $C$ and a functor $F : C \to D$ such that $F(f) = e$ for some morphism $f$ in $C$. It is probably a good idea to gauge your current level of ability by trying to construct such a functor!