Are Fully Faithful Functors Injective?

Question

Suppose $F : C \rightarrow D$ is a full and faithful functor from $C$ to $D$.

Clearly, this does not imply that $F : \text{Ob}(C) \rightarrow \text{Ob}(D)$ is surjective.

Question: But does this imply that $F : \text{Ob}(C) \rightarrow \text{Ob}(D)$ is injective?

Answer 1

The answer is no. A minimal example would be the following: let $D$ be the category with only one object $X$, and only one morphism (the identity of $X$). Now, let $C$ be the category with two objects $Y$ and $Z$, and only four morphisms: the identity of $Y$, that of $Z$, a morphism $f:Y\to Z$ and a morphism $g:Z\to Y$, such that $f$ and $g$ are inverse to each other.

Define a functor $F:C\to D$ by $F(Y)=F(Z) = X$, and all morphisms are sent by $F$ to the identity of $X$. Then $F$ is fully faithful (it is even an equivalence of categories), but is not injective on the sets of objects.

Answer 2

Generalizing Pierre-Guy's answer, as a rule of thumb every "categorical" property is invariant with respect to isomorphism of objects (or more generally/precisely is preserved by equivalence of categories, or, even more generally, higher equivalences). "Categorical" can be but usually isn't formalized. Since "injective-on-objects" requires equality of objects, it is not a "categorical" property and thus it is often called "evil". A functor being fully faithful is a categorical property though, and thus does allow identifying of isomorphic objects. For our purposes, this means a fully faithful functor can collapse together isomorphic objects in general, which is to say if $X\neq Y$ but $X \cong Y$ then $FX = FY$ cannot break the property of $F$ being fully faithful. (Another way of saying this is that we can identify isomorphic objects.) Sure enough, it's trivial to show that identifying objects in this way always coherent. By assumption $$\text{Hom}(X,Z) \cong \text{Hom}(FX,FZ) = \text{Hom}(FY,FZ)$$ but $\text{Hom}(X,Z) \cong \text{Hom}(Y,Z)$ by Yoneda and the fact that $X\cong Y$. Therefore, $\text{Hom}(Y,Z)\cong\text{Hom}(FY,FZ)$ too.

Answer 3

An important feature of fully faithful functors is that they are conservative: if $FX\xrightarrow{Ff}FY$ is an isomorphism, then $X\xrightarrow{f}Y$ is an isomorphism; this is because fullness implies there is a moprhism $X\xleftarrow{g}Y$ such that $FX\xleftarrow{Fg}FY$ is the inverse of $FX\xrightarrow{Ff}FY$, and faithfulness guarantees that $F(g\circ f)=Fg\circ Ff=\mathrm{id}$ implies $g\circ f=\mathrm{id}$, and similarly $f\circ g=\mathrm{id}$.

Any full conservative functor then has the property that $FX_1=FX_2$ implies $X_1\cong X_2$ since by fullness there must be a morphism $X_1\xrightarrow{f}X_2$ so that $Ff=\mathrm{id}$, and by conservation $f$ must be an isomorphism.

In particular, fully faithful functors are always injective when the domain category is skeletal, i.e. when all isomorphisms are identities, the most frequently seen cases of skeletal categories are thin categories (at most one morphisms between any two objects) such as discrete categories (only morphisms are identities) and posets (small thin categories).