# Understanding full and faithful functors

A functor $$F:\mathscr A\to\mathscr B$$ is faithful (resp., full) if for each $$A,A'\in Ob(\mathscr A)$$, the function $$\mathscr A(A,A')\to \mathscr B(F(A),F(A' ))\\ f\mapsto F(f)$$ is injective (resp., surjective).

In the situation of the figure below, $$F$$ is faithful if for each $$A, A'$$, and $$g$$ as shown, there is at most one dotted arrow that $$F$$ sends to $$g$$. It is full if for each such $$A, A'$$, and $$g$$, there is at least one dotted arrow that $$F$$ sends to $$g$$.

1) I don't quite understand why this explanation with the figure is the same as injectiveness. Injectiveness says that for all $$A,A'$$, if there are two arrows $$F(f):F(A)\to F(A')$$ and $$F(g):F(A)\to F(A')$$, then $$f=g$$, right? The explanation cited above includes the case when no arrow maps to $$g$$ under $$F$$, I don't see how this is accounted for in the definition of injectiveness I stated.

1') Is there an easy example when the map mentioned above is injective but there exists $$g$$ that doesn't come (via $$F$$) from any arrow in $$\mathscr A$$?

2) Is there an easy example when $$F$$ is faithful and there are distinct arrows $$f_1,f_2$$ in $$\mathscr A$$ with $$F(f_1)=F(f_2)$$?

• For 2, take $f_2=f_1$. – Lord Shark the Unknown Jun 21 at 4:34
• Sorry, I meant distinct arrows. – user634426 Jun 21 at 4:36
• For the current version of 2, take two different objects each with only the identity arrow and send both to a single object with a single identity arrow. The two distinct identity arrows then map to the single identity arrow, but the functor is faithful. – jgon Jun 21 at 4:41
• @jgon Great example! I think one can modify it to get an example for 1': Replace the target category (which was the discrete category on 1 object) to the category with one object $c$ and one non-trivial arrow (and keep the first category -- the discrete category with 2 objects $a,b$). Use the same functor $F$ as you described. Now there is an arrow (the unique non-trivial arrow) $F(a)\to F(a)$ in the target category, and it doesn't come (by means of $F$) from any arrow in the "domain category". But the functor is still faithful. Am I right? – user634426 Jun 21 at 23:18

First: The figure is exactly the same, as your description of faithful and full. The functor $$F$$ sends the dotted arrow to $$g$$, thus, if there are two dotted arrows, sending to the same $$g$$, $$F$$ is not faithful. Analoguos for full $$F$$.
One easy exampel for a faithfull but not full functor ist the Forget-Fuctor from one category to another. More explicitly consider the fuctor $$F\colon k\text{-vector spaces} \rightarrow Sets$$, which sends every vector space onto its underlying set and every $$k$$-linear map to the map of sets. Then $$F$$ is faithful, but not full.
On the other hand you can assert to every group $$G$$ a category, consisting of one element $$X$$ and $$Hom(X,X)=G$$, than any group homomorphism gives rise to a functor of these categories. This functor is full iff the group homomorphism is onto.