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 $g$.

enter image description here

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 $F(f_1)=F(f_2)$?

  • $\begingroup$ For 2, take $f_2=f_1$. $\endgroup$ – Lord Shark the Unknown Jun 21 at 4:34
  • $\begingroup$ Sorry, I meant distinct arrows. $\endgroup$ – user634426 Jun 21 at 4:36
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    $\begingroup$ 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. $\endgroup$ – jgon Jun 21 at 4:41
  • $\begingroup$ @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? $\endgroup$ – 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.


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