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