Image of a Functor

The naive image of a functor need not be a category.

Let $S=0,1,2,3$ Let $A=\{(0,0),(1,1),(2,2),(3,3),(0,1),(2,3)\}$

$$\begin{array}{rl}F:& 0\mapsto 0\\ & 1\mapsto 0\\ & 2\mapsto 0\\ & 3\mapsto 0\\ & (0,1)\mapsto A\\ & (2,3)\mapsto B\end{array}$$

Say composition in the destination category is defined as, then $C$ need not be in the image of $F$

$$A\circ B=C$$