Can it happen that the image of a functor is not a category? On Hilton and Stammbach's homological algebra book, end of chap. 2, they wrote in general $F(\mathfrak{C})$ is not a category at all in general. But I don't quite get it. I checked the axioms of a category for the image, and I think they are all satisfied. Am I missing something? Thanks.
 A: Consider the category $C$ with four objects, $a,b,c,d$ and, other than identity arrows, a single arrow $a\to b$ and a single arrow $c\to d$. Now consider the category $D$ with three objects $x,y,z$, and, aside from identity arrows, the arrows $x\to y$, $y\to z$, and $x\to z$. Now, consider the functor $F:C\to D$ with $F(a)=x$, $F(b)=F(c)=y$, and $F(d)=z$ (extended uniquely to arrows). Its image is not a category. 
This business is related to the fact that epis in $Cat$ are not so simple at all. In work of Isbell epis in $Cat$ are characterized. It's worth noting that regular epis, split epis, etc. in $Cat$ are quite different, attesting again to the subtlety of epis.
A: Drawing a picture of Ittay Weiss's answer:
\begin{align*}
    a \underset{f}{\longrightarrow}\, &b
    \\                &c \underset{g}{\longrightarrow} d
    \\
    \\ &\Big{\downarrow}F
    \\
    \\
    x \underset{F(f)}{\longrightarrow}\,&y \underset{F(g)}{\longrightarrow} z
\end{align*}
Here, the objects of the image are $x$, $y$, and $z$,
and the arrows are $F(f)$ and $F(g)$. Categories are closed under composition of arrows, but $F(g)F(f)$ is not in the image of $F$ (there are no arrows $a \to d$), so the image of $F$ cannot be a category.
