Graphs of Categories Exercise I have completed the following introductory exercise from The Joy of Cats, which I am self-studying. There are no easily available online solution sets, but I am looking to check my work. The exercise is as follows (G(A) is the graph of the category A):

I have concluded that every one of these graphs is realizable as a category, but it seems unlikely the question would have been written this way in that case.

Am I mistaken that each of these graphs represents a category?

The categories I chose are as follows:
(1) $\{0\}\leftrightarrows\{1\}\rightarrow\{2\}$ where the arrows represent the only possible functions between these singleton sets.
(2) $\{0,1\}\leftleftarrows\{1\}\rightarrow\{2\}$ where the two left-pointing arrows are the functions $f:1\mapsto 0$ and $g:1 \mapsto 1$, and the remaining arrow is the unique function $h: \{1\} \to \{2\}$.
(3) Let the three vertices be singleton sets and the arrows be the unique maps between them.
(4) Let the vertices be the sets $A=\{0,1\}$ and $B=\{2,3\}$. Let the functions be
$f:B\to A$ sends $2\mapsto 0$ and $3\mapsto 1$.
$g:A\to B$ sends $0\mapsto 2$ and $1\mapsto 3$.
$h:B\to A$ sends $2\mapsto 1$ and $3\mapsto 0$.
Note that $f$ and $g$ are inverses.

Is each of my examples a category, and do they each correctly determine the graphs shown?

Edit:
It's clear now that (1) cannot be a category, because there is no arrow from the leftmost object to the rightmost object, but that should exist to obey composition. However, (4) is a bit less clear.

Thanks to the commenters, I have a clear solution. (2) and (3) are correct realizations of the graph as a category (though it's not necessary that we represent them as concrete categories).
For (1) call the vertices from left to right $A$, $B$, and $C$. There's no arrow from $A\to C$ but there is from $A\to B$ and $B\to C$, so this cannot be a category since it is not closed under composition.
For (4), call the vertices $X$ and $Y$ from left to right. Then $gh:Y \to Y$, and since there are no loops, $gh$ is the identity on $Y$. Similarly, $hg:X\to X$ is the identity on $X$, so $g$ and $h$ are inverses. But that means $g$ and $f$ must be inverses for the same reason. Thus $f=h$ by uniqueness of inverses, which is a contradiction. Thus (4) cannot be the graph a category.
