Let $\mathbf{Grf}$ be the category whose objects are small graphs and whose arrows are graph homomorphisms.
A small graph is tuple $G = \langle V, E, \operatorname{src}, \operatorname{trg}\rangle$ where $V$ and $E$ are sets, and $\operatorname{src},\operatorname{trg}\colon E \to V$ are functions.
Let $G_1 = \langle V_1, E_1, \operatorname{src}_1, \operatorname{trg}_1\rangle$ and $G_2 = \langle V_2, E_2, \operatorname{src}_2, \operatorname{trg}_2\rangle$ be two graphs. A graph homomorphism $h\colon G_1 \to G_2$ is a pair of functions $h_0\colon V_1 \to V_2$ and $h_1\colon E_1 \to E_2$ such that
$$ \operatorname{src}_2 \circ h_1 = h_0 \circ \operatorname{src}_1 \qquad\text{and}\qquad\operatorname{trg}_2 \circ h_1 = h_0 \circ \operatorname{trg}_1 $$
Is $\mathbf{Grf}$ a concrete category?
I would say yes,and here it is my solution: Let $U\colon\mathbf{Grf}\to\mathbf{Set}$ be the functor defined by
$$ UG = V\uplus E = (V \times \{0\}) \cup (E \times \{1\}) $$ and
$$ (Uh)(x,i) = \begin{cases} h_0(x) & \text{if $i = 0$}\\ h_1(x) & \text{if $i = 1$} \end{cases} $$
The functor $U$ is faithful: Let $f,g\colon G_1 \to G_2$ be two graph homomorphisms. If $Uf = Ug$, then $(Uf)(v, 0) = f_0(v) = g_0(v) = (Ug)(v, 0)$ for each $v \in V$ and $(Uf)(e, 1) = f_1(e) = g_1(e) = (Ug)(e, 1)$ for each $e \in E$, and therefore $f = g$.
Is this solution correct?
If yes, there are other simpler solutions?
