# Is the category of small graphs a concrete category?

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?

• Yes, it's correct. This is basically what has to be done, I don't see any simpler way.. – Berci Feb 21 at 15:12