Maybe it is a trivial question, but I'm having a hard time in truly understanding this answer, so I believe I need to clarify to myself some probably obvious things first.
Given this description
- it contains two objects
- $a$
- $b$
- it contains 3 morphisms
- $id_a: a \to a$
- $id_b: b \to b$
- $f: a \to b$
Can it represent a category?
I can see the following
- it has objects
- it has morphisms
- it has identity morphisms for each object
- composition of morphims are morphisms
So I'd say it is.
Note that composition is part of the data of a category. So in this case we are given a graph consisting of two vertices $A,B$ and three conveniently labeled directed edges $1_A: A \rightarrow A$, $f:A\rightarrow B$, $1_B:B\rightarrow B$. We define composition to be that $$\begin{align*} 1_A \circ 1_A &= 1_A\\ f \circ 1_A &= f\\ 1_B \circ f &= f\\ 1_B \circ 1_B &= 1_B \end{align*}$$ From here on it is straightforward to check that this data indeed satisfies all the axioms of a category (associativity and identity).
This category is important, because it is the free category with an arrow in the following sense. Given any category $\mathcal{C}$ an arrow $\widetilde{f}: \widetilde{A} \rightarrow \widetilde{B}$ in $\mathcal{C}$ induces a unique functor $$\begin{array}{rcl} \{A \xrightarrow{f} B\} & \longrightarrow & \mathcal C\\ A & \longmapsto & \widetilde A\\ B & \longmapsto & \widetilde B\\ f & \longmapsto & \widetilde f \end{array}$$ Conversely any functor $\{A \xrightarrow{f} B\} \longrightarrow \mathcal{C}$ defines a unique morphism $\widetilde f: \widetilde A \rightarrow \widetilde B$ in $\mathcal{C}$ as the image of $f$. These assignments are mutually inverse, so we can deduce that there is a canonical bijection $$\operatorname{Fun}(\{A\xrightarrow{f}B\},\mathcal{C}) \cong \operatorname{Mor}(\mathcal{C})$$ This makes $\operatorname{Mor}: \mathsf{Cat} \rightarrow \mathsf{Set}$ a representable functor, a concept, which you will either already know or learn soon and which is of utter most importance for category theory.