Lemma
The identity functor is a functor.
References: Definition~Functor, Definition~.
Proof
Let $f : X \to Y$ and $g : Y \to Z$.
Then
$$
\operatorname{Id}_{\mathcal{C}}(g \circ f)
=
g \circ f
=
\operatorname{Id}_{\mathcal{C}}(g) \circ \operatorname{Id}_{\mathcal{C}}(f).
$$
Also
$$
\operatorname{Id}_{\mathcal{C}}(\operatorname{id}_X)
=
\operatorname{id}_X.
$$
Thus the functor axioms hold.
Depends on
Dependency Graph
flowchart LR
classDef current fill:#6366f1,color:#fff,stroke:#4f46e5
n39e872fb["Functor"]
n297b3591["Definition"]
n76f014fe["Lemma"]:::current
n39e872fb --> n76f014fe
n297b3591 --> n76f014fe
n39e872fb --> n76f014fe
n297b3591 --> n76f014fe
click n39e872fb "../objects/39e872fb.html" "_self"
click n297b3591 "../objects/297b3591.html" "_self"