Identity morphism in an opposite category

Question

This is most likely a silly question, but it's probably because I'm not very used to thinking about opposite categories, but how do I show that for every object $C$ of a category $\mathscr{C}$ one has $(\text{id}_C)^{\text{op}} = \text{id}_C$?

Answer 1

Fix $c \in C^{op}$. By definition, it is $c \in C$. Now, since $id_c : c \to c$ is an arrow in $C(c,c)$, once again by definition it has to be an arrow in $C^{op}(c,c)$. If $f \in C^{op}(c,d) = C(d,c)$, then

$$ f \circ_{op} id_c \stackrel{(\text{def})}{=} id_cf = f. $$

Similarly we obtain $id_c \circ_{op} g = g$ for any $g \in C^{op}(d,c)$.