Why is the post-composing map of f dual to the pre-composing map of f? I was reading in Riehl's book "Category Theory in Context (2016)" and I couldnot understand the proof of Lemma 1.2.3 on page 11
Let $C$ be a category, $f:x \rightarrow y$ is a morphism in $C$, and $c \in C$ is an object in $C$. We can define the post-composition map of $f$:
$$f_{\ast} : C(c,x) \rightarrow C(c,y) : \alpha \mapsto f \circ \alpha$$
and the pre-composition map of $f$ :
$$f^{\ast} : C(y,c) \rightarrow C(x,c) : \alpha \mapsto \alpha \circ f$$
We can also define the post-composition map of the dual $f^{op} : y \rightarrow x$ in $C^{op}$ :
$$(f^{op})_{\ast} : C^{op}(c,y) \rightarrow C^{op}(c,x) : \alpha^{op} \mapsto f^{op} \circ \alpha^{op} = (\alpha \circ f)^{op}$$
Now, Riehl claims that $(f^{op})_{\ast} = f^{\ast}$, because $C^{op}(c,x) = C(x,c)$ and $C^{op}(c,y) = C(y,c)$, and "post-composition with $f^{op}$ translates to pre-composition with $f$ in the opposite category $C$." I do not understand this. Can anyone clarify this to me and explain why $(f^{op})_{\ast} = f^{\ast}$ ?
I have a second question related to this. We also have the dual of the post-composing map $f_{\ast}$:
$$(f_{\ast})^{op} : C^{op}(c,y) \rightarrow C^{op}(c,x) : \alpha^{op} \mapsto ???$$
What is $(f_{\ast})^{op}(\alpha^{op})$ and can we conclude that $(f_{\ast})^{op} = (f^{op})_{\ast}$ ?
I hope anyone can help me with these questions and can give me a detailed answer. Thank you.
 A: $-^{op} : C \to C^{op}$ can be applied on objects and maps. It is the identity on objects and flips the direction of maps. It also reverses composition $(f \circ g)^{op} = g^{op} \circ f^{op}$
The proof first shows that (i) $f : x \to y \in C$ is an isomorphism and (ii) for all $c \in C$, $f_* : C(c,x) \to C(c,y)$ is a bijection are equivalent.
We can apply this fact with the category $C^{op}$ and the map $f^{op} : y \to x \in C^{op}$ to get that:

*

*(a) $f^{op} : y \to x \in C^{op}$ is an isomorphism.

*(b) for all $c \in C^{op}$, ${f^{op}}_* : C^{op}(c,y) \to C^{op}(c,x)$ is a bijection.

are equivalent.
What we can now do is expand out the opposites to understand what these two statements say in terms of the non-opposite category $C$.
(a) is equivalent to $f : x \to y \in C$.
To see what (b) is equivalent to, let's expand out ${f^{op}}_*$.
Given $\alpha^{op} : c \to y \in C^{op}$

*

*${f^{op}}_*(\alpha^{op}) : c \to x \in C^{op}$

*${f^{op}}_*(\alpha^{op}) = f^{op} \circ \alpha$
This is the same as taking $\alpha : y \to c \in C$ and we can apply the opposite to the whole term:

*

*$({f^{op}}_*(\alpha^{op}))^{op} : x \to c \in C$

*$({f^{op}}_*(\alpha^{op}))^{op} = (f^{op} \circ \alpha^{op})^{op} = {\alpha^{op}}^{op} \circ {f^{op}}^{op} = \alpha \circ f = f^*(\alpha)$
Which is pre-composition.
So we have seen that ${f^{op}}_* : C^{op}(c,y) \to C^{op}(c,x)$ being a bijection is exactly the same as ${f}^* : C(y,c) \to C(x,c)$ being a bijection.
So applying the fact that (i) <=> (ii) with a carefully chosen map from the opposite category ended up proving exactly the fact that (i) <=> (iii) in the non-opposite category.
