Why do we take opposite category when stating the Yoneda Lemma?

Question

For any function $F: C^{op} \to \text{Set}$ and any object $X$ in $C$, natural transformations $\text{Hom}(-,X) \to F$ are in bijection with the elements in the set $F(x)$. That is,

$\text{Nat} (\text{hom}(-,X),F ) \cong F(X)$

Why is that we consider the opposite category in the statement of the lemma?

Answer 1

If you replace $C$ with $C^{\text{op}}$ in the given statement, the fact that $(C^{\text{op}})^{\text{op}}=C$ gives you the Yoneda lemma for functors $C\to\text{Set}$ instead of $C^{\text{op}}\to\text{Set}$.

For any functor $C\to\text{Set}$, natural transformations $\text{Hom}(X,-)\to F$ are in bijection with the elements of the set $F(x)$.

One change is that the relevant "Hom" functors are now $\text{Hom}(X,-)$ instead of $\text{Hom}(-,X)$. This switch happens because morphisms in the opposite category go the opposite direction. This is discusssed on pages 9 and 10 of your text.

The formulation for functors $C^{\text{op}}\to\text{Set}$ is preferred because it defines the Yoneda embedding: replacing objects and morphisms of $C$ with functors $\text{Hom}(-,X)$ and natural transformations between them gives a fully faithful functor from $C$ into the category of functors from $C^\text{op}\to\text{Set}$. This is pages 11-13 of your text at https://topology.mitpress.mit.edu/.