What are the objects of Hom$(y,x)$ in the contra-variant Yoneda functor

Question

I am reading about the contra-variant Yoneda functor and I am bit confused about what objects actually are in the Hom set.

More specifically, if $Hom(-,x):\mathcal{C}^{op}\rightarrow Set$ is the contra-variant Yoneda functor, then are the elements of $Hom(-,x)(y) = Hom(y,x)$ morphisms in $\mathcal{C}$ between $y$ and $x$? Or are they morphisms in $\mathcal{C}^{op}$ from $y$ to $x$, which then would correspond to morphism $x$ to $y$ in $\mathcal{C}$?

Answer 1

The confusion comes from the notation $C^{op}$, which is often used only to indicate that the functor is contravariant.

The covariant Yoneda lemma uses the covariant functor $Hom(x,-):C\to Set$, while the contravariant Yoneda lemma uses the contravariant functor $Hom(-,x):C\to Set$. This contravariant functor is equivalent to the covariant functor $Hom(x,-):C^{op}\to Set$ if we want to be strict with the notation, but as I said, usually one only means by $C^{op}\to Set$ that the functor is contravariant.

To sum up, $Hom(-,x)(y)=Hom(y,x)$ is the set of morphism $y\to x$ in $C$, which is the same as morphisms $x\to y$ in $C^{op}$.

Answer 2

Indeed, the elements of $Hom(-,x)(y)=Hom(y,x)$ are the morphisms from $y$ to $x$ in $\mathcal{C}$. For this reason, this functor is often written $\mathcal{C}(-,x)$ as well.

The reason why $\mathcal{C}^{op}$ gets involved is that this functor is contravariant. Given a morphism $f:y \rightarrow z$ in $\mathcal{C}$, the only natural way to understand $\mathcal{C}(-,x)(f)$ is as a function $\mathcal{C}(z,x) \rightarrow \mathcal{C}(y,x)$ given by composing i.e. $g:z \rightarrow x$ is sent to $g \circ f: y \rightarrow z \rightarrow x$.

This makes the functor $\mathcal{C}(-,x)$ contravariant on $\mathcal{C}$, or covariant on $\mathcal{C}^{op}$.