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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 $\mathcal{C}$?

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  • $\begingroup$ Minor nit. Usually the Yoneda embedding is the functor that produces $\mathsf{Hom}(-,X)$, i.e. it is a functor $\mathcal C\to[\mathcal C^{op},\mathbf{Set}]$ were $[\mathcal D,\mathcal E]$ stands for the category of functors from $\mathcal D$ to $\mathcal E$. The idea is that $\mathcal C$ is embedded into $[\mathcal C^{op},\mathbf{Set}]$. Since this overall functor is covariant, it is the (covariant) Yoneda embedding. The other way, $\mathcal C^{op}\to[\mathcal C,\mathbf{Set}]$ is sometimes called the contravariant Yoneda embedding, though in reality they are the same thing. $\endgroup$ – Derek Elkins Feb 2 at 21:15
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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 $C^{op}$.

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    $\begingroup$ Great. Thanks for your answer. I will accept it when I am allowed to. $\endgroup$ – fosho Feb 2 at 13:17
  • $\begingroup$ Thanks, I'm gald it helped :) $\endgroup$ – Javi Feb 2 at 13:17
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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 $\mathcal{C}^{op}$.

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