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What are the objects of Hom$(y,x)$ in the contra-variant Yoneda functor
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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 $mathcal{C}$?
category-theory
asked Feb 2 at 13:06
foshofosho
4,7811033
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add a comment |
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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 $mathcal{C}$?
category-theory
asked Feb 2 at 13:06
foshofosho
4,7811033
$endgroup$
$begingroup$
Minor nit. Usually the Yoneda embedding is the functor that produces $mathsf{Hom}(-,X)$, i.e. it is a functor $mathcal Cto[mathcal C^{op},mathbf{Set}]$ 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
add a comment |
$begingroup$
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 $mathcal{C}$?
category-theory
asked Feb 2 at 13:06
foshofosho
4,7811033
$endgroup$
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 $mathcal{C}$?
category-theory
category-theory
asked Feb 2 at 13:06
foshofosho
4,7811033
asked Feb 2 at 13:06
foshofosho
4,7811033
asked Feb 2 at 13:06
foshofosho
4,7811033
asked Feb 2 at 13:06
foshofosho
4,7811033
asked Feb 2 at 13:06
foshofosho
4,7811033
4,7811033
$begingroup$
Minor nit. Usually the Yoneda embedding is the functor that produces $mathsf{Hom}(-,X)$, i.e. it is a functor $mathcal Cto[mathcal C^{op},mathbf{Set}]$ 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
add a comment |
$begingroup$
Minor nit. Usually the Yoneda embedding is the functor that produces $mathsf{Hom}(-,X)$, i.e. it is a functor $mathcal Cto[mathcal C^{op},mathbf{Set}]$ 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
$begingroup$
Minor nit. Usually the Yoneda embedding is the functor that produces $mathsf{Hom}(-,X)$, i.e. it is a functor $mathcal Cto[mathcal C^{op},mathbf{Set}]$ 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
$begingroup$
Minor nit. Usually the Yoneda embedding is the functor that produces $mathsf{Hom}(-,X)$, i.e. it is a functor $mathcal Cto[mathcal C^{op},mathbf{Set}]$ 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
add a comment |
2 Answers
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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 $Hom(x,-):Cto Set$, while the contravariant Yoneda lemma uses the contravariant functor $Hom(-,x):Cto Set$. 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 $yto x$ in $C$, which is the same as morphisms $xto y$ in $C^{op}$.
answered Feb 2 at 13:15
JaviJavi
3,15321032
$endgroup$
1
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Great. Thanks for your answer. I will accept it when I am allowed to.
$endgroup$
– fosho
Feb 2 at 13:17
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Thanks, I'm gald it helped :)
$endgroup$
– Javi
Feb 2 at 13:17
add a comment |
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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 $mathcal{C}^{op}$.
answered Feb 2 at 13:19
ChessanatorChessanator
2,3751412
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2 Answers
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2 Answers
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$begingroup$
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 $Hom(x,-):Cto Set$, while the contravariant Yoneda lemma uses the contravariant functor $Hom(-,x):Cto Set$. 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 $yto x$ in $C$, which is the same as morphisms $xto y$ in $C^{op}$.
answered Feb 2 at 13:15
JaviJavi
3,15321032
$endgroup$
1
$begingroup$
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
add a comment |
$begingroup$
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 $Hom(x,-):Cto Set$, while the contravariant Yoneda lemma uses the contravariant functor $Hom(-,x):Cto Set$. 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 $yto x$ in $C$, which is the same as morphisms $xto y$ in $C^{op}$.
answered Feb 2 at 13:15
JaviJavi
3,15321032
$endgroup$
1
$begingroup$
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
add a comment |
$begingroup$
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 $Hom(x,-):Cto Set$, while the contravariant Yoneda lemma uses the contravariant functor $Hom(-,x):Cto Set$. 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 $yto x$ in $C$, which is the same as morphisms $xto y$ in $C^{op}$.
answered Feb 2 at 13:15
JaviJavi
3,15321032
$endgroup$
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 $Hom(x,-):Cto Set$, while the contravariant Yoneda lemma uses the contravariant functor $Hom(-,x):Cto Set$. 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 $yto x$ in $C$, which is the same as morphisms $xto y$ in $C^{op}$.
answered Feb 2 at 13:15
JaviJavi
3,15321032
edited Feb 2 at 13:18
answered Feb 2 at 13:15
JaviJavi
3,15321032
answered Feb 2 at 13:15
JaviJavi
3,15321032
answered Feb 2 at 13:15
JaviJavi
3,15321032
3,15321032
1
$begingroup$
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
add a comment |
1
$begingroup$
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
1
1
$begingroup$
Great. Thanks for your answer. I will accept it when I am allowed to.
$endgroup$
– fosho
Feb 2 at 13:17
$begingroup$
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
$begingroup$
Thanks, I'm gald it helped :)
$endgroup$
– Javi
Feb 2 at 13:17
add a comment |
$begingroup$
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 $mathcal{C}^{op}$.
answered Feb 2 at 13:19
ChessanatorChessanator
2,3751412
$endgroup$
add a comment |
$begingroup$
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 $mathcal{C}^{op}$.
answered Feb 2 at 13:19
ChessanatorChessanator
2,3751412
$endgroup$
add a comment |
$begingroup$
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 $mathcal{C}^{op}$.
answered Feb 2 at 13:19
ChessanatorChessanator
2,3751412
$endgroup$
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 $mathcal{C}^{op}$.
answered Feb 2 at 13:19
ChessanatorChessanator
2,3751412
answered Feb 2 at 13:19
ChessanatorChessanator
2,3751412
answered Feb 2 at 13:19
ChessanatorChessanator
2,3751412
answered Feb 2 at 13:19
ChessanatorChessanator
2,3751412
2,3751412
add a comment |
add a comment |
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Minor nit. Usually the Yoneda embedding is the functor that produces $mathsf{Hom}(-,X)$, i.e. it is a functor $mathcal Cto[mathcal C^{op},mathbf{Set}]$ 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