Mixing
Object of a Category $C$ acts as Functor
$begingroup$
I have a question about a notation used in following paper: https://etale.site/writing/stax-seminar-talk.pdf (see page 4):
We take a category $C$ and consider a pair $(X_0,X_1)$ of two objects $X_1, X_1 in C$ satisfying identities between maps $s, t , epsilon, i $ and $m$ as described in the article above.
Then we define a new category in following way:
We take a $U in C$ and define a category ${X_0(U)/X_1(U)}$ such that:
The objects are defined via $ob({X_0(U)/X_1(U)}):= X_0(U)$.
And exactly this is the problem: What is exactly $X_0(U)$? Especially how $X_0$ "acts" on $U in C$?
Is $X_0$ concretely an element in $C$ or a functor $X_0:C to (Set)$, $U mapsto X_0(U)$?
algebraic-geometry category-theory groupoids algebraic-stacks
asked Feb 2 at 17:02
KarlPeterKarlPeter
6841416
$endgroup$
add a comment |
$begingroup$
I have a question about a notation used in following paper: https://etale.site/writing/stax-seminar-talk.pdf (see page 4):
We take a category $C$ and consider a pair $(X_0,X_1)$ of two objects $X_1, X_1 in C$ satisfying identities between maps $s, t , epsilon, i $ and $m$ as described in the article above.
Then we define a new category in following way:
We take a $U in C$ and define a category ${X_0(U)/X_1(U)}$ such that:
The objects are defined via $ob({X_0(U)/X_1(U)}):= X_0(U)$.
And exactly this is the problem: What is exactly $X_0(U)$? Especially how $X_0$ "acts" on $U in C$?
Is $X_0$ concretely an element in $C$ or a functor $X_0:C to (Set)$, $U mapsto X_0(U)$?
algebraic-geometry category-theory groupoids algebraic-stacks
asked Feb 2 at 17:02
KarlPeterKarlPeter
6841416
$endgroup$
2
$begingroup$
I assume that $X_0(U)=operatorname{Hom}(U,X_0)$. That's fairly typical notation in AG (identify an object with its functor of points).
$endgroup$
– jgon
Feb 2 at 17:23
add a comment |
$begingroup$
I have a question about a notation used in following paper: https://etale.site/writing/stax-seminar-talk.pdf (see page 4):
We take a category $C$ and consider a pair $(X_0,X_1)$ of two objects $X_1, X_1 in C$ satisfying identities between maps $s, t , epsilon, i $ and $m$ as described in the article above.
Then we define a new category in following way:
We take a $U in C$ and define a category ${X_0(U)/X_1(U)}$ such that:
The objects are defined via $ob({X_0(U)/X_1(U)}):= X_0(U)$.
And exactly this is the problem: What is exactly $X_0(U)$? Especially how $X_0$ "acts" on $U in C$?
Is $X_0$ concretely an element in $C$ or a functor $X_0:C to (Set)$, $U mapsto X_0(U)$?
algebraic-geometry category-theory groupoids algebraic-stacks
asked Feb 2 at 17:02
KarlPeterKarlPeter
6841416
$endgroup$
I have a question about a notation used in following paper: https://etale.site/writing/stax-seminar-talk.pdf (see page 4):
We take a category $C$ and consider a pair $(X_0,X_1)$ of two objects $X_1, X_1 in C$ satisfying identities between maps $s, t , epsilon, i $ and $m$ as described in the article above.
Then we define a new category in following way:
We take a $U in C$ and define a category ${X_0(U)/X_1(U)}$ such that:
The objects are defined via $ob({X_0(U)/X_1(U)}):= X_0(U)$.
And exactly this is the problem: What is exactly $X_0(U)$? Especially how $X_0$ "acts" on $U in C$?
Is $X_0$ concretely an element in $C$ or a functor $X_0:C to (Set)$, $U mapsto X_0(U)$?
algebraic-geometry category-theory groupoids algebraic-stacks
algebraic-geometry category-theory groupoids algebraic-stacks
asked Feb 2 at 17:02
KarlPeterKarlPeter
6841416
asked Feb 2 at 17:02
KarlPeterKarlPeter
6841416
asked Feb 2 at 17:02
KarlPeterKarlPeter
6841416
asked Feb 2 at 17:02
KarlPeterKarlPeter
6841416
asked Feb 2 at 17:02
KarlPeterKarlPeter
6841416
6841416
2
$begingroup$
I assume that $X_0(U)=operatorname{Hom}(U,X_0)$. That's fairly typical notation in AG (identify an object with its functor of points).
$endgroup$
– jgon
Feb 2 at 17:23
add a comment |
2
$begingroup$
I assume that $X_0(U)=operatorname{Hom}(U,X_0)$. That's fairly typical notation in AG (identify an object with its functor of points).
$endgroup$
– jgon
Feb 2 at 17:23
2
2
$begingroup$
I assume that $X_0(U)=operatorname{Hom}(U,X_0)$. That's fairly typical notation in AG (identify an object with its functor of points).
$endgroup$
– jgon
Feb 2 at 17:23
$begingroup$
I assume that $X_0(U)=operatorname{Hom}(U,X_0)$. That's fairly typical notation in AG (identify an object with its functor of points).
$endgroup$
– jgon
Feb 2 at 17:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To expand on my comment, $X_0(U)=newcommandHom{operatorname{Hom}}Hom(U,X_0)$. That is, we identify objects of a category with their functors of points.
To rephrase the definition here, $X_0$ and $X_1$ are objects in $C$ parametrizing objects and morphisms of the groupoid respectively. Applying the functor $Hom(U,-)$ to our groupoid, we see that for all $U$, $X_0(U)$ and $X_1(U)$ together with $s,t,epsilon,i,m$ determine a groupoid in $mathbf{Set}$. This is the category ${X_0(U)/X_1(U)}$. Since $Hom(U,-)$ is also functorial in $U$, for morphisms $f:Vto U$, we get pullback maps $f^* : {X_0(U)/X_1(U)}to {X_0(V)/X_1(V)}$.
answered Feb 2 at 17:33
jgonjgon
16.6k32143
$endgroup$
$begingroup$
so essentially this is the Yoneda embedding?
$endgroup$
– KarlPeter
Feb 4 at 0:45
$begingroup$
@KarlPeter, yes exactly
$endgroup$
– jgon
Feb 4 at 0:55
add a comment |
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$begingroup$
To expand on my comment, $X_0(U)=newcommandHom{operatorname{Hom}}Hom(U,X_0)$. That is, we identify objects of a category with their functors of points.
To rephrase the definition here, $X_0$ and $X_1$ are objects in $C$ parametrizing objects and morphisms of the groupoid respectively. Applying the functor $Hom(U,-)$ to our groupoid, we see that for all $U$, $X_0(U)$ and $X_1(U)$ together with $s,t,epsilon,i,m$ determine a groupoid in $mathbf{Set}$. This is the category ${X_0(U)/X_1(U)}$. Since $Hom(U,-)$ is also functorial in $U$, for morphisms $f:Vto U$, we get pullback maps $f^* : {X_0(U)/X_1(U)}to {X_0(V)/X_1(V)}$.
answered Feb 2 at 17:33
jgonjgon
16.6k32143
$endgroup$
$begingroup$
so essentially this is the Yoneda embedding?
$endgroup$
– KarlPeter
Feb 4 at 0:45
$begingroup$
@KarlPeter, yes exactly
$endgroup$
– jgon
Feb 4 at 0:55
add a comment |
$begingroup$
To expand on my comment, $X_0(U)=newcommandHom{operatorname{Hom}}Hom(U,X_0)$. That is, we identify objects of a category with their functors of points.
To rephrase the definition here, $X_0$ and $X_1$ are objects in $C$ parametrizing objects and morphisms of the groupoid respectively. Applying the functor $Hom(U,-)$ to our groupoid, we see that for all $U$, $X_0(U)$ and $X_1(U)$ together with $s,t,epsilon,i,m$ determine a groupoid in $mathbf{Set}$. This is the category ${X_0(U)/X_1(U)}$. Since $Hom(U,-)$ is also functorial in $U$, for morphisms $f:Vto U$, we get pullback maps $f^* : {X_0(U)/X_1(U)}to {X_0(V)/X_1(V)}$.
answered Feb 2 at 17:33
jgonjgon
16.6k32143
$endgroup$
$begingroup$
so essentially this is the Yoneda embedding?
$endgroup$
– KarlPeter
Feb 4 at 0:45
$begingroup$
@KarlPeter, yes exactly
$endgroup$
– jgon
Feb 4 at 0:55
add a comment |
$begingroup$
To expand on my comment, $X_0(U)=newcommandHom{operatorname{Hom}}Hom(U,X_0)$. That is, we identify objects of a category with their functors of points.
To rephrase the definition here, $X_0$ and $X_1$ are objects in $C$ parametrizing objects and morphisms of the groupoid respectively. Applying the functor $Hom(U,-)$ to our groupoid, we see that for all $U$, $X_0(U)$ and $X_1(U)$ together with $s,t,epsilon,i,m$ determine a groupoid in $mathbf{Set}$. This is the category ${X_0(U)/X_1(U)}$. Since $Hom(U,-)$ is also functorial in $U$, for morphisms $f:Vto U$, we get pullback maps $f^* : {X_0(U)/X_1(U)}to {X_0(V)/X_1(V)}$.
answered Feb 2 at 17:33
jgonjgon
16.6k32143
$endgroup$
To expand on my comment, $X_0(U)=newcommandHom{operatorname{Hom}}Hom(U,X_0)$. That is, we identify objects of a category with their functors of points.
To rephrase the definition here, $X_0$ and $X_1$ are objects in $C$ parametrizing objects and morphisms of the groupoid respectively. Applying the functor $Hom(U,-)$ to our groupoid, we see that for all $U$, $X_0(U)$ and $X_1(U)$ together with $s,t,epsilon,i,m$ determine a groupoid in $mathbf{Set}$. This is the category ${X_0(U)/X_1(U)}$. Since $Hom(U,-)$ is also functorial in $U$, for morphisms $f:Vto U$, we get pullback maps $f^* : {X_0(U)/X_1(U)}to {X_0(V)/X_1(V)}$.
answered Feb 2 at 17:33
jgonjgon
16.6k32143
answered Feb 2 at 17:33
jgonjgon
16.6k32143
answered Feb 2 at 17:33
jgonjgon
16.6k32143
answered Feb 2 at 17:33
jgonjgon
16.6k32143
16.6k32143
$begingroup$
so essentially this is the Yoneda embedding?
$endgroup$
– KarlPeter
Feb 4 at 0:45
$begingroup$
@KarlPeter, yes exactly
$endgroup$
– jgon
Feb 4 at 0:55
add a comment |
$begingroup$
so essentially this is the Yoneda embedding?
$endgroup$
– KarlPeter
Feb 4 at 0:45
$begingroup$
@KarlPeter, yes exactly
$endgroup$
– jgon
Feb 4 at 0:55
$begingroup$
so essentially this is the Yoneda embedding?
$endgroup$
– KarlPeter
Feb 4 at 0:45
$begingroup$
so essentially this is the Yoneda embedding?
$endgroup$
– KarlPeter
Feb 4 at 0:45
$begingroup$
@KarlPeter, yes exactly
$endgroup$
– jgon
Feb 4 at 0:55
$begingroup$
@KarlPeter, yes exactly
$endgroup$
– jgon
Feb 4 at 0:55
add a comment |
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$begingroup$
I assume that $X_0(U)=operatorname{Hom}(U,X_0)$. That's fairly typical notation in AG (identify an object with its functor of points).
$endgroup$
– jgon
Feb 2 at 17:23