How is it obvious that $times : C times C to C$ is right adjoint to the diagonal functor?
$begingroup$
This is from "Sheaves in Geometry & Logic".
$times : C times C to C$ is the cartesian product of two objects. So assume that finite products exist in $C$ the above is a functor.
To say that it is right adjoint to a functor $L$, is to say that for all $a times b in C times C$ and all $d in C$ we have a bijection of sets:
$$
text{Hom}_{C}(d, a times b) simeq text{Hom}_{Ctimes C}((d, d), (a,b))
$$
where we've used $(x,y)$ to distinguish elements in $C times C$ from those in $C$.
This is supposed to be obvious, but it isn't to me.
Further, we have to show that the above is natural in $d, (a,b)$. Is there some sort of composition of natural maps and functors we can find here?
Is it:
$text{Hom}_{Ctimes C}((d,d), (a,b)) = text{Hom}_C(d, a) times text{Hom}_C(d, b) simeq text{Hom}_C(d, atimes b)$.
?
But how do I show naturality in $(a,b)$?
category-theory products adjoint-functors functors natural-transformations
asked Jan 2 at 22:05
BananaCatsBananaCats
9,26252558
$endgroup$
add a comment |
$begingroup$
This is from "Sheaves in Geometry & Logic".
$times : C times C to C$ is the cartesian product of two objects. So assume that finite products exist in $C$ the above is a functor.
To say that it is right adjoint to a functor $L$, is to say that for all $a times b in C times C$ and all $d in C$ we have a bijection of sets:
$$
text{Hom}_{C}(d, a times b) simeq text{Hom}_{Ctimes C}((d, d), (a,b))
$$
where we've used $(x,y)$ to distinguish elements in $C times C$ from those in $C$.
This is supposed to be obvious, but it isn't to me.
Further, we have to show that the above is natural in $d, (a,b)$. Is there some sort of composition of natural maps and functors we can find here?
Is it:
$text{Hom}_{Ctimes C}((d,d), (a,b)) = text{Hom}_C(d, a) times text{Hom}_C(d, b) simeq text{Hom}_C(d, atimes b)$.
?
But how do I show naturality in $(a,b)$?
category-theory products adjoint-functors functors natural-transformations
asked Jan 2 at 22:05
BananaCatsBananaCats
9,26252558
$endgroup$
add a comment |
$begingroup$
This is from "Sheaves in Geometry & Logic".
$times : C times C to C$ is the cartesian product of two objects. So assume that finite products exist in $C$ the above is a functor.
To say that it is right adjoint to a functor $L$, is to say that for all $a times b in C times C$ and all $d in C$ we have a bijection of sets:
$$
text{Hom}_{C}(d, a times b) simeq text{Hom}_{Ctimes C}((d, d), (a,b))
$$
where we've used $(x,y)$ to distinguish elements in $C times C$ from those in $C$.
This is supposed to be obvious, but it isn't to me.
Further, we have to show that the above is natural in $d, (a,b)$. Is there some sort of composition of natural maps and functors we can find here?
Is it:
$text{Hom}_{Ctimes C}((d,d), (a,b)) = text{Hom}_C(d, a) times text{Hom}_C(d, b) simeq text{Hom}_C(d, atimes b)$.
?
But how do I show naturality in $(a,b)$?
category-theory products adjoint-functors functors natural-transformations
asked Jan 2 at 22:05
BananaCatsBananaCats
9,26252558
$endgroup$
This is from "Sheaves in Geometry & Logic".
$times : C times C to C$ is the cartesian product of two objects. So assume that finite products exist in $C$ the above is a functor.
To say that it is right adjoint to a functor $L$, is to say that for all $a times b in C times C$ and all $d in C$ we have a bijection of sets:
$$
text{Hom}_{C}(d, a times b) simeq text{Hom}_{Ctimes C}((d, d), (a,b))
$$
where we've used $(x,y)$ to distinguish elements in $C times C$ from those in $C$.
This is supposed to be obvious, but it isn't to me.
Further, we have to show that the above is natural in $d, (a,b)$. Is there some sort of composition of natural maps and functors we can find here?
Is it:
$text{Hom}_{Ctimes C}((d,d), (a,b)) = text{Hom}_C(d, a) times text{Hom}_C(d, b) simeq text{Hom}_C(d, atimes b)$.
?
But how do I show naturality in $(a,b)$?
category-theory products adjoint-functors functors natural-transformations
category-theory products adjoint-functors functors natural-transformations
asked Jan 2 at 22:05
BananaCatsBananaCats
9,26252558
asked Jan 2 at 22:05
BananaCatsBananaCats
9,26252558
edited Jan 2 at 22:11
BananaCats
asked Jan 2 at 22:05
BananaCatsBananaCats
9,26252558
asked Jan 2 at 22:05
BananaCatsBananaCats
9,26252558
asked Jan 2 at 22:05
BananaCatsBananaCats
9,26252558
9,26252558
add a comment |
add a comment |
1 Answer
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$begingroup$
You are on the right track. The diagonal functor is defined by $ Delta: Cto Ctimes C$. On objects: $cmapsto ctimes c$. On arrows: $(f,f):(c, c)to (c', c')$.
From this, we can construct a right adjoint simply by enforcing the rules:
A right adjoint is a functor $G:Ctimes Cto C$ such that
$text{Hom}_C(c,G(a,b))cong text{Hom}_{Ctimes C}(Delta c,(a,b)).$
The left-hand side of this is clear. The right-hand side is a set of arrows of the form $(c,c)to (a,b)$ in the category $Ctimes C.$ But these arrows, by definition, are $pairs (f,g)$ where $f:cto a$ and $g:cto b$, so there is the obvious bijection $text{Hom}_{Ctimes C}(Delta c,(a,b))cong text{Hom}_{C}( c,a)times text{Hom}_{C}(c,b)$. This is natural as well, but that's not important here. It just gives us an idea of what $G should $ be.
That is, $G=times$, the product functor. On objects: $(a,b)to atimes b$ and on arrows, the unique morphism induced by the UMP of the product.
As long as $C$ has products and small hom-sets, this will work.
The obvious unit is $eta_c=langle 1_c,1_crangle .$
All that remains is to check that $GDelta fcirc eta_c=f$ for $f:cto atimes b.$
I guess you can take it from here.
answered Jan 2 at 23:00
MatematletaMatematleta
11.5k2920
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Congrats on reaching 10k rep
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– BananaCats
Jan 2 at 23:05
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Thank you!-----
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– Matematleta
Jan 2 at 23:11
add a comment |
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$begingroup$
You are on the right track. The diagonal functor is defined by $ Delta: Cto Ctimes C$. On objects: $cmapsto ctimes c$. On arrows: $(f,f):(c, c)to (c', c')$.
From this, we can construct a right adjoint simply by enforcing the rules:
A right adjoint is a functor $G:Ctimes Cto C$ such that
$text{Hom}_C(c,G(a,b))cong text{Hom}_{Ctimes C}(Delta c,(a,b)).$
The left-hand side of this is clear. The right-hand side is a set of arrows of the form $(c,c)to (a,b)$ in the category $Ctimes C.$ But these arrows, by definition, are $pairs (f,g)$ where $f:cto a$ and $g:cto b$, so there is the obvious bijection $text{Hom}_{Ctimes C}(Delta c,(a,b))cong text{Hom}_{C}( c,a)times text{Hom}_{C}(c,b)$. This is natural as well, but that's not important here. It just gives us an idea of what $G should $ be.
That is, $G=times$, the product functor. On objects: $(a,b)to atimes b$ and on arrows, the unique morphism induced by the UMP of the product.
As long as $C$ has products and small hom-sets, this will work.
The obvious unit is $eta_c=langle 1_c,1_crangle .$
All that remains is to check that $GDelta fcirc eta_c=f$ for $f:cto atimes b.$
I guess you can take it from here.
answered Jan 2 at 23:00
MatematletaMatematleta
11.5k2920
$endgroup$
$begingroup$
Congrats on reaching 10k rep
$endgroup$
– BananaCats
Jan 2 at 23:05
$begingroup$
Thank you!-----
$endgroup$
– Matematleta
Jan 2 at 23:11
add a comment |
$begingroup$
You are on the right track. The diagonal functor is defined by $ Delta: Cto Ctimes C$. On objects: $cmapsto ctimes c$. On arrows: $(f,f):(c, c)to (c', c')$.
From this, we can construct a right adjoint simply by enforcing the rules:
A right adjoint is a functor $G:Ctimes Cto C$ such that
$text{Hom}_C(c,G(a,b))cong text{Hom}_{Ctimes C}(Delta c,(a,b)).$
The left-hand side of this is clear. The right-hand side is a set of arrows of the form $(c,c)to (a,b)$ in the category $Ctimes C.$ But these arrows, by definition, are $pairs (f,g)$ where $f:cto a$ and $g:cto b$, so there is the obvious bijection $text{Hom}_{Ctimes C}(Delta c,(a,b))cong text{Hom}_{C}( c,a)times text{Hom}_{C}(c,b)$. This is natural as well, but that's not important here. It just gives us an idea of what $G should $ be.
That is, $G=times$, the product functor. On objects: $(a,b)to atimes b$ and on arrows, the unique morphism induced by the UMP of the product.
As long as $C$ has products and small hom-sets, this will work.
The obvious unit is $eta_c=langle 1_c,1_crangle .$
All that remains is to check that $GDelta fcirc eta_c=f$ for $f:cto atimes b.$
I guess you can take it from here.
answered Jan 2 at 23:00
MatematletaMatematleta
11.5k2920
$endgroup$
$begingroup$
Congrats on reaching 10k rep
$endgroup$
– BananaCats
Jan 2 at 23:05
$begingroup$
Thank you!-----
$endgroup$
– Matematleta
Jan 2 at 23:11
add a comment |
$begingroup$
You are on the right track. The diagonal functor is defined by $ Delta: Cto Ctimes C$. On objects: $cmapsto ctimes c$. On arrows: $(f,f):(c, c)to (c', c')$.
From this, we can construct a right adjoint simply by enforcing the rules:
A right adjoint is a functor $G:Ctimes Cto C$ such that
$text{Hom}_C(c,G(a,b))cong text{Hom}_{Ctimes C}(Delta c,(a,b)).$
The left-hand side of this is clear. The right-hand side is a set of arrows of the form $(c,c)to (a,b)$ in the category $Ctimes C.$ But these arrows, by definition, are $pairs (f,g)$ where $f:cto a$ and $g:cto b$, so there is the obvious bijection $text{Hom}_{Ctimes C}(Delta c,(a,b))cong text{Hom}_{C}( c,a)times text{Hom}_{C}(c,b)$. This is natural as well, but that's not important here. It just gives us an idea of what $G should $ be.
That is, $G=times$, the product functor. On objects: $(a,b)to atimes b$ and on arrows, the unique morphism induced by the UMP of the product.
As long as $C$ has products and small hom-sets, this will work.
The obvious unit is $eta_c=langle 1_c,1_crangle .$
All that remains is to check that $GDelta fcirc eta_c=f$ for $f:cto atimes b.$
I guess you can take it from here.
answered Jan 2 at 23:00
MatematletaMatematleta
11.5k2920
$endgroup$
You are on the right track. The diagonal functor is defined by $ Delta: Cto Ctimes C$. On objects: $cmapsto ctimes c$. On arrows: $(f,f):(c, c)to (c', c')$.
From this, we can construct a right adjoint simply by enforcing the rules:
A right adjoint is a functor $G:Ctimes Cto C$ such that
$text{Hom}_C(c,G(a,b))cong text{Hom}_{Ctimes C}(Delta c,(a,b)).$
The left-hand side of this is clear. The right-hand side is a set of arrows of the form $(c,c)to (a,b)$ in the category $Ctimes C.$ But these arrows, by definition, are $pairs (f,g)$ where $f:cto a$ and $g:cto b$, so there is the obvious bijection $text{Hom}_{Ctimes C}(Delta c,(a,b))cong text{Hom}_{C}( c,a)times text{Hom}_{C}(c,b)$. This is natural as well, but that's not important here. It just gives us an idea of what $G should $ be.
That is, $G=times$, the product functor. On objects: $(a,b)to atimes b$ and on arrows, the unique morphism induced by the UMP of the product.
As long as $C$ has products and small hom-sets, this will work.
The obvious unit is $eta_c=langle 1_c,1_crangle .$
All that remains is to check that $GDelta fcirc eta_c=f$ for $f:cto atimes b.$
I guess you can take it from here.
answered Jan 2 at 23:00
MatematletaMatematleta
11.5k2920
edited Jan 2 at 23:20
answered Jan 2 at 23:00
MatematletaMatematleta
11.5k2920
answered Jan 2 at 23:00
MatematletaMatematleta
11.5k2920
answered Jan 2 at 23:00
MatematletaMatematleta
11.5k2920
11.5k2920
$begingroup$
Congrats on reaching 10k rep
$endgroup$
– BananaCats
Jan 2 at 23:05
$begingroup$
Thank you!-----
$endgroup$
– Matematleta
Jan 2 at 23:11
add a comment |
$begingroup$
Congrats on reaching 10k rep
$endgroup$
– BananaCats
Jan 2 at 23:05
$begingroup$
Thank you!-----
$endgroup$
– Matematleta
Jan 2 at 23:11
$begingroup$
Congrats on reaching 10k rep
$endgroup$
– BananaCats
Jan 2 at 23:05
$begingroup$
Congrats on reaching 10k rep
$endgroup$
– BananaCats
Jan 2 at 23:05
$begingroup$
Thank you!-----
$endgroup$
– Matematleta
Jan 2 at 23:11
$begingroup$
Thank you!-----
$endgroup$
– Matematleta
Jan 2 at 23:11
add a comment |
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