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
$endgroup$
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
$endgroup$
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
$endgroup$
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
edited Jan 2 at 22:11
Roll up and smoke Adjoint
asked Jan 2 at 22:05
Roll up and smoke AdjointRoll up and smoke Adjoint
9,10052458
9,10052458
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
Congrats on reaching 10k rep
$endgroup$
– Roll up and smoke Adjoint
Jan 2 at 23:05
$begingroup$
Thank you!-----
$endgroup$
– Matematleta
Jan 2 at 23:11
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060033%2fhow-is-it-obvious-that-times-c-times-c-to-c-is-right-adjoint-to-the-diago%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
Congrats on reaching 10k rep
$endgroup$
– Roll up and smoke Adjoint
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.
$endgroup$
$begingroup$
Congrats on reaching 10k rep
$endgroup$
– Roll up and smoke Adjoint
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.
$endgroup$
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.
edited Jan 2 at 23:20
answered Jan 2 at 23:00
MatematletaMatematleta
10.2k2918
10.2k2918
$begingroup$
Congrats on reaching 10k rep
$endgroup$
– Roll up and smoke Adjoint
Jan 2 at 23:05
$begingroup$
Thank you!-----
$endgroup$
– Matematleta
Jan 2 at 23:11
add a comment |
$begingroup$
Congrats on reaching 10k rep
$endgroup$
– Roll up and smoke Adjoint
Jan 2 at 23:05
$begingroup$
Thank you!-----
$endgroup$
– Matematleta
Jan 2 at 23:11
$begingroup$
Congrats on reaching 10k rep
$endgroup$
– Roll up and smoke Adjoint
Jan 2 at 23:05
$begingroup$
Congrats on reaching 10k rep
$endgroup$
– Roll up and smoke Adjoint
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060033%2fhow-is-it-obvious-that-times-c-times-c-to-c-is-right-adjoint-to-the-diago%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown