Mixing
Is it meaningful to distinguish a one-object category and its opposite?
$begingroup$
As in the title, for a category with only one object $mathcal C$, $mathcal C^{op}$ seems the same with the original one but I still believe there is a difference between them, can someone give an example where $mathcal C$ and $mathcal C^{op}$ play different roles?
Thank you in advance.
category-theory
asked Jan 28 at 20:01
Display NameDisplay Name
454313
$endgroup$
add a comment |
$begingroup$
As in the title, for a category with only one object $mathcal C$, $mathcal C^{op}$ seems the same with the original one but I still believe there is a difference between them, can someone give an example where $mathcal C$ and $mathcal C^{op}$ play different roles?
Thank you in advance.
category-theory
asked Jan 28 at 20:01
Display NameDisplay Name
454313
$endgroup$
$begingroup$
Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
$endgroup$
– Musa Al-hassy
Jan 31 at 0:31
add a comment |
$begingroup$
As in the title, for a category with only one object $mathcal C$, $mathcal C^{op}$ seems the same with the original one but I still believe there is a difference between them, can someone give an example where $mathcal C$ and $mathcal C^{op}$ play different roles?
Thank you in advance.
category-theory
asked Jan 28 at 20:01
Display NameDisplay Name
454313
$endgroup$
As in the title, for a category with only one object $mathcal C$, $mathcal C^{op}$ seems the same with the original one but I still believe there is a difference between them, can someone give an example where $mathcal C$ and $mathcal C^{op}$ play different roles?
Thank you in advance.
category-theory
category-theory
asked Jan 28 at 20:01
Display NameDisplay Name
454313
asked Jan 28 at 20:01
Display NameDisplay Name
454313
asked Jan 28 at 20:01
Display NameDisplay Name
454313
asked Jan 28 at 20:01
Display NameDisplay Name
454313
asked Jan 28 at 20:01
Display NameDisplay Name
454313
454313
$begingroup$
Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
$endgroup$
– Musa Al-hassy
Jan 31 at 0:31
add a comment |
$begingroup$
Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
$endgroup$
– Musa Al-hassy
Jan 31 at 0:31
$begingroup$
Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
$endgroup$
– Musa Al-hassy
Jan 31 at 0:31
$begingroup$
Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
$endgroup$
– Musa Al-hassy
Jan 31 at 0:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$mathcal{C}$ corresponds to a monoid $M$ which is the set of morphisms $hom(cdot,cdot)$. Let us write the multiplication by $circ$. Then $mathcal{C}^{op}$ corresponds to $M^{op}$ which is the same set of $M$ but with multiplication $xcirc' y=ycirc x$.
This is useful sometimes. For example, if $R$ is a ring, then $R^{op}$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^{op}$-module.
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,9851438
$endgroup$
add a comment |
$begingroup$
Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BMto Set$ can be identified with a left $M$-set, while a functor $BM^{mathrm{op}}to Set$ is similarly identified with a right $M$-set.
The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is
$$M=langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=brangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^{mathrm{op}}$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.8k23372
$endgroup$
$begingroup$
Perhaps providing an example where $M$ and $M^{op}$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
$endgroup$
– Max
Jan 28 at 20:32
1
$begingroup$
@Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
$endgroup$
– Kevin Carlson
Jan 28 at 21:56
$begingroup$
Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
$endgroup$
– Max
Jan 28 at 21:58
1
$begingroup$
Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
$endgroup$
– Kevin Carlson
Jan 28 at 22:17
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%2f3091320%2fis-it-meaningful-to-distinguish-a-one-object-category-and-its-opposite%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$mathcal{C}$ corresponds to a monoid $M$ which is the set of morphisms $hom(cdot,cdot)$. Let us write the multiplication by $circ$. Then $mathcal{C}^{op}$ corresponds to $M^{op}$ which is the same set of $M$ but with multiplication $xcirc' y=ycirc x$.
This is useful sometimes. For example, if $R$ is a ring, then $R^{op}$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^{op}$-module.
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,9851438
$endgroup$
add a comment |
$begingroup$
$mathcal{C}$ corresponds to a monoid $M$ which is the set of morphisms $hom(cdot,cdot)$. Let us write the multiplication by $circ$. Then $mathcal{C}^{op}$ corresponds to $M^{op}$ which is the same set of $M$ but with multiplication $xcirc' y=ycirc x$.
This is useful sometimes. For example, if $R$ is a ring, then $R^{op}$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^{op}$-module.
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,9851438
$endgroup$
add a comment |
$begingroup$
$mathcal{C}$ corresponds to a monoid $M$ which is the set of morphisms $hom(cdot,cdot)$. Let us write the multiplication by $circ$. Then $mathcal{C}^{op}$ corresponds to $M^{op}$ which is the same set of $M$ but with multiplication $xcirc' y=ycirc x$.
This is useful sometimes. For example, if $R$ is a ring, then $R^{op}$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^{op}$-module.
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,9851438
$endgroup$
$mathcal{C}$ corresponds to a monoid $M$ which is the set of morphisms $hom(cdot,cdot)$. Let us write the multiplication by $circ$. Then $mathcal{C}^{op}$ corresponds to $M^{op}$ which is the same set of $M$ but with multiplication $xcirc' y=ycirc x$.
This is useful sometimes. For example, if $R$ is a ring, then $R^{op}$ is also a ring, known as its opposite ring. And a left $R$-module becomes a right $R^{op}$-module.
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,9851438
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,9851438
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,9851438
answered Jan 28 at 20:24
Eclipse SunEclipse Sun
7,9851438
7,9851438
add a comment |
add a comment |
$begingroup$
Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BMto Set$ can be identified with a left $M$-set, while a functor $BM^{mathrm{op}}to Set$ is similarly identified with a right $M$-set.
The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is
$$M=langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=brangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^{mathrm{op}}$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.8k23372
$endgroup$
$begingroup$
Perhaps providing an example where $M$ and $M^{op}$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
$endgroup$
– Max
Jan 28 at 20:32
1
$begingroup$
@Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
$endgroup$
– Kevin Carlson
Jan 28 at 21:56
$begingroup$
Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
$endgroup$
– Max
Jan 28 at 21:58
1
$begingroup$
Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
$endgroup$
– Kevin Carlson
Jan 28 at 22:17
add a comment |
$begingroup$
Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BMto Set$ can be identified with a left $M$-set, while a functor $BM^{mathrm{op}}to Set$ is similarly identified with a right $M$-set.
The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is
$$M=langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=brangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^{mathrm{op}}$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.8k23372
$endgroup$
$begingroup$
Perhaps providing an example where $M$ and $M^{op}$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
$endgroup$
– Max
Jan 28 at 20:32
1
$begingroup$
@Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
$endgroup$
– Kevin Carlson
Jan 28 at 21:56
$begingroup$
Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
$endgroup$
– Max
Jan 28 at 21:58
1
$begingroup$
Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
$endgroup$
– Kevin Carlson
Jan 28 at 22:17
add a comment |
$begingroup$
Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BMto Set$ can be identified with a left $M$-set, while a functor $BM^{mathrm{op}}to Set$ is similarly identified with a right $M$-set.
The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is
$$M=langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=brangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^{mathrm{op}}$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.8k23372
$endgroup$
Sure they're different. The have the same object and arrows, but composition is reversed. If $BM$ is a one-object category whose monoid of morphisms is $M$, then a functor $BMto Set$ can be identified with a left $M$-set, while a functor $BM^{mathrm{op}}to Set$ is similarly identified with a right $M$-set.
The simplest example I can think of (although I think there should exist a four-element example) of a monoid non-isomorphic to its opposite is
$$M=langle a,b,c, z: ab=a, cb=c, ac=ca=z, ba=b, bc=brangle$$, and such that the generators are all idempotent and $z$ is an absorbing element. Any isomorphism between $M$ and $M^{mathrm{op}}$ must preserve $z$, the unique absorbing element, and $b$, the unique element absorbing exactly two other elements of $M$ from the right, and so is either the identity or the 2-cycle exchanging $a$ and $c$. But neither function is a homomorphism. I think it should be easy to
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.8k23372
edited Jan 28 at 22:16
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.8k23372
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.8k23372
answered Jan 28 at 20:21
Kevin CarlsonKevin Carlson
33.8k23372
33.8k23372
$begingroup$
Perhaps providing an example where $M$ and $M^{op}$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
$endgroup$
– Max
Jan 28 at 20:32
1
$begingroup$
@Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
$endgroup$
– Kevin Carlson
Jan 28 at 21:56
$begingroup$
Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
$endgroup$
– Max
Jan 28 at 21:58
1
$begingroup$
Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
$endgroup$
– Kevin Carlson
Jan 28 at 22:17
add a comment |
$begingroup$
Perhaps providing an example where $M$ and $M^{op}$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
$endgroup$
– Max
Jan 28 at 20:32
1
$begingroup$
@Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
$endgroup$
– Kevin Carlson
Jan 28 at 21:56
$begingroup$
Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
$endgroup$
– Max
Jan 28 at 21:58
1
$begingroup$
Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
$endgroup$
– Kevin Carlson
Jan 28 at 22:17
$begingroup$
Perhaps providing an example where $M$ and $M^{op}$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
$endgroup$
– Max
Jan 28 at 20:32
$begingroup$
Perhaps providing an example where $M$ and $M^{op}$ aren't isomorphic would be helpful (equivalently where the categories of actions aren't equivalent)
$endgroup$
– Max
Jan 28 at 20:32
1
1
$begingroup$
@Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
$endgroup$
– Kevin Carlson
Jan 28 at 21:56
$begingroup$
@Max Actually, Morita equivalent monoids need not be isomorphic, unlike the case of groups. Examples can be found here; the core difference is that monoids often have non-identity idempotents, so interesting identifications can happen under idempotent completion. That said, thanks for the suggestion.
$endgroup$
– Kevin Carlson
Jan 28 at 21:56
$begingroup$
Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
$endgroup$
– Max
Jan 28 at 21:58
$begingroup$
Oh I went too fast in my head when reviewing the proof for groups, thanks for correcting that !
$endgroup$
– Max
Jan 28 at 21:58
1
1
$begingroup$
Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
$endgroup$
– Kevin Carlson
Jan 28 at 22:17
$begingroup$
Oops, forgot to include the link: link.springer.com/article/10.1007/BF02572973
$endgroup$
– Kevin Carlson
Jan 28 at 22:17
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%2f3091320%2fis-it-meaningful-to-distinguish-a-one-object-category-and-its-opposite%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
$begingroup$
Consider the natural numbers as a semigroup with binary operation $x ⊕ y = x$. Then $0 ⊕ 1 = 0$ but in the opposite semigroup $0 ⊕ 1 = 1$!
$endgroup$
– Musa Al-hassy
Jan 31 at 0:31