Can we say that a diagram commutes if we don't define a category?












0












$begingroup$


Suppose we have a commutative diagram which commutes, but there is no predefined category, for example, if we don't define the identity morphisms.



Can we say so? The only reason that I can think that this cannot be done is that maybe the notion of a commutative diagram only makes sense when talking about a category. Am I right?










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    It would be helpful if you could give an example.
    $endgroup$
    – Chris Culter
    Dec 12 '18 at 21:24






  • 6




    $begingroup$
    You don't need identity morphisms to talk about a diagram commuting; you just need objects, morphisms, and composition. The resulting object is called a semicategory. Examples include some cobordism categories which by default don't have identity morphisms unless you add them in explicitly.
    $endgroup$
    – Qiaochu Yuan
    Dec 12 '18 at 21:35






  • 1




    $begingroup$
    A category without identities is also known as a ``semigroupoid'' --- just as a monoid without unit is known as a semigroup ;-)
    $endgroup$
    – Musa Al-hassy
    Dec 13 '18 at 10:19










  • $begingroup$
    Thanks @QiaochuYuan
    $endgroup$
    – Garmekain
    Dec 13 '18 at 13:03






  • 1




    $begingroup$
    @ChrisCulter The example would be a set of partial functions as objects and certain sets as morphisms, where the composition operator is defined as the intersection.
    $endgroup$
    – Garmekain
    Dec 13 '18 at 14:01


















0












$begingroup$


Suppose we have a commutative diagram which commutes, but there is no predefined category, for example, if we don't define the identity morphisms.



Can we say so? The only reason that I can think that this cannot be done is that maybe the notion of a commutative diagram only makes sense when talking about a category. Am I right?










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    It would be helpful if you could give an example.
    $endgroup$
    – Chris Culter
    Dec 12 '18 at 21:24






  • 6




    $begingroup$
    You don't need identity morphisms to talk about a diagram commuting; you just need objects, morphisms, and composition. The resulting object is called a semicategory. Examples include some cobordism categories which by default don't have identity morphisms unless you add them in explicitly.
    $endgroup$
    – Qiaochu Yuan
    Dec 12 '18 at 21:35






  • 1




    $begingroup$
    A category without identities is also known as a ``semigroupoid'' --- just as a monoid without unit is known as a semigroup ;-)
    $endgroup$
    – Musa Al-hassy
    Dec 13 '18 at 10:19










  • $begingroup$
    Thanks @QiaochuYuan
    $endgroup$
    – Garmekain
    Dec 13 '18 at 13:03






  • 1




    $begingroup$
    @ChrisCulter The example would be a set of partial functions as objects and certain sets as morphisms, where the composition operator is defined as the intersection.
    $endgroup$
    – Garmekain
    Dec 13 '18 at 14:01
















0












0








0





$begingroup$


Suppose we have a commutative diagram which commutes, but there is no predefined category, for example, if we don't define the identity morphisms.



Can we say so? The only reason that I can think that this cannot be done is that maybe the notion of a commutative diagram only makes sense when talking about a category. Am I right?










share|cite|improve this question









$endgroup$




Suppose we have a commutative diagram which commutes, but there is no predefined category, for example, if we don't define the identity morphisms.



Can we say so? The only reason that I can think that this cannot be done is that maybe the notion of a commutative diagram only makes sense when talking about a category. Am I right?







category-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 21:13









GarmekainGarmekain

1,386720




1,386720








  • 2




    $begingroup$
    It would be helpful if you could give an example.
    $endgroup$
    – Chris Culter
    Dec 12 '18 at 21:24






  • 6




    $begingroup$
    You don't need identity morphisms to talk about a diagram commuting; you just need objects, morphisms, and composition. The resulting object is called a semicategory. Examples include some cobordism categories which by default don't have identity morphisms unless you add them in explicitly.
    $endgroup$
    – Qiaochu Yuan
    Dec 12 '18 at 21:35






  • 1




    $begingroup$
    A category without identities is also known as a ``semigroupoid'' --- just as a monoid without unit is known as a semigroup ;-)
    $endgroup$
    – Musa Al-hassy
    Dec 13 '18 at 10:19










  • $begingroup$
    Thanks @QiaochuYuan
    $endgroup$
    – Garmekain
    Dec 13 '18 at 13:03






  • 1




    $begingroup$
    @ChrisCulter The example would be a set of partial functions as objects and certain sets as morphisms, where the composition operator is defined as the intersection.
    $endgroup$
    – Garmekain
    Dec 13 '18 at 14:01
















  • 2




    $begingroup$
    It would be helpful if you could give an example.
    $endgroup$
    – Chris Culter
    Dec 12 '18 at 21:24






  • 6




    $begingroup$
    You don't need identity morphisms to talk about a diagram commuting; you just need objects, morphisms, and composition. The resulting object is called a semicategory. Examples include some cobordism categories which by default don't have identity morphisms unless you add them in explicitly.
    $endgroup$
    – Qiaochu Yuan
    Dec 12 '18 at 21:35






  • 1




    $begingroup$
    A category without identities is also known as a ``semigroupoid'' --- just as a monoid without unit is known as a semigroup ;-)
    $endgroup$
    – Musa Al-hassy
    Dec 13 '18 at 10:19










  • $begingroup$
    Thanks @QiaochuYuan
    $endgroup$
    – Garmekain
    Dec 13 '18 at 13:03






  • 1




    $begingroup$
    @ChrisCulter The example would be a set of partial functions as objects and certain sets as morphisms, where the composition operator is defined as the intersection.
    $endgroup$
    – Garmekain
    Dec 13 '18 at 14:01










2




2




$begingroup$
It would be helpful if you could give an example.
$endgroup$
– Chris Culter
Dec 12 '18 at 21:24




$begingroup$
It would be helpful if you could give an example.
$endgroup$
– Chris Culter
Dec 12 '18 at 21:24




6




6




$begingroup$
You don't need identity morphisms to talk about a diagram commuting; you just need objects, morphisms, and composition. The resulting object is called a semicategory. Examples include some cobordism categories which by default don't have identity morphisms unless you add them in explicitly.
$endgroup$
– Qiaochu Yuan
Dec 12 '18 at 21:35




$begingroup$
You don't need identity morphisms to talk about a diagram commuting; you just need objects, morphisms, and composition. The resulting object is called a semicategory. Examples include some cobordism categories which by default don't have identity morphisms unless you add them in explicitly.
$endgroup$
– Qiaochu Yuan
Dec 12 '18 at 21:35




1




1




$begingroup$
A category without identities is also known as a ``semigroupoid'' --- just as a monoid without unit is known as a semigroup ;-)
$endgroup$
– Musa Al-hassy
Dec 13 '18 at 10:19




$begingroup$
A category without identities is also known as a ``semigroupoid'' --- just as a monoid without unit is known as a semigroup ;-)
$endgroup$
– Musa Al-hassy
Dec 13 '18 at 10:19












$begingroup$
Thanks @QiaochuYuan
$endgroup$
– Garmekain
Dec 13 '18 at 13:03




$begingroup$
Thanks @QiaochuYuan
$endgroup$
– Garmekain
Dec 13 '18 at 13:03




1




1




$begingroup$
@ChrisCulter The example would be a set of partial functions as objects and certain sets as morphisms, where the composition operator is defined as the intersection.
$endgroup$
– Garmekain
Dec 13 '18 at 14:01






$begingroup$
@ChrisCulter The example would be a set of partial functions as objects and certain sets as morphisms, where the composition operator is defined as the intersection.
$endgroup$
– Garmekain
Dec 13 '18 at 14:01












0






active

oldest

votes











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037228%2fcan-we-say-that-a-diagram-commutes-if-we-dont-define-a-category%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037228%2fcan-we-say-that-a-diagram-commutes-if-we-dont-define-a-category%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Ellipse (mathématiques)

Quarter-circle Tiles

Mont Emei