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












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$begingroup$


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?






category-theory







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?






category-theory







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?






category-theory







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




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












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