Can we say that a diagram commutes if we don't define a category?
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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
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add a comment |
$begingroup$
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
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2
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It would be helpful if you could give an example.
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– Chris Culter
Dec 12 '18 at 21:24
6
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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.
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– Qiaochu Yuan
Dec 12 '18 at 21:35
1
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A category without identities is also known as a ``semigroupoid'' --- just as a monoid without unit is known as a semigroup ;-)
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– Musa Al-hassy
Dec 13 '18 at 10:19
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Thanks @QiaochuYuan
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– Garmekain
Dec 13 '18 at 13:03
1
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@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.
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– Garmekain
Dec 13 '18 at 14:01
add a comment |
$begingroup$
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
$endgroup$
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
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
add a comment |
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
add a comment |
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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