Term for this concept in category theory?
Suppose we have three objects $X,Y,Z$, and a morphism $m:Xto Y$.
Moreover, this morphism has the following property:
For any morphism $f:Zto Y$, there exists a morphism $f_X:Zto X$, such that $mcirc f_X=f$.
Intuitively, this seems to me to capture the notion that “any information we need to pick an element of $X$, there is enough information in Y to do so”
Essentially, it seems to me that this generalizes the idea of a surjective function, but this concept is already generalized by “epimorphism”, whose definition is different.
Is my definition equivalent to that of epimorphism? If not, is there a term for my definition?
category-theory
asked Nov 21 '18 at 12:31
user56834
3,18821149
add a comment |
Suppose we have three objects $X,Y,Z$, and a morphism $m:Xto Y$.
Moreover, this morphism has the following property:
For any morphism $f:Zto Y$, there exists a morphism $f_X:Zto X$, such that $mcirc f_X=f$.
Intuitively, this seems to me to capture the notion that “any information we need to pick an element of $X$, there is enough information in Y to do so”
Essentially, it seems to me that this generalizes the idea of a surjective function, but this concept is already generalized by “epimorphism”, whose definition is different.
Is my definition equivalent to that of epimorphism? If not, is there a term for my definition?
category-theory
asked Nov 21 '18 at 12:31
user56834
3,18821149
If $Z$ can be any object, then what you describe is just a split epimorphism; see this question
– Arnaud D.
Nov 21 '18 at 13:18
add a comment |
Suppose we have three objects $X,Y,Z$, and a morphism $m:Xto Y$.
Moreover, this morphism has the following property:
For any morphism $f:Zto Y$, there exists a morphism $f_X:Zto X$, such that $mcirc f_X=f$.
Intuitively, this seems to me to capture the notion that “any information we need to pick an element of $X$, there is enough information in Y to do so”
Essentially, it seems to me that this generalizes the idea of a surjective function, but this concept is already generalized by “epimorphism”, whose definition is different.
Is my definition equivalent to that of epimorphism? If not, is there a term for my definition?
category-theory
asked Nov 21 '18 at 12:31
user56834
3,18821149
Suppose we have three objects $X,Y,Z$, and a morphism $m:Xto Y$.
Moreover, this morphism has the following property:
For any morphism $f:Zto Y$, there exists a morphism $f_X:Zto X$, such that $mcirc f_X=f$.
Intuitively, this seems to me to capture the notion that “any information we need to pick an element of $X$, there is enough information in Y to do so”
Essentially, it seems to me that this generalizes the idea of a surjective function, but this concept is already generalized by “epimorphism”, whose definition is different.
Is my definition equivalent to that of epimorphism? If not, is there a term for my definition?
category-theory
category-theory
asked Nov 21 '18 at 12:31
user56834
3,18821149
asked Nov 21 '18 at 12:31
user56834
3,18821149
asked Nov 21 '18 at 12:31
user56834
3,18821149
asked Nov 21 '18 at 12:31
user56834
3,18821149
asked Nov 21 '18 at 12:31
user56834
3,18821149
3,18821149
If $Z$ can be any object, then what you describe is just a split epimorphism; see this question
– Arnaud D.
Nov 21 '18 at 13:18
add a comment |
If $Z$ can be any object, then what you describe is just a split epimorphism; see this question
– Arnaud D.
Nov 21 '18 at 13:18
If $Z$ can be any object, then what you describe is just a split epimorphism; see this question
– Arnaud D.
Nov 21 '18 at 13:18
If $Z$ can be any object, then what you describe is just a split epimorphism; see this question
– Arnaud D.
Nov 21 '18 at 13:18
add a comment |
2 Answers
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Take $f=id_Y$ : this gives $i$ such that $mcirc i = id_Y$. Thus $m$ is not only an epimorphism, it is a split epimorphism.
Conversely, given $m:Xto Y$ a split epimorphism, i.e. an epimorphism with a section $i:Yto X$ such that $mcirc i =id_Y$, let $f:Zto Y$, and consider $f_X :=icirc f : Zto X$. Then $mcirc f_X = mcirc i circ f= id_Ycirc f = f$
answered Nov 21 '18 at 13:18
Max
12.9k11040
add a comment |
The natural way to name this property is "the object $Z$ has the left lifting property with respect to the morphism $m$". Indeed, if the category has an initial object, then the property you mentioned is equivalent to the left lifting property between $i_Z$ and $m$, where $i_Z$ is the unique morphism from the initial object to $Z$. If, in addition, the lifting morphism $f_X$ is unique for every $f$, then this property is called "the object $Z$ is orthogonal to $m$" (denoted by $Zperp m$, see definition 5.4.2 in F.Borceux, "Handbook of Categorical Algebra 1").
answered Nov 21 '18 at 13:59
Oskar
2,8331719
Wouldn't that only apply if there was a unique $f_X$?
– Arnaud D.
Nov 21 '18 at 14:07
@ArnaudD. You are right, I overlooked it in Borceux, thank you. Considering it, I edited the answer.
– Oskar
Nov 21 '18 at 14:43
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
oldest
votes
Take $f=id_Y$ : this gives $i$ such that $mcirc i = id_Y$. Thus $m$ is not only an epimorphism, it is a split epimorphism.
Conversely, given $m:Xto Y$ a split epimorphism, i.e. an epimorphism with a section $i:Yto X$ such that $mcirc i =id_Y$, let $f:Zto Y$, and consider $f_X :=icirc f : Zto X$. Then $mcirc f_X = mcirc i circ f= id_Ycirc f = f$
answered Nov 21 '18 at 13:18
Max
12.9k11040
add a comment |
Take $f=id_Y$ : this gives $i$ such that $mcirc i = id_Y$. Thus $m$ is not only an epimorphism, it is a split epimorphism.
Conversely, given $m:Xto Y$ a split epimorphism, i.e. an epimorphism with a section $i:Yto X$ such that $mcirc i =id_Y$, let $f:Zto Y$, and consider $f_X :=icirc f : Zto X$. Then $mcirc f_X = mcirc i circ f= id_Ycirc f = f$
answered Nov 21 '18 at 13:18
Max
12.9k11040
add a comment |
Take $f=id_Y$ : this gives $i$ such that $mcirc i = id_Y$. Thus $m$ is not only an epimorphism, it is a split epimorphism.
Conversely, given $m:Xto Y$ a split epimorphism, i.e. an epimorphism with a section $i:Yto X$ such that $mcirc i =id_Y$, let $f:Zto Y$, and consider $f_X :=icirc f : Zto X$. Then $mcirc f_X = mcirc i circ f= id_Ycirc f = f$
answered Nov 21 '18 at 13:18
Max
12.9k11040
Take $f=id_Y$ : this gives $i$ such that $mcirc i = id_Y$. Thus $m$ is not only an epimorphism, it is a split epimorphism.
Conversely, given $m:Xto Y$ a split epimorphism, i.e. an epimorphism with a section $i:Yto X$ such that $mcirc i =id_Y$, let $f:Zto Y$, and consider $f_X :=icirc f : Zto X$. Then $mcirc f_X = mcirc i circ f= id_Ycirc f = f$
answered Nov 21 '18 at 13:18
Max
12.9k11040
answered Nov 21 '18 at 13:18
Max
12.9k11040
answered Nov 21 '18 at 13:18
Max
12.9k11040
answered Nov 21 '18 at 13:18
Max
12.9k11040
12.9k11040
add a comment |
add a comment |
The natural way to name this property is "the object $Z$ has the left lifting property with respect to the morphism $m$". Indeed, if the category has an initial object, then the property you mentioned is equivalent to the left lifting property between $i_Z$ and $m$, where $i_Z$ is the unique morphism from the initial object to $Z$. If, in addition, the lifting morphism $f_X$ is unique for every $f$, then this property is called "the object $Z$ is orthogonal to $m$" (denoted by $Zperp m$, see definition 5.4.2 in F.Borceux, "Handbook of Categorical Algebra 1").
answered Nov 21 '18 at 13:59
Oskar
2,8331719
Wouldn't that only apply if there was a unique $f_X$?
– Arnaud D.
Nov 21 '18 at 14:07
@ArnaudD. You are right, I overlooked it in Borceux, thank you. Considering it, I edited the answer.
– Oskar
Nov 21 '18 at 14:43
add a comment |
The natural way to name this property is "the object $Z$ has the left lifting property with respect to the morphism $m$". Indeed, if the category has an initial object, then the property you mentioned is equivalent to the left lifting property between $i_Z$ and $m$, where $i_Z$ is the unique morphism from the initial object to $Z$. If, in addition, the lifting morphism $f_X$ is unique for every $f$, then this property is called "the object $Z$ is orthogonal to $m$" (denoted by $Zperp m$, see definition 5.4.2 in F.Borceux, "Handbook of Categorical Algebra 1").
answered Nov 21 '18 at 13:59
Oskar
2,8331719
Wouldn't that only apply if there was a unique $f_X$?
– Arnaud D.
Nov 21 '18 at 14:07
@ArnaudD. You are right, I overlooked it in Borceux, thank you. Considering it, I edited the answer.
– Oskar
Nov 21 '18 at 14:43
add a comment |
The natural way to name this property is "the object $Z$ has the left lifting property with respect to the morphism $m$". Indeed, if the category has an initial object, then the property you mentioned is equivalent to the left lifting property between $i_Z$ and $m$, where $i_Z$ is the unique morphism from the initial object to $Z$. If, in addition, the lifting morphism $f_X$ is unique for every $f$, then this property is called "the object $Z$ is orthogonal to $m$" (denoted by $Zperp m$, see definition 5.4.2 in F.Borceux, "Handbook of Categorical Algebra 1").
answered Nov 21 '18 at 13:59
Oskar
2,8331719
The natural way to name this property is "the object $Z$ has the left lifting property with respect to the morphism $m$". Indeed, if the category has an initial object, then the property you mentioned is equivalent to the left lifting property between $i_Z$ and $m$, where $i_Z$ is the unique morphism from the initial object to $Z$. If, in addition, the lifting morphism $f_X$ is unique for every $f$, then this property is called "the object $Z$ is orthogonal to $m$" (denoted by $Zperp m$, see definition 5.4.2 in F.Borceux, "Handbook of Categorical Algebra 1").
answered Nov 21 '18 at 13:59
Oskar
2,8331719
edited Nov 21 '18 at 14:36
answered Nov 21 '18 at 13:59
Oskar
2,8331719
answered Nov 21 '18 at 13:59
Oskar
2,8331719
answered Nov 21 '18 at 13:59
Oskar
2,8331719
2,8331719
Wouldn't that only apply if there was a unique $f_X$?
– Arnaud D.
Nov 21 '18 at 14:07
@ArnaudD. You are right, I overlooked it in Borceux, thank you. Considering it, I edited the answer.
– Oskar
Nov 21 '18 at 14:43
add a comment |
Wouldn't that only apply if there was a unique $f_X$?
– Arnaud D.
Nov 21 '18 at 14:07
@ArnaudD. You are right, I overlooked it in Borceux, thank you. Considering it, I edited the answer.
– Oskar
Nov 21 '18 at 14:43
Wouldn't that only apply if there was a unique $f_X$?
– Arnaud D.
Nov 21 '18 at 14:07
Wouldn't that only apply if there was a unique $f_X$?
– Arnaud D.
Nov 21 '18 at 14:07
@ArnaudD. You are right, I overlooked it in Borceux, thank you. Considering it, I edited the answer.
– Oskar
Nov 21 '18 at 14:43
@ArnaudD. You are right, I overlooked it in Borceux, thank you. Considering it, I edited the answer.
– Oskar
Nov 21 '18 at 14:43
add a comment |
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If $Z$ can be any object, then what you describe is just a split epimorphism; see this question
– Arnaud D.
Nov 21 '18 at 13:18