Term for this concept in category theory?











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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?










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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 at 13:18















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2
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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?










share|cite|improve this question






















  • If $Z$ can be any object, then what you describe is just a split epimorphism; see this question
    – Arnaud D.
    Nov 21 at 13:18













up vote
2
down vote

favorite









up vote
2
down vote

favorite











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?










share|cite|improve this question













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






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asked Nov 21 at 12:31









user56834

3,14321149




3,14321149












  • If $Z$ can be any object, then what you describe is just a split epimorphism; see this question
    – Arnaud D.
    Nov 21 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 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 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 at 13:18










2 Answers
2






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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 $mcirc f_X = mcirc i circ f= id_Ycirc f = f$






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    1
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    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").






    share|cite|improve this answer























    • Wouldn't that only apply if there was a unique $f_X$?
      – Arnaud D.
      Nov 21 at 14:07










    • @ArnaudD. You are right, I overlooked it in Borceux, thank you. Considering it, I edited the answer.
      – Oskar
      Nov 21 at 14:43











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    2 Answers
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    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 $mcirc f_X = mcirc i circ f= id_Ycirc f = f$






    share|cite|improve this answer

























      up vote
      3
      down vote













      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 $mcirc f_X = mcirc i circ f= id_Ycirc f = f$






      share|cite|improve this answer























        up vote
        3
        down vote










        up vote
        3
        down vote









        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 $mcirc f_X = mcirc i circ f= id_Ycirc f = f$






        share|cite|improve this answer












        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 $mcirc f_X = mcirc i circ f= id_Ycirc f = f$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 21 at 13:18









        Max

        12.5k11040




        12.5k11040






















            up vote
            1
            down vote













            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").






            share|cite|improve this answer























            • Wouldn't that only apply if there was a unique $f_X$?
              – Arnaud D.
              Nov 21 at 14:07










            • @ArnaudD. You are right, I overlooked it in Borceux, thank you. Considering it, I edited the answer.
              – Oskar
              Nov 21 at 14:43















            up vote
            1
            down vote













            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").






            share|cite|improve this answer























            • Wouldn't that only apply if there was a unique $f_X$?
              – Arnaud D.
              Nov 21 at 14:07










            • @ArnaudD. You are right, I overlooked it in Borceux, thank you. Considering it, I edited the answer.
              – Oskar
              Nov 21 at 14:43













            up vote
            1
            down vote










            up vote
            1
            down vote









            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").






            share|cite|improve this answer














            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").







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 21 at 14:36

























            answered Nov 21 at 13:59









            Oskar

            2,6961718




            2,6961718












            • Wouldn't that only apply if there was a unique $f_X$?
              – Arnaud D.
              Nov 21 at 14:07










            • @ArnaudD. You are right, I overlooked it in Borceux, thank you. Considering it, I edited the answer.
              – Oskar
              Nov 21 at 14:43


















            • Wouldn't that only apply if there was a unique $f_X$?
              – Arnaud D.
              Nov 21 at 14:07










            • @ArnaudD. You are right, I overlooked it in Borceux, thank you. Considering it, I edited the answer.
              – Oskar
              Nov 21 at 14:43
















            Wouldn't that only apply if there was a unique $f_X$?
            – Arnaud D.
            Nov 21 at 14:07




            Wouldn't that only apply if there was a unique $f_X$?
            – Arnaud D.
            Nov 21 at 14:07












            @ArnaudD. You are right, I overlooked it in Borceux, thank you. Considering it, I edited the answer.
            – Oskar
            Nov 21 at 14:43




            @ArnaudD. You are right, I overlooked it in Borceux, thank you. Considering it, I edited the answer.
            – Oskar
            Nov 21 at 14:43


















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