Commuting diagrams











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Reading "P. Smith, Category Theory - A Gentle Introduction". There it is defined




A category diagram commutes if for any two directed paths along edges
in the diagram from a node X to the node Y, the composite arrow along
the first path is equal to the composite arrow along the second path.




Let's assume we have the following diagram:



A ---f---> B
|
|
k g
|
|
/ /
C


Now, let's check if this diagram commutes. I will instantiate the above definition with the nodes A and C:




A category diagram commutes if for any two directed paths along edges in the diagram from the node A to the node C, the composite arrow along the first path is equal to the composite arrow along the second path.




We have two paths:



path_1 := A --f--> B --g--> C
path_2 := A --k--> C


And the corresponding composites:



composite along path_1 := g . f
composite along path_2 := k


So we have to show, that those two composites are equal, i.e., g . f = k.



I'd like to understand intuitively, why can I reason that those two composites are equal?



In the article it is stated, "the diagram commutes by the definition of composition".



OK, I know the fact, that due to cod f = dom g we know that there must exist the composite of f with g: g . f :: A -> C but how do I know that this composite is equal to k (which is the condition for the diagram to commute). In other words, why can't there be a "function" k from A to C not being equal to the "function" g .f?





Update as reply to Matthew Towers and Peter Smith (2018-11-25):



Matthew Towers and Peter Smith, please don't get me wrong here. I guess all this is very trivial to all of you. I just wanted to have a pragmatic (algorithmic) way of testing if a diagram commutes.



So, is the following correct:



If one wants to check if a diagram commutes,
then for /all/ paths from a node X to a node Y the
respective composites have to be checked for equality.



Right?



If yes, let me try to apply this to the first triangle of




the triangle made by the arrows $g circ fcolon A to C$, and $fcolon Ato B$ and $gcolon B to C$, and then also the triangle made by the arrows $h circ gcolon B to D$, and $gcolon B to C$ and $hcolon C to G$.




Please let me also re-ref your (adapted) definition:




A category diagram commutes if for any two directed paths along edges in the
diagram from a node X to a node Y, the composite arrow along the first path
is equal to the composite arrow along the second path.




So, with $X = A$ and $Y = C$ we have the following paths:




p1 := A --f--> B --g--> C
p2 := A --g∘f--> C



And the corresponding composites along those paths:




c1 along p1 := g∘f
c2 along p2 := g∘f



(Matthew Towers and Peter Smith: Is c2, the correct composite along the second path (even the path consists of only one edge)?)



So, in order to check if the first triangle commutes, for /all/ paths from A to C we have to check the respective composites are equal. In this case, there are only two paths, so we have to check $c1 = c2$, which is true.










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  • 1




    It's really unclear what you're asking here. Of course there could be lots of functions $A to C$ not equal to $g circ f$ in general, so it's certainly not true that every such diagram commutes - in fact it commutes if and only if $gf=k$, and that's really all you can say. Where exactly in the article is the sentence you are confused about?
    – Matthew Towers
    Nov 24 at 15:38






  • 2




    Hmmmm: not sure why I wrote "the node $Y$" rather than "a node $Y$". I'll mark that for correction!
    – Peter Smith
    Nov 24 at 20:18















up vote
0
down vote

favorite












Reading "P. Smith, Category Theory - A Gentle Introduction". There it is defined




A category diagram commutes if for any two directed paths along edges
in the diagram from a node X to the node Y, the composite arrow along
the first path is equal to the composite arrow along the second path.




Let's assume we have the following diagram:



A ---f---> B
|
|
k g
|
|
/ /
C


Now, let's check if this diagram commutes. I will instantiate the above definition with the nodes A and C:




A category diagram commutes if for any two directed paths along edges in the diagram from the node A to the node C, the composite arrow along the first path is equal to the composite arrow along the second path.




We have two paths:



path_1 := A --f--> B --g--> C
path_2 := A --k--> C


And the corresponding composites:



composite along path_1 := g . f
composite along path_2 := k


So we have to show, that those two composites are equal, i.e., g . f = k.



I'd like to understand intuitively, why can I reason that those two composites are equal?



In the article it is stated, "the diagram commutes by the definition of composition".



OK, I know the fact, that due to cod f = dom g we know that there must exist the composite of f with g: g . f :: A -> C but how do I know that this composite is equal to k (which is the condition for the diagram to commute). In other words, why can't there be a "function" k from A to C not being equal to the "function" g .f?





Update as reply to Matthew Towers and Peter Smith (2018-11-25):



Matthew Towers and Peter Smith, please don't get me wrong here. I guess all this is very trivial to all of you. I just wanted to have a pragmatic (algorithmic) way of testing if a diagram commutes.



So, is the following correct:



If one wants to check if a diagram commutes,
then for /all/ paths from a node X to a node Y the
respective composites have to be checked for equality.



Right?



If yes, let me try to apply this to the first triangle of




the triangle made by the arrows $g circ fcolon A to C$, and $fcolon Ato B$ and $gcolon B to C$, and then also the triangle made by the arrows $h circ gcolon B to D$, and $gcolon B to C$ and $hcolon C to G$.




Please let me also re-ref your (adapted) definition:




A category diagram commutes if for any two directed paths along edges in the
diagram from a node X to a node Y, the composite arrow along the first path
is equal to the composite arrow along the second path.




So, with $X = A$ and $Y = C$ we have the following paths:




p1 := A --f--> B --g--> C
p2 := A --g∘f--> C



And the corresponding composites along those paths:




c1 along p1 := g∘f
c2 along p2 := g∘f



(Matthew Towers and Peter Smith: Is c2, the correct composite along the second path (even the path consists of only one edge)?)



So, in order to check if the first triangle commutes, for /all/ paths from A to C we have to check the respective composites are equal. In this case, there are only two paths, so we have to check $c1 = c2$, which is true.










share|cite|improve this question




















  • 1




    It's really unclear what you're asking here. Of course there could be lots of functions $A to C$ not equal to $g circ f$ in general, so it's certainly not true that every such diagram commutes - in fact it commutes if and only if $gf=k$, and that's really all you can say. Where exactly in the article is the sentence you are confused about?
    – Matthew Towers
    Nov 24 at 15:38






  • 2




    Hmmmm: not sure why I wrote "the node $Y$" rather than "a node $Y$". I'll mark that for correction!
    – Peter Smith
    Nov 24 at 20:18













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Reading "P. Smith, Category Theory - A Gentle Introduction". There it is defined




A category diagram commutes if for any two directed paths along edges
in the diagram from a node X to the node Y, the composite arrow along
the first path is equal to the composite arrow along the second path.




Let's assume we have the following diagram:



A ---f---> B
|
|
k g
|
|
/ /
C


Now, let's check if this diagram commutes. I will instantiate the above definition with the nodes A and C:




A category diagram commutes if for any two directed paths along edges in the diagram from the node A to the node C, the composite arrow along the first path is equal to the composite arrow along the second path.




We have two paths:



path_1 := A --f--> B --g--> C
path_2 := A --k--> C


And the corresponding composites:



composite along path_1 := g . f
composite along path_2 := k


So we have to show, that those two composites are equal, i.e., g . f = k.



I'd like to understand intuitively, why can I reason that those two composites are equal?



In the article it is stated, "the diagram commutes by the definition of composition".



OK, I know the fact, that due to cod f = dom g we know that there must exist the composite of f with g: g . f :: A -> C but how do I know that this composite is equal to k (which is the condition for the diagram to commute). In other words, why can't there be a "function" k from A to C not being equal to the "function" g .f?





Update as reply to Matthew Towers and Peter Smith (2018-11-25):



Matthew Towers and Peter Smith, please don't get me wrong here. I guess all this is very trivial to all of you. I just wanted to have a pragmatic (algorithmic) way of testing if a diagram commutes.



So, is the following correct:



If one wants to check if a diagram commutes,
then for /all/ paths from a node X to a node Y the
respective composites have to be checked for equality.



Right?



If yes, let me try to apply this to the first triangle of




the triangle made by the arrows $g circ fcolon A to C$, and $fcolon Ato B$ and $gcolon B to C$, and then also the triangle made by the arrows $h circ gcolon B to D$, and $gcolon B to C$ and $hcolon C to G$.




Please let me also re-ref your (adapted) definition:




A category diagram commutes if for any two directed paths along edges in the
diagram from a node X to a node Y, the composite arrow along the first path
is equal to the composite arrow along the second path.




So, with $X = A$ and $Y = C$ we have the following paths:




p1 := A --f--> B --g--> C
p2 := A --g∘f--> C



And the corresponding composites along those paths:




c1 along p1 := g∘f
c2 along p2 := g∘f



(Matthew Towers and Peter Smith: Is c2, the correct composite along the second path (even the path consists of only one edge)?)



So, in order to check if the first triangle commutes, for /all/ paths from A to C we have to check the respective composites are equal. In this case, there are only two paths, so we have to check $c1 = c2$, which is true.










share|cite|improve this question















Reading "P. Smith, Category Theory - A Gentle Introduction". There it is defined




A category diagram commutes if for any two directed paths along edges
in the diagram from a node X to the node Y, the composite arrow along
the first path is equal to the composite arrow along the second path.




Let's assume we have the following diagram:



A ---f---> B
|
|
k g
|
|
/ /
C


Now, let's check if this diagram commutes. I will instantiate the above definition with the nodes A and C:




A category diagram commutes if for any two directed paths along edges in the diagram from the node A to the node C, the composite arrow along the first path is equal to the composite arrow along the second path.




We have two paths:



path_1 := A --f--> B --g--> C
path_2 := A --k--> C


And the corresponding composites:



composite along path_1 := g . f
composite along path_2 := k


So we have to show, that those two composites are equal, i.e., g . f = k.



I'd like to understand intuitively, why can I reason that those two composites are equal?



In the article it is stated, "the diagram commutes by the definition of composition".



OK, I know the fact, that due to cod f = dom g we know that there must exist the composite of f with g: g . f :: A -> C but how do I know that this composite is equal to k (which is the condition for the diagram to commute). In other words, why can't there be a "function" k from A to C not being equal to the "function" g .f?





Update as reply to Matthew Towers and Peter Smith (2018-11-25):



Matthew Towers and Peter Smith, please don't get me wrong here. I guess all this is very trivial to all of you. I just wanted to have a pragmatic (algorithmic) way of testing if a diagram commutes.



So, is the following correct:



If one wants to check if a diagram commutes,
then for /all/ paths from a node X to a node Y the
respective composites have to be checked for equality.



Right?



If yes, let me try to apply this to the first triangle of




the triangle made by the arrows $g circ fcolon A to C$, and $fcolon Ato B$ and $gcolon B to C$, and then also the triangle made by the arrows $h circ gcolon B to D$, and $gcolon B to C$ and $hcolon C to G$.




Please let me also re-ref your (adapted) definition:




A category diagram commutes if for any two directed paths along edges in the
diagram from a node X to a node Y, the composite arrow along the first path
is equal to the composite arrow along the second path.




So, with $X = A$ and $Y = C$ we have the following paths:




p1 := A --f--> B --g--> C
p2 := A --g∘f--> C



And the corresponding composites along those paths:




c1 along p1 := g∘f
c2 along p2 := g∘f



(Matthew Towers and Peter Smith: Is c2, the correct composite along the second path (even the path consists of only one edge)?)



So, in order to check if the first triangle commutes, for /all/ paths from A to C we have to check the respective composites are equal. In this case, there are only two paths, so we have to check $c1 = c2$, which is true.







category-theory






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edited Nov 27 at 7:34

























asked Nov 24 at 14:56









Normen Müller

1012




1012








  • 1




    It's really unclear what you're asking here. Of course there could be lots of functions $A to C$ not equal to $g circ f$ in general, so it's certainly not true that every such diagram commutes - in fact it commutes if and only if $gf=k$, and that's really all you can say. Where exactly in the article is the sentence you are confused about?
    – Matthew Towers
    Nov 24 at 15:38






  • 2




    Hmmmm: not sure why I wrote "the node $Y$" rather than "a node $Y$". I'll mark that for correction!
    – Peter Smith
    Nov 24 at 20:18














  • 1




    It's really unclear what you're asking here. Of course there could be lots of functions $A to C$ not equal to $g circ f$ in general, so it's certainly not true that every such diagram commutes - in fact it commutes if and only if $gf=k$, and that's really all you can say. Where exactly in the article is the sentence you are confused about?
    – Matthew Towers
    Nov 24 at 15:38






  • 2




    Hmmmm: not sure why I wrote "the node $Y$" rather than "a node $Y$". I'll mark that for correction!
    – Peter Smith
    Nov 24 at 20:18








1




1




It's really unclear what you're asking here. Of course there could be lots of functions $A to C$ not equal to $g circ f$ in general, so it's certainly not true that every such diagram commutes - in fact it commutes if and only if $gf=k$, and that's really all you can say. Where exactly in the article is the sentence you are confused about?
– Matthew Towers
Nov 24 at 15:38




It's really unclear what you're asking here. Of course there could be lots of functions $A to C$ not equal to $g circ f$ in general, so it's certainly not true that every such diagram commutes - in fact it commutes if and only if $gf=k$, and that's really all you can say. Where exactly in the article is the sentence you are confused about?
– Matthew Towers
Nov 24 at 15:38




2




2




Hmmmm: not sure why I wrote "the node $Y$" rather than "a node $Y$". I'll mark that for correction!
– Peter Smith
Nov 24 at 20:18




Hmmmm: not sure why I wrote "the node $Y$" rather than "a node $Y$". I'll mark that for correction!
– Peter Smith
Nov 24 at 20:18










4 Answers
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You are simply misreading.



What is clearly said to "commute[s] by definition of composition" is NOT the diagram you give, but rather each of the two triangles in the next diagram in the notes -- the triangle made by the arrows $g circ fcolon A to C$, and $fcolon A to B$ and $gcolon B to C$, and then also the triangle made by the arrows $h circ gcolon B to D$, and $gcolon B to C$ and $hcolon C to G$.



As others have pointed out, the diagram you give may or may not commute, depending on what $k$ is. But you didn't really need others to tell you that, as five lines after the words you quote I wrote, just to block this sort of misunderstanding, "And note too that merely drawing a diagram with different routes from e.g. $A$ to $D$ in the relevant category doesn't always mean that we have a commutative diagram -- the identity of the composites along the paths in each case has to be argued for!"






share|cite|improve this answer























  • Thanks for pointing out "[...] merely drawing a diagram with different routes [...] doesn't always mean that we have a commutative diagram -- the identity of the composites along the paths in each case has to be argued for!" I guess my phrase at the very top of this comment is equal to yours but more lax.
    – Normen Müller
    Nov 25 at 18:29










  • As suggest by other users, I removed my top level comment as it is not an answer. I rather updated my initial question. Could you have a look, please?
    – Normen Müller
    Nov 26 at 12:52


















up vote
3
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Indeed, there can be a $k:Ato C$ such that $kne gcirc f$, only that in such case the diagram is not called commutative.






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    up vote
    0
    down vote













    The set of arrows $text{hom}(A,C)$ is a set and asking if two elements of this set are equal or not has sense. But of course a priori you don't know if two elements are equal or not, it depends on the particular case.






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      up vote
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      down vote













      One of the purposes of category theory is to abstract the data of "set and maps". Set functions associate, so category morphisms are defined to. And when there are two compatible morphisms you can compose them, just like set maps.



      Saying $f = g$ as functions is the same as saying:



      $$
      begin{matrix}
      x & xrightarrow{f} & y \
      mid{} & & mid{} \
      x & xrightarrow{g} & y
      end{matrix}
      $$

      where the vertical arrows are identity maps. If you drew just two objects, $x, y$ and two arrows between them, you can also say that $f = g$ if and only if that diagram commutes.



      So whenever you have two compatible arrows: $a xrightarrow{f} b xrightarrow{g} c$ (think "set maps") you should always have a composition of them defined so you immediately get a commuting triangle.



      These commutative diagrams seem very important because they occur on every page of advanced math books. They are a very visual way to do math and should be used as much as possible. Long compositions look unreadable on a line of text, but in CD form they look elegant.






      share|cite|improve this answer





















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        4 Answers
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        You are simply misreading.



        What is clearly said to "commute[s] by definition of composition" is NOT the diagram you give, but rather each of the two triangles in the next diagram in the notes -- the triangle made by the arrows $g circ fcolon A to C$, and $fcolon A to B$ and $gcolon B to C$, and then also the triangle made by the arrows $h circ gcolon B to D$, and $gcolon B to C$ and $hcolon C to G$.



        As others have pointed out, the diagram you give may or may not commute, depending on what $k$ is. But you didn't really need others to tell you that, as five lines after the words you quote I wrote, just to block this sort of misunderstanding, "And note too that merely drawing a diagram with different routes from e.g. $A$ to $D$ in the relevant category doesn't always mean that we have a commutative diagram -- the identity of the composites along the paths in each case has to be argued for!"






        share|cite|improve this answer























        • Thanks for pointing out "[...] merely drawing a diagram with different routes [...] doesn't always mean that we have a commutative diagram -- the identity of the composites along the paths in each case has to be argued for!" I guess my phrase at the very top of this comment is equal to yours but more lax.
          – Normen Müller
          Nov 25 at 18:29










        • As suggest by other users, I removed my top level comment as it is not an answer. I rather updated my initial question. Could you have a look, please?
          – Normen Müller
          Nov 26 at 12:52















        up vote
        4
        down vote













        You are simply misreading.



        What is clearly said to "commute[s] by definition of composition" is NOT the diagram you give, but rather each of the two triangles in the next diagram in the notes -- the triangle made by the arrows $g circ fcolon A to C$, and $fcolon A to B$ and $gcolon B to C$, and then also the triangle made by the arrows $h circ gcolon B to D$, and $gcolon B to C$ and $hcolon C to G$.



        As others have pointed out, the diagram you give may or may not commute, depending on what $k$ is. But you didn't really need others to tell you that, as five lines after the words you quote I wrote, just to block this sort of misunderstanding, "And note too that merely drawing a diagram with different routes from e.g. $A$ to $D$ in the relevant category doesn't always mean that we have a commutative diagram -- the identity of the composites along the paths in each case has to be argued for!"






        share|cite|improve this answer























        • Thanks for pointing out "[...] merely drawing a diagram with different routes [...] doesn't always mean that we have a commutative diagram -- the identity of the composites along the paths in each case has to be argued for!" I guess my phrase at the very top of this comment is equal to yours but more lax.
          – Normen Müller
          Nov 25 at 18:29










        • As suggest by other users, I removed my top level comment as it is not an answer. I rather updated my initial question. Could you have a look, please?
          – Normen Müller
          Nov 26 at 12:52













        up vote
        4
        down vote










        up vote
        4
        down vote









        You are simply misreading.



        What is clearly said to "commute[s] by definition of composition" is NOT the diagram you give, but rather each of the two triangles in the next diagram in the notes -- the triangle made by the arrows $g circ fcolon A to C$, and $fcolon A to B$ and $gcolon B to C$, and then also the triangle made by the arrows $h circ gcolon B to D$, and $gcolon B to C$ and $hcolon C to G$.



        As others have pointed out, the diagram you give may or may not commute, depending on what $k$ is. But you didn't really need others to tell you that, as five lines after the words you quote I wrote, just to block this sort of misunderstanding, "And note too that merely drawing a diagram with different routes from e.g. $A$ to $D$ in the relevant category doesn't always mean that we have a commutative diagram -- the identity of the composites along the paths in each case has to be argued for!"






        share|cite|improve this answer














        You are simply misreading.



        What is clearly said to "commute[s] by definition of composition" is NOT the diagram you give, but rather each of the two triangles in the next diagram in the notes -- the triangle made by the arrows $g circ fcolon A to C$, and $fcolon A to B$ and $gcolon B to C$, and then also the triangle made by the arrows $h circ gcolon B to D$, and $gcolon B to C$ and $hcolon C to G$.



        As others have pointed out, the diagram you give may or may not commute, depending on what $k$ is. But you didn't really need others to tell you that, as five lines after the words you quote I wrote, just to block this sort of misunderstanding, "And note too that merely drawing a diagram with different routes from e.g. $A$ to $D$ in the relevant category doesn't always mean that we have a commutative diagram -- the identity of the composites along the paths in each case has to be argued for!"







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 24 at 20:36

























        answered Nov 24 at 20:28









        Peter Smith

        40.3k339118




        40.3k339118












        • Thanks for pointing out "[...] merely drawing a diagram with different routes [...] doesn't always mean that we have a commutative diagram -- the identity of the composites along the paths in each case has to be argued for!" I guess my phrase at the very top of this comment is equal to yours but more lax.
          – Normen Müller
          Nov 25 at 18:29










        • As suggest by other users, I removed my top level comment as it is not an answer. I rather updated my initial question. Could you have a look, please?
          – Normen Müller
          Nov 26 at 12:52


















        • Thanks for pointing out "[...] merely drawing a diagram with different routes [...] doesn't always mean that we have a commutative diagram -- the identity of the composites along the paths in each case has to be argued for!" I guess my phrase at the very top of this comment is equal to yours but more lax.
          – Normen Müller
          Nov 25 at 18:29










        • As suggest by other users, I removed my top level comment as it is not an answer. I rather updated my initial question. Could you have a look, please?
          – Normen Müller
          Nov 26 at 12:52
















        Thanks for pointing out "[...] merely drawing a diagram with different routes [...] doesn't always mean that we have a commutative diagram -- the identity of the composites along the paths in each case has to be argued for!" I guess my phrase at the very top of this comment is equal to yours but more lax.
        – Normen Müller
        Nov 25 at 18:29




        Thanks for pointing out "[...] merely drawing a diagram with different routes [...] doesn't always mean that we have a commutative diagram -- the identity of the composites along the paths in each case has to be argued for!" I guess my phrase at the very top of this comment is equal to yours but more lax.
        – Normen Müller
        Nov 25 at 18:29












        As suggest by other users, I removed my top level comment as it is not an answer. I rather updated my initial question. Could you have a look, please?
        – Normen Müller
        Nov 26 at 12:52




        As suggest by other users, I removed my top level comment as it is not an answer. I rather updated my initial question. Could you have a look, please?
        – Normen Müller
        Nov 26 at 12:52










        up vote
        3
        down vote













        Indeed, there can be a $k:Ato C$ such that $kne gcirc f$, only that in such case the diagram is not called commutative.






        share|cite|improve this answer

























          up vote
          3
          down vote













          Indeed, there can be a $k:Ato C$ such that $kne gcirc f$, only that in such case the diagram is not called commutative.






          share|cite|improve this answer























            up vote
            3
            down vote










            up vote
            3
            down vote









            Indeed, there can be a $k:Ato C$ such that $kne gcirc f$, only that in such case the diagram is not called commutative.






            share|cite|improve this answer












            Indeed, there can be a $k:Ato C$ such that $kne gcirc f$, only that in such case the diagram is not called commutative.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 24 at 16:05









            Berci

            59.2k23671




            59.2k23671






















                up vote
                0
                down vote













                The set of arrows $text{hom}(A,C)$ is a set and asking if two elements of this set are equal or not has sense. But of course a priori you don't know if two elements are equal or not, it depends on the particular case.






                share|cite|improve this answer

























                  up vote
                  0
                  down vote













                  The set of arrows $text{hom}(A,C)$ is a set and asking if two elements of this set are equal or not has sense. But of course a priori you don't know if two elements are equal or not, it depends on the particular case.






                  share|cite|improve this answer























                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    The set of arrows $text{hom}(A,C)$ is a set and asking if two elements of this set are equal or not has sense. But of course a priori you don't know if two elements are equal or not, it depends on the particular case.






                    share|cite|improve this answer












                    The set of arrows $text{hom}(A,C)$ is a set and asking if two elements of this set are equal or not has sense. But of course a priori you don't know if two elements are equal or not, it depends on the particular case.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 24 at 15:32









                    Dante Grevino

                    7547




                    7547






















                        up vote
                        0
                        down vote













                        One of the purposes of category theory is to abstract the data of "set and maps". Set functions associate, so category morphisms are defined to. And when there are two compatible morphisms you can compose them, just like set maps.



                        Saying $f = g$ as functions is the same as saying:



                        $$
                        begin{matrix}
                        x & xrightarrow{f} & y \
                        mid{} & & mid{} \
                        x & xrightarrow{g} & y
                        end{matrix}
                        $$

                        where the vertical arrows are identity maps. If you drew just two objects, $x, y$ and two arrows between them, you can also say that $f = g$ if and only if that diagram commutes.



                        So whenever you have two compatible arrows: $a xrightarrow{f} b xrightarrow{g} c$ (think "set maps") you should always have a composition of them defined so you immediately get a commuting triangle.



                        These commutative diagrams seem very important because they occur on every page of advanced math books. They are a very visual way to do math and should be used as much as possible. Long compositions look unreadable on a line of text, but in CD form they look elegant.






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          One of the purposes of category theory is to abstract the data of "set and maps". Set functions associate, so category morphisms are defined to. And when there are two compatible morphisms you can compose them, just like set maps.



                          Saying $f = g$ as functions is the same as saying:



                          $$
                          begin{matrix}
                          x & xrightarrow{f} & y \
                          mid{} & & mid{} \
                          x & xrightarrow{g} & y
                          end{matrix}
                          $$

                          where the vertical arrows are identity maps. If you drew just two objects, $x, y$ and two arrows between them, you can also say that $f = g$ if and only if that diagram commutes.



                          So whenever you have two compatible arrows: $a xrightarrow{f} b xrightarrow{g} c$ (think "set maps") you should always have a composition of them defined so you immediately get a commuting triangle.



                          These commutative diagrams seem very important because they occur on every page of advanced math books. They are a very visual way to do math and should be used as much as possible. Long compositions look unreadable on a line of text, but in CD form they look elegant.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            One of the purposes of category theory is to abstract the data of "set and maps". Set functions associate, so category morphisms are defined to. And when there are two compatible morphisms you can compose them, just like set maps.



                            Saying $f = g$ as functions is the same as saying:



                            $$
                            begin{matrix}
                            x & xrightarrow{f} & y \
                            mid{} & & mid{} \
                            x & xrightarrow{g} & y
                            end{matrix}
                            $$

                            where the vertical arrows are identity maps. If you drew just two objects, $x, y$ and two arrows between them, you can also say that $f = g$ if and only if that diagram commutes.



                            So whenever you have two compatible arrows: $a xrightarrow{f} b xrightarrow{g} c$ (think "set maps") you should always have a composition of them defined so you immediately get a commuting triangle.



                            These commutative diagrams seem very important because they occur on every page of advanced math books. They are a very visual way to do math and should be used as much as possible. Long compositions look unreadable on a line of text, but in CD form they look elegant.






                            share|cite|improve this answer












                            One of the purposes of category theory is to abstract the data of "set and maps". Set functions associate, so category morphisms are defined to. And when there are two compatible morphisms you can compose them, just like set maps.



                            Saying $f = g$ as functions is the same as saying:



                            $$
                            begin{matrix}
                            x & xrightarrow{f} & y \
                            mid{} & & mid{} \
                            x & xrightarrow{g} & y
                            end{matrix}
                            $$

                            where the vertical arrows are identity maps. If you drew just two objects, $x, y$ and two arrows between them, you can also say that $f = g$ if and only if that diagram commutes.



                            So whenever you have two compatible arrows: $a xrightarrow{f} b xrightarrow{g} c$ (think "set maps") you should always have a composition of them defined so you immediately get a commuting triangle.



                            These commutative diagrams seem very important because they occur on every page of advanced math books. They are a very visual way to do math and should be used as much as possible. Long compositions look unreadable on a line of text, but in CD form they look elegant.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 24 at 19:13









                            Roll up and smoke Adjoint

                            8,96252357




                            8,96252357






























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