Commutative diagram question












7












$begingroup$


enter image description here



Suppose outer square in above diagram is comutative i.e., $ccirc qcirc p=scirc rcirc a$



Further, suppose right side square is commutative i.e., $ccirc q=scirc b$?



Does it imply left side square is commutative i.e., $bcirc p=rcirc a$??



If $s$ has an inverse, it follows trivially. Are there other conditions that confirm this?










share|cite|improve this question











$endgroup$

















    7












    $begingroup$


    enter image description here



    Suppose outer square in above diagram is comutative i.e., $ccirc qcirc p=scirc rcirc a$



    Further, suppose right side square is commutative i.e., $ccirc q=scirc b$?



    Does it imply left side square is commutative i.e., $bcirc p=rcirc a$??



    If $s$ has an inverse, it follows trivially. Are there other conditions that confirm this?










    share|cite|improve this question











    $endgroup$















      7












      7








      7


      1



      $begingroup$


      enter image description here



      Suppose outer square in above diagram is comutative i.e., $ccirc qcirc p=scirc rcirc a$



      Further, suppose right side square is commutative i.e., $ccirc q=scirc b$?



      Does it imply left side square is commutative i.e., $bcirc p=rcirc a$??



      If $s$ has an inverse, it follows trivially. Are there other conditions that confirm this?










      share|cite|improve this question











      $endgroup$




      enter image description here



      Suppose outer square in above diagram is comutative i.e., $ccirc qcirc p=scirc rcirc a$



      Further, suppose right side square is commutative i.e., $ccirc q=scirc b$?



      Does it imply left side square is commutative i.e., $bcirc p=rcirc a$??



      If $s$ has an inverse, it follows trivially. Are there other conditions that confirm this?







      category-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 5 at 18:40







      Praphulla Koushik

















      asked Jan 5 at 15:37









      Praphulla KoushikPraphulla Koushik

      26917




      26917






















          1 Answer
          1






          active

          oldest

          votes


















          8












          $begingroup$

          In general, no. If $F$ is a terminal object, then the whole diagram and the right square commute automatically (since all composites to $F$ are the unique morphism to the terminal object), and then the left-hand square can be anything at all, commutative or otherwise.



          If $s$ is monic, then the left-hand square commutes, since
          $$s circ r circ a = c circ q circ p = s circ b circ p quad Rightarrow quad r circ a = b circ p$$



          In fact, $s$ being monic is, in a sense, equivalent to the condition in your question. Indeed, suppose $s$ is fixed and, for all such diagrams with $s$ in the bottom-right, commutativity of the outer and right squares implies commutativity of the left-hand square. We prove that $s$ is monic.



          So let $f,g : B to E$ and suppose $s circ f = s circ g$. Form the diagram with $a=p=q=mathrm{id}_B$, $b=f$, $r=g$ and $c = s circ f$. Then the right-hand square commutes trivially and the outer square commutes since $s circ f = s circ g$. Hence the left-hand square commutes, and so $f=g$. So $s$ is monic.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            :) If $s$ is monic, it follows trivially just because of the definition.. I should have added this along with "If $s$ is invertible, it follows immediately"...Thanks anyways :)
            $endgroup$
            – Praphulla Koushik
            Jan 5 at 15:58










          • $begingroup$
            @PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a left-inverse—that's split monic). I've updated my answer to prove that, relative to fixed $s$, the condition in your question is equivalent to the assertion that $s$ is monic.
            $endgroup$
            – Clive Newstead
            Jan 5 at 16:00












          • $begingroup$
            I dont know whats split monic.. I only know monic and that means it has left inverse.. So, I said for monic it follows... what is monic for you if not left invertible arrow?
            $endgroup$
            – Praphulla Koushik
            Jan 5 at 16:50






          • 1




            $begingroup$
            @PraphullaKoushik: Having a left inverse implies monic, but being monic does not imply having a left inverse. See monic versus split monic on nLab, and also on Wikipedia. In some categories (e.g. the category of sets) all monomorphisms are split, but that is not the case in all categories. I don't know of any credible source on category theory that defines 'monomorphism' to mean 'has a left inverse'.
            $endgroup$
            – Clive Newstead
            Jan 5 at 16:52












          • $begingroup$
            Ok Ok... An arrow $g$ is monic if $gcirc f=gcirc h$ implies $f=h$... An arrow $g$ is called split monic if there exists an arrow $t$ such that $tcirc g=1$... Suppose $g$ is split monic and $gcirc f=gcirc h$.. We have $tcirc gcirc f=tcirc gcirc h$.. As $tcirc g=1$, we have $f=h$... YOu are only using monic in your answer... Though this is not what I expected, this is interesting :) :) Thanks, I learned some new terminology.. :) In some sense, having a section of $s$ confirms what I asked.. :)
            $endgroup$
            – Praphulla Koushik
            Jan 5 at 16:59













          Your Answer





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          1 Answer
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          active

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          8












          $begingroup$

          In general, no. If $F$ is a terminal object, then the whole diagram and the right square commute automatically (since all composites to $F$ are the unique morphism to the terminal object), and then the left-hand square can be anything at all, commutative or otherwise.



          If $s$ is monic, then the left-hand square commutes, since
          $$s circ r circ a = c circ q circ p = s circ b circ p quad Rightarrow quad r circ a = b circ p$$



          In fact, $s$ being monic is, in a sense, equivalent to the condition in your question. Indeed, suppose $s$ is fixed and, for all such diagrams with $s$ in the bottom-right, commutativity of the outer and right squares implies commutativity of the left-hand square. We prove that $s$ is monic.



          So let $f,g : B to E$ and suppose $s circ f = s circ g$. Form the diagram with $a=p=q=mathrm{id}_B$, $b=f$, $r=g$ and $c = s circ f$. Then the right-hand square commutes trivially and the outer square commutes since $s circ f = s circ g$. Hence the left-hand square commutes, and so $f=g$. So $s$ is monic.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            :) If $s$ is monic, it follows trivially just because of the definition.. I should have added this along with "If $s$ is invertible, it follows immediately"...Thanks anyways :)
            $endgroup$
            – Praphulla Koushik
            Jan 5 at 15:58










          • $begingroup$
            @PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a left-inverse—that's split monic). I've updated my answer to prove that, relative to fixed $s$, the condition in your question is equivalent to the assertion that $s$ is monic.
            $endgroup$
            – Clive Newstead
            Jan 5 at 16:00












          • $begingroup$
            I dont know whats split monic.. I only know monic and that means it has left inverse.. So, I said for monic it follows... what is monic for you if not left invertible arrow?
            $endgroup$
            – Praphulla Koushik
            Jan 5 at 16:50






          • 1




            $begingroup$
            @PraphullaKoushik: Having a left inverse implies monic, but being monic does not imply having a left inverse. See monic versus split monic on nLab, and also on Wikipedia. In some categories (e.g. the category of sets) all monomorphisms are split, but that is not the case in all categories. I don't know of any credible source on category theory that defines 'monomorphism' to mean 'has a left inverse'.
            $endgroup$
            – Clive Newstead
            Jan 5 at 16:52












          • $begingroup$
            Ok Ok... An arrow $g$ is monic if $gcirc f=gcirc h$ implies $f=h$... An arrow $g$ is called split monic if there exists an arrow $t$ such that $tcirc g=1$... Suppose $g$ is split monic and $gcirc f=gcirc h$.. We have $tcirc gcirc f=tcirc gcirc h$.. As $tcirc g=1$, we have $f=h$... YOu are only using monic in your answer... Though this is not what I expected, this is interesting :) :) Thanks, I learned some new terminology.. :) In some sense, having a section of $s$ confirms what I asked.. :)
            $endgroup$
            – Praphulla Koushik
            Jan 5 at 16:59


















          8












          $begingroup$

          In general, no. If $F$ is a terminal object, then the whole diagram and the right square commute automatically (since all composites to $F$ are the unique morphism to the terminal object), and then the left-hand square can be anything at all, commutative or otherwise.



          If $s$ is monic, then the left-hand square commutes, since
          $$s circ r circ a = c circ q circ p = s circ b circ p quad Rightarrow quad r circ a = b circ p$$



          In fact, $s$ being monic is, in a sense, equivalent to the condition in your question. Indeed, suppose $s$ is fixed and, for all such diagrams with $s$ in the bottom-right, commutativity of the outer and right squares implies commutativity of the left-hand square. We prove that $s$ is monic.



          So let $f,g : B to E$ and suppose $s circ f = s circ g$. Form the diagram with $a=p=q=mathrm{id}_B$, $b=f$, $r=g$ and $c = s circ f$. Then the right-hand square commutes trivially and the outer square commutes since $s circ f = s circ g$. Hence the left-hand square commutes, and so $f=g$. So $s$ is monic.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            :) If $s$ is monic, it follows trivially just because of the definition.. I should have added this along with "If $s$ is invertible, it follows immediately"...Thanks anyways :)
            $endgroup$
            – Praphulla Koushik
            Jan 5 at 15:58










          • $begingroup$
            @PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a left-inverse—that's split monic). I've updated my answer to prove that, relative to fixed $s$, the condition in your question is equivalent to the assertion that $s$ is monic.
            $endgroup$
            – Clive Newstead
            Jan 5 at 16:00












          • $begingroup$
            I dont know whats split monic.. I only know monic and that means it has left inverse.. So, I said for monic it follows... what is monic for you if not left invertible arrow?
            $endgroup$
            – Praphulla Koushik
            Jan 5 at 16:50






          • 1




            $begingroup$
            @PraphullaKoushik: Having a left inverse implies monic, but being monic does not imply having a left inverse. See monic versus split monic on nLab, and also on Wikipedia. In some categories (e.g. the category of sets) all monomorphisms are split, but that is not the case in all categories. I don't know of any credible source on category theory that defines 'monomorphism' to mean 'has a left inverse'.
            $endgroup$
            – Clive Newstead
            Jan 5 at 16:52












          • $begingroup$
            Ok Ok... An arrow $g$ is monic if $gcirc f=gcirc h$ implies $f=h$... An arrow $g$ is called split monic if there exists an arrow $t$ such that $tcirc g=1$... Suppose $g$ is split monic and $gcirc f=gcirc h$.. We have $tcirc gcirc f=tcirc gcirc h$.. As $tcirc g=1$, we have $f=h$... YOu are only using monic in your answer... Though this is not what I expected, this is interesting :) :) Thanks, I learned some new terminology.. :) In some sense, having a section of $s$ confirms what I asked.. :)
            $endgroup$
            – Praphulla Koushik
            Jan 5 at 16:59
















          8












          8








          8





          $begingroup$

          In general, no. If $F$ is a terminal object, then the whole diagram and the right square commute automatically (since all composites to $F$ are the unique morphism to the terminal object), and then the left-hand square can be anything at all, commutative or otherwise.



          If $s$ is monic, then the left-hand square commutes, since
          $$s circ r circ a = c circ q circ p = s circ b circ p quad Rightarrow quad r circ a = b circ p$$



          In fact, $s$ being monic is, in a sense, equivalent to the condition in your question. Indeed, suppose $s$ is fixed and, for all such diagrams with $s$ in the bottom-right, commutativity of the outer and right squares implies commutativity of the left-hand square. We prove that $s$ is monic.



          So let $f,g : B to E$ and suppose $s circ f = s circ g$. Form the diagram with $a=p=q=mathrm{id}_B$, $b=f$, $r=g$ and $c = s circ f$. Then the right-hand square commutes trivially and the outer square commutes since $s circ f = s circ g$. Hence the left-hand square commutes, and so $f=g$. So $s$ is monic.






          share|cite|improve this answer











          $endgroup$



          In general, no. If $F$ is a terminal object, then the whole diagram and the right square commute automatically (since all composites to $F$ are the unique morphism to the terminal object), and then the left-hand square can be anything at all, commutative or otherwise.



          If $s$ is monic, then the left-hand square commutes, since
          $$s circ r circ a = c circ q circ p = s circ b circ p quad Rightarrow quad r circ a = b circ p$$



          In fact, $s$ being monic is, in a sense, equivalent to the condition in your question. Indeed, suppose $s$ is fixed and, for all such diagrams with $s$ in the bottom-right, commutativity of the outer and right squares implies commutativity of the left-hand square. We prove that $s$ is monic.



          So let $f,g : B to E$ and suppose $s circ f = s circ g$. Form the diagram with $a=p=q=mathrm{id}_B$, $b=f$, $r=g$ and $c = s circ f$. Then the right-hand square commutes trivially and the outer square commutes since $s circ f = s circ g$. Hence the left-hand square commutes, and so $f=g$. So $s$ is monic.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 5 at 16:00

























          answered Jan 5 at 15:53









          Clive NewsteadClive Newstead

          50.9k474133




          50.9k474133












          • $begingroup$
            :) If $s$ is monic, it follows trivially just because of the definition.. I should have added this along with "If $s$ is invertible, it follows immediately"...Thanks anyways :)
            $endgroup$
            – Praphulla Koushik
            Jan 5 at 15:58










          • $begingroup$
            @PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a left-inverse—that's split monic). I've updated my answer to prove that, relative to fixed $s$, the condition in your question is equivalent to the assertion that $s$ is monic.
            $endgroup$
            – Clive Newstead
            Jan 5 at 16:00












          • $begingroup$
            I dont know whats split monic.. I only know monic and that means it has left inverse.. So, I said for monic it follows... what is monic for you if not left invertible arrow?
            $endgroup$
            – Praphulla Koushik
            Jan 5 at 16:50






          • 1




            $begingroup$
            @PraphullaKoushik: Having a left inverse implies monic, but being monic does not imply having a left inverse. See monic versus split monic on nLab, and also on Wikipedia. In some categories (e.g. the category of sets) all monomorphisms are split, but that is not the case in all categories. I don't know of any credible source on category theory that defines 'monomorphism' to mean 'has a left inverse'.
            $endgroup$
            – Clive Newstead
            Jan 5 at 16:52












          • $begingroup$
            Ok Ok... An arrow $g$ is monic if $gcirc f=gcirc h$ implies $f=h$... An arrow $g$ is called split monic if there exists an arrow $t$ such that $tcirc g=1$... Suppose $g$ is split monic and $gcirc f=gcirc h$.. We have $tcirc gcirc f=tcirc gcirc h$.. As $tcirc g=1$, we have $f=h$... YOu are only using monic in your answer... Though this is not what I expected, this is interesting :) :) Thanks, I learned some new terminology.. :) In some sense, having a section of $s$ confirms what I asked.. :)
            $endgroup$
            – Praphulla Koushik
            Jan 5 at 16:59




















          • $begingroup$
            :) If $s$ is monic, it follows trivially just because of the definition.. I should have added this along with "If $s$ is invertible, it follows immediately"...Thanks anyways :)
            $endgroup$
            – Praphulla Koushik
            Jan 5 at 15:58










          • $begingroup$
            @PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a left-inverse—that's split monic). I've updated my answer to prove that, relative to fixed $s$, the condition in your question is equivalent to the assertion that $s$ is monic.
            $endgroup$
            – Clive Newstead
            Jan 5 at 16:00












          • $begingroup$
            I dont know whats split monic.. I only know monic and that means it has left inverse.. So, I said for monic it follows... what is monic for you if not left invertible arrow?
            $endgroup$
            – Praphulla Koushik
            Jan 5 at 16:50






          • 1




            $begingroup$
            @PraphullaKoushik: Having a left inverse implies monic, but being monic does not imply having a left inverse. See monic versus split monic on nLab, and also on Wikipedia. In some categories (e.g. the category of sets) all monomorphisms are split, but that is not the case in all categories. I don't know of any credible source on category theory that defines 'monomorphism' to mean 'has a left inverse'.
            $endgroup$
            – Clive Newstead
            Jan 5 at 16:52












          • $begingroup$
            Ok Ok... An arrow $g$ is monic if $gcirc f=gcirc h$ implies $f=h$... An arrow $g$ is called split monic if there exists an arrow $t$ such that $tcirc g=1$... Suppose $g$ is split monic and $gcirc f=gcirc h$.. We have $tcirc gcirc f=tcirc gcirc h$.. As $tcirc g=1$, we have $f=h$... YOu are only using monic in your answer... Though this is not what I expected, this is interesting :) :) Thanks, I learned some new terminology.. :) In some sense, having a section of $s$ confirms what I asked.. :)
            $endgroup$
            – Praphulla Koushik
            Jan 5 at 16:59


















          $begingroup$
          :) If $s$ is monic, it follows trivially just because of the definition.. I should have added this along with "If $s$ is invertible, it follows immediately"...Thanks anyways :)
          $endgroup$
          – Praphulla Koushik
          Jan 5 at 15:58




          $begingroup$
          :) If $s$ is monic, it follows trivially just because of the definition.. I should have added this along with "If $s$ is invertible, it follows immediately"...Thanks anyways :)
          $endgroup$
          – Praphulla Koushik
          Jan 5 at 15:58












          $begingroup$
          @PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a left-inverse—that's split monic). I've updated my answer to prove that, relative to fixed $s$, the condition in your question is equivalent to the assertion that $s$ is monic.
          $endgroup$
          – Clive Newstead
          Jan 5 at 16:00






          $begingroup$
          @PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a left-inverse—that's split monic). I've updated my answer to prove that, relative to fixed $s$, the condition in your question is equivalent to the assertion that $s$ is monic.
          $endgroup$
          – Clive Newstead
          Jan 5 at 16:00














          $begingroup$
          I dont know whats split monic.. I only know monic and that means it has left inverse.. So, I said for monic it follows... what is monic for you if not left invertible arrow?
          $endgroup$
          – Praphulla Koushik
          Jan 5 at 16:50




          $begingroup$
          I dont know whats split monic.. I only know monic and that means it has left inverse.. So, I said for monic it follows... what is monic for you if not left invertible arrow?
          $endgroup$
          – Praphulla Koushik
          Jan 5 at 16:50




          1




          1




          $begingroup$
          @PraphullaKoushik: Having a left inverse implies monic, but being monic does not imply having a left inverse. See monic versus split monic on nLab, and also on Wikipedia. In some categories (e.g. the category of sets) all monomorphisms are split, but that is not the case in all categories. I don't know of any credible source on category theory that defines 'monomorphism' to mean 'has a left inverse'.
          $endgroup$
          – Clive Newstead
          Jan 5 at 16:52






          $begingroup$
          @PraphullaKoushik: Having a left inverse implies monic, but being monic does not imply having a left inverse. See monic versus split monic on nLab, and also on Wikipedia. In some categories (e.g. the category of sets) all monomorphisms are split, but that is not the case in all categories. I don't know of any credible source on category theory that defines 'monomorphism' to mean 'has a left inverse'.
          $endgroup$
          – Clive Newstead
          Jan 5 at 16:52














          $begingroup$
          Ok Ok... An arrow $g$ is monic if $gcirc f=gcirc h$ implies $f=h$... An arrow $g$ is called split monic if there exists an arrow $t$ such that $tcirc g=1$... Suppose $g$ is split monic and $gcirc f=gcirc h$.. We have $tcirc gcirc f=tcirc gcirc h$.. As $tcirc g=1$, we have $f=h$... YOu are only using monic in your answer... Though this is not what I expected, this is interesting :) :) Thanks, I learned some new terminology.. :) In some sense, having a section of $s$ confirms what I asked.. :)
          $endgroup$
          – Praphulla Koushik
          Jan 5 at 16:59






          $begingroup$
          Ok Ok... An arrow $g$ is monic if $gcirc f=gcirc h$ implies $f=h$... An arrow $g$ is called split monic if there exists an arrow $t$ such that $tcirc g=1$... Suppose $g$ is split monic and $gcirc f=gcirc h$.. We have $tcirc gcirc f=tcirc gcirc h$.. As $tcirc g=1$, we have $f=h$... YOu are only using monic in your answer... Though this is not what I expected, this is interesting :) :) Thanks, I learned some new terminology.. :) In some sense, having a section of $s$ confirms what I asked.. :)
          $endgroup$
          – Praphulla Koushik
          Jan 5 at 16:59




















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