Surjective module homomorphism? $0$ module homomorphism?












2












$begingroup$


I am trying to resolve an exercise and there are 2 point that are missing in order to finalize:



Suppose $A$, $B$, $C$, and $P$ are $R$-modules, and $f:A rightarrow B$ and $g:Brightarrow C$ are both $R$-module morphisms.



1) $forall phi : C rightarrow P$ morphism, if $phi circ g = 0 Rightarrow phi = 0$, for a morphism $phi : C rightarrow P$, does this imply that $g$ is surjective? Why?



2) If $phi circ g circ f = 0$ $ forall phi : C rightarrow P$ morphism does this mean that $g circ f = 0$? Why?






modules







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    In the first point, should the implication hold for some morphism $phi$ or for all morphisms $phi$? And is $P$ some fixed $R$-module, or should the implication hold for all $R$-modules $P$?
    $endgroup$
    – Servaes
    Dec 5 '18 at 19:30












  • $begingroup$
    for all! I will edit, thanks
    $endgroup$
    – roi_saumon
    Dec 5 '18 at 19:31


















2












$begingroup$


I am trying to resolve an exercise and there are 2 point that are missing in order to finalize:



Suppose $A$, $B$, $C$, and $P$ are $R$-modules, and $f:A rightarrow B$ and $g:Brightarrow C$ are both $R$-module morphisms.



1) $forall phi : C rightarrow P$ morphism, if $phi circ g = 0 Rightarrow phi = 0$, for a morphism $phi : C rightarrow P$, does this imply that $g$ is surjective? Why?



2) If $phi circ g circ f = 0$ $ forall phi : C rightarrow P$ morphism does this mean that $g circ f = 0$? Why?






modules







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    In the first point, should the implication hold for some morphism $phi$ or for all morphisms $phi$? And is $P$ some fixed $R$-module, or should the implication hold for all $R$-modules $P$?
    $endgroup$
    – Servaes
    Dec 5 '18 at 19:30












  • $begingroup$
    for all! I will edit, thanks
    $endgroup$
    – roi_saumon
    Dec 5 '18 at 19:31
















2












2








2





$begingroup$


I am trying to resolve an exercise and there are 2 point that are missing in order to finalize:



Suppose $A$, $B$, $C$, and $P$ are $R$-modules, and $f:A rightarrow B$ and $g:Brightarrow C$ are both $R$-module morphisms.



1) $forall phi : C rightarrow P$ morphism, if $phi circ g = 0 Rightarrow phi = 0$, for a morphism $phi : C rightarrow P$, does this imply that $g$ is surjective? Why?



2) If $phi circ g circ f = 0$ $ forall phi : C rightarrow P$ morphism does this mean that $g circ f = 0$? Why?






modules







share|cite|improve this question











$endgroup$




I am trying to resolve an exercise and there are 2 point that are missing in order to finalize:



Suppose $A$, $B$, $C$, and $P$ are $R$-modules, and $f:A rightarrow B$ and $g:Brightarrow C$ are both $R$-module morphisms.



1) $forall phi : C rightarrow P$ morphism, if $phi circ g = 0 Rightarrow phi = 0$, for a morphism $phi : C rightarrow P$, does this imply that $g$ is surjective? Why?



2) If $phi circ g circ f = 0$ $ forall phi : C rightarrow P$ morphism does this mean that $g circ f = 0$? Why?






modules




modules






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 19:35









zipirovich

11.3k11731




11.3k11731










asked Dec 5 '18 at 19:21









roi_saumonroi_saumon

57238




57238








  • 1




    $begingroup$
    In the first point, should the implication hold for some morphism $phi$ or for all morphisms $phi$? And is $P$ some fixed $R$-module, or should the implication hold for all $R$-modules $P$?
    $endgroup$
    – Servaes
    Dec 5 '18 at 19:30












  • $begingroup$
    for all! I will edit, thanks
    $endgroup$
    – roi_saumon
    Dec 5 '18 at 19:31
















  • 1




    $begingroup$
    In the first point, should the implication hold for some morphism $phi$ or for all morphisms $phi$? And is $P$ some fixed $R$-module, or should the implication hold for all $R$-modules $P$?
    $endgroup$
    – Servaes
    Dec 5 '18 at 19:30












  • $begingroup$
    for all! I will edit, thanks
    $endgroup$
    – roi_saumon
    Dec 5 '18 at 19:31










1




1




$begingroup$
In the first point, should the implication hold for some morphism $phi$ or for all morphisms $phi$? And is $P$ some fixed $R$-module, or should the implication hold for all $R$-modules $P$?
$endgroup$
– Servaes
Dec 5 '18 at 19:30






$begingroup$
In the first point, should the implication hold for some morphism $phi$ or for all morphisms $phi$? And is $P$ some fixed $R$-module, or should the implication hold for all $R$-modules $P$?
$endgroup$
– Servaes
Dec 5 '18 at 19:30














$begingroup$
for all! I will edit, thanks
$endgroup$
– roi_saumon
Dec 5 '18 at 19:31






$begingroup$
for all! I will edit, thanks
$endgroup$
– roi_saumon
Dec 5 '18 at 19:31












2 Answers
2






active

oldest

votes


















1












$begingroup$

For the first point consider for $phi$ the quotient morphism $pi: C longrightarrow operatorname{coker}g$.



For the second point consider for $phi$ the identity morphism $operatorname{id}: C longrightarrow C$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh, so nice! Especially the one with the cokernel. How did you think about that? Is it a standard trick?
    $endgroup$
    – roi_saumon
    Dec 5 '18 at 21:27










  • $begingroup$
    Well if $phicirc g=0$ then $operatorname{im}gsubsetkerphi$ and so $phi$ factors over the morphism $C longrightarrow operatorname{coker}g$. So the implication holds for all $phi$ if and only if it holds for the factor map $pi: C longrightarrow operatorname{coker}g$.
    $endgroup$
    – Servaes
    Dec 5 '18 at 21:42





















0












$begingroup$

1) No, not necessarily. Here's a counterexample. Let $B=mathbb{Z}$ and $C=P=mathbb{Q}$ as $R=mathbb{Z}$-modules. Further, let $g:Bto C$, i.e. $g:mathbb{Z}tomathbb{Q}$, be the inclusion map $g(n)=n$. Then for any $varphi:Cto P$, i.e. for any $varphi:mathbb{Q}tomathbb{Q}$, $varphicirc g=0$ implies $varphi=0$ (basically, because $varphicirc g=0$ implies $varphi(1)=0$ implies $varphi=0$). And yet, $gneq0$.



2) No, not necessarily. Here's a counterexample. Let $A=B=C=mathbb{Z}$ and $P=mathbb{Z}_2$ as $R=mathbb{Z}$-modules. Further, let $f:mathbb{Z}tomathbb{Z}$ be the identity map $f(n)=n$ and $g:mathbb{Z}tomathbb{Z}$ be multiplication by two map $g(n)=2n$. Then for any $varphi:mathbb{Z}tomathbb{Z}_2$. we have $varphicirc gcirc f=0$, even though $gcirc fneq0$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    For the first point consider for $phi$ the quotient morphism $pi: C longrightarrow operatorname{coker}g$.



    For the second point consider for $phi$ the identity morphism $operatorname{id}: C longrightarrow C$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Oh, so nice! Especially the one with the cokernel. How did you think about that? Is it a standard trick?
      $endgroup$
      – roi_saumon
      Dec 5 '18 at 21:27










    • $begingroup$
      Well if $phicirc g=0$ then $operatorname{im}gsubsetkerphi$ and so $phi$ factors over the morphism $C longrightarrow operatorname{coker}g$. So the implication holds for all $phi$ if and only if it holds for the factor map $pi: C longrightarrow operatorname{coker}g$.
      $endgroup$
      – Servaes
      Dec 5 '18 at 21:42


















    1












    $begingroup$

    For the first point consider for $phi$ the quotient morphism $pi: C longrightarrow operatorname{coker}g$.



    For the second point consider for $phi$ the identity morphism $operatorname{id}: C longrightarrow C$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Oh, so nice! Especially the one with the cokernel. How did you think about that? Is it a standard trick?
      $endgroup$
      – roi_saumon
      Dec 5 '18 at 21:27










    • $begingroup$
      Well if $phicirc g=0$ then $operatorname{im}gsubsetkerphi$ and so $phi$ factors over the morphism $C longrightarrow operatorname{coker}g$. So the implication holds for all $phi$ if and only if it holds for the factor map $pi: C longrightarrow operatorname{coker}g$.
      $endgroup$
      – Servaes
      Dec 5 '18 at 21:42
















    1












    1








    1





    $begingroup$

    For the first point consider for $phi$ the quotient morphism $pi: C longrightarrow operatorname{coker}g$.



    For the second point consider for $phi$ the identity morphism $operatorname{id}: C longrightarrow C$.






    share|cite|improve this answer











    $endgroup$



    For the first point consider for $phi$ the quotient morphism $pi: C longrightarrow operatorname{coker}g$.



    For the second point consider for $phi$ the identity morphism $operatorname{id}: C longrightarrow C$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 5 '18 at 21:42

























    answered Dec 5 '18 at 19:33









    ServaesServaes

    26.8k34098




    26.8k34098












    • $begingroup$
      Oh, so nice! Especially the one with the cokernel. How did you think about that? Is it a standard trick?
      $endgroup$
      – roi_saumon
      Dec 5 '18 at 21:27










    • $begingroup$
      Well if $phicirc g=0$ then $operatorname{im}gsubsetkerphi$ and so $phi$ factors over the morphism $C longrightarrow operatorname{coker}g$. So the implication holds for all $phi$ if and only if it holds for the factor map $pi: C longrightarrow operatorname{coker}g$.
      $endgroup$
      – Servaes
      Dec 5 '18 at 21:42




















    • $begingroup$
      Oh, so nice! Especially the one with the cokernel. How did you think about that? Is it a standard trick?
      $endgroup$
      – roi_saumon
      Dec 5 '18 at 21:27










    • $begingroup$
      Well if $phicirc g=0$ then $operatorname{im}gsubsetkerphi$ and so $phi$ factors over the morphism $C longrightarrow operatorname{coker}g$. So the implication holds for all $phi$ if and only if it holds for the factor map $pi: C longrightarrow operatorname{coker}g$.
      $endgroup$
      – Servaes
      Dec 5 '18 at 21:42


















    $begingroup$
    Oh, so nice! Especially the one with the cokernel. How did you think about that? Is it a standard trick?
    $endgroup$
    – roi_saumon
    Dec 5 '18 at 21:27




    $begingroup$
    Oh, so nice! Especially the one with the cokernel. How did you think about that? Is it a standard trick?
    $endgroup$
    – roi_saumon
    Dec 5 '18 at 21:27












    $begingroup$
    Well if $phicirc g=0$ then $operatorname{im}gsubsetkerphi$ and so $phi$ factors over the morphism $C longrightarrow operatorname{coker}g$. So the implication holds for all $phi$ if and only if it holds for the factor map $pi: C longrightarrow operatorname{coker}g$.
    $endgroup$
    – Servaes
    Dec 5 '18 at 21:42






    $begingroup$
    Well if $phicirc g=0$ then $operatorname{im}gsubsetkerphi$ and so $phi$ factors over the morphism $C longrightarrow operatorname{coker}g$. So the implication holds for all $phi$ if and only if it holds for the factor map $pi: C longrightarrow operatorname{coker}g$.
    $endgroup$
    – Servaes
    Dec 5 '18 at 21:42













    0












    $begingroup$

    1) No, not necessarily. Here's a counterexample. Let $B=mathbb{Z}$ and $C=P=mathbb{Q}$ as $R=mathbb{Z}$-modules. Further, let $g:Bto C$, i.e. $g:mathbb{Z}tomathbb{Q}$, be the inclusion map $g(n)=n$. Then for any $varphi:Cto P$, i.e. for any $varphi:mathbb{Q}tomathbb{Q}$, $varphicirc g=0$ implies $varphi=0$ (basically, because $varphicirc g=0$ implies $varphi(1)=0$ implies $varphi=0$). And yet, $gneq0$.



    2) No, not necessarily. Here's a counterexample. Let $A=B=C=mathbb{Z}$ and $P=mathbb{Z}_2$ as $R=mathbb{Z}$-modules. Further, let $f:mathbb{Z}tomathbb{Z}$ be the identity map $f(n)=n$ and $g:mathbb{Z}tomathbb{Z}$ be multiplication by two map $g(n)=2n$. Then for any $varphi:mathbb{Z}tomathbb{Z}_2$. we have $varphicirc gcirc f=0$, even though $gcirc fneq0$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      1) No, not necessarily. Here's a counterexample. Let $B=mathbb{Z}$ and $C=P=mathbb{Q}$ as $R=mathbb{Z}$-modules. Further, let $g:Bto C$, i.e. $g:mathbb{Z}tomathbb{Q}$, be the inclusion map $g(n)=n$. Then for any $varphi:Cto P$, i.e. for any $varphi:mathbb{Q}tomathbb{Q}$, $varphicirc g=0$ implies $varphi=0$ (basically, because $varphicirc g=0$ implies $varphi(1)=0$ implies $varphi=0$). And yet, $gneq0$.



      2) No, not necessarily. Here's a counterexample. Let $A=B=C=mathbb{Z}$ and $P=mathbb{Z}_2$ as $R=mathbb{Z}$-modules. Further, let $f:mathbb{Z}tomathbb{Z}$ be the identity map $f(n)=n$ and $g:mathbb{Z}tomathbb{Z}$ be multiplication by two map $g(n)=2n$. Then for any $varphi:mathbb{Z}tomathbb{Z}_2$. we have $varphicirc gcirc f=0$, even though $gcirc fneq0$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        1) No, not necessarily. Here's a counterexample. Let $B=mathbb{Z}$ and $C=P=mathbb{Q}$ as $R=mathbb{Z}$-modules. Further, let $g:Bto C$, i.e. $g:mathbb{Z}tomathbb{Q}$, be the inclusion map $g(n)=n$. Then for any $varphi:Cto P$, i.e. for any $varphi:mathbb{Q}tomathbb{Q}$, $varphicirc g=0$ implies $varphi=0$ (basically, because $varphicirc g=0$ implies $varphi(1)=0$ implies $varphi=0$). And yet, $gneq0$.



        2) No, not necessarily. Here's a counterexample. Let $A=B=C=mathbb{Z}$ and $P=mathbb{Z}_2$ as $R=mathbb{Z}$-modules. Further, let $f:mathbb{Z}tomathbb{Z}$ be the identity map $f(n)=n$ and $g:mathbb{Z}tomathbb{Z}$ be multiplication by two map $g(n)=2n$. Then for any $varphi:mathbb{Z}tomathbb{Z}_2$. we have $varphicirc gcirc f=0$, even though $gcirc fneq0$.






        share|cite|improve this answer









        $endgroup$



        1) No, not necessarily. Here's a counterexample. Let $B=mathbb{Z}$ and $C=P=mathbb{Q}$ as $R=mathbb{Z}$-modules. Further, let $g:Bto C$, i.e. $g:mathbb{Z}tomathbb{Q}$, be the inclusion map $g(n)=n$. Then for any $varphi:Cto P$, i.e. for any $varphi:mathbb{Q}tomathbb{Q}$, $varphicirc g=0$ implies $varphi=0$ (basically, because $varphicirc g=0$ implies $varphi(1)=0$ implies $varphi=0$). And yet, $gneq0$.



        2) No, not necessarily. Here's a counterexample. Let $A=B=C=mathbb{Z}$ and $P=mathbb{Z}_2$ as $R=mathbb{Z}$-modules. Further, let $f:mathbb{Z}tomathbb{Z}$ be the identity map $f(n)=n$ and $g:mathbb{Z}tomathbb{Z}$ be multiplication by two map $g(n)=2n$. Then for any $varphi:mathbb{Z}tomathbb{Z}_2$. we have $varphicirc gcirc f=0$, even though $gcirc fneq0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 21:54









        zipirovichzipirovich

        11.3k11731




        11.3k11731






























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