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-2

$begingroup$

Let $R$ be a ring, $M, N$ left $R$-modules, $phi:Mrightarrow N$ a
homomorphism.

Prove that $phi$ is bijective $Leftrightarrow$ exists a homomorphism $psi: Nrightarrow M$ such that $phicircpsi=id_N, psicircphi=id_M$.

abstract-algebra category-theory modules

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asked Mar 9 at 16:20

ToidiToidi

43

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put on hold as off-topic by José Carlos Santos, John Omielan, Jyrki Lahtonen, user26857, Peter Foreman Mar 9 at 23:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, John Omielan, Jyrki Lahtonen, user26857, Peter Foreman

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The 2 given equations are the definitions of injections and surjections.
  $endgroup$
  – UnexpectedExpectation
  Mar 9 at 16:42
  
  
  

  
  
  
  

add a comment | 

-2

$begingroup$

Let $R$ be a ring, $M, N$ left $R$-modules, $phi:Mrightarrow N$ a
homomorphism.

Prove that $phi$ is bijective $Leftrightarrow$ exists a homomorphism $psi: Nrightarrow M$ such that $phicircpsi=id_N, psicircphi=id_M$.

abstract-algebra category-theory modules

share|cite|improve this question

asked Mar 9 at 16:20

ToidiToidi

43

New contributor

Toidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by José Carlos Santos, John Omielan, Jyrki Lahtonen, user26857, Peter Foreman Mar 9 at 23:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, John Omielan, Jyrki Lahtonen, user26857, Peter Foreman

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The 2 given equations are the definitions of injections and surjections.
  $endgroup$
  – UnexpectedExpectation
  Mar 9 at 16:42
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $R$ be a ring, $M, N$ left $R$-modules, $phi:Mrightarrow N$ a
homomorphism.

Prove that $phi$ is bijective $Leftrightarrow$ exists a homomorphism $psi: Nrightarrow M$ such that $phicircpsi=id_N, psicircphi=id_M$.

abstract-algebra category-theory modules

share|cite|improve this question

asked Mar 9 at 16:20

ToidiToidi

43

New contributor

Toidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked Mar 9 at 16:20

ToidiToidi

43

New contributor

Toidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked Mar 9 at 16:20

ToidiToidi

43

New contributor

Toidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 9 at 16:20

ToidiToidi

43

New contributor

Toidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 9 at 16:20

ToidiToidi

43

1 Answer
1

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0

$begingroup$

$$phi:Mrightarrow N\psi:Nrightarrow M$$

• $(Rightarrow)$ Define $psi$ in this way: Take $nin N$, then since $phi$ is surjective there is some $min M$ such that $phi(m)=n$. Therefore, let $psi(n)=m$. This $m$ is unique since $phi$ is injective, because if there were two different values $$psi(n)=mneq m'=psi(n')$$ then $phi(m)neq phi(m')$ and therefore $psi$ is well-defined.
  You can now check that indeed $phicircpsi=text{id}_N$, $psicircphi=text{id}_M$ and that $psi$ is an $R$-module homomorphism.




• $(Leftarrow)$ for this part you need to show that $phi$ is both injective (that is $phi(m)=phi(m')Rightarrow m=m'$) and surjective (that is given $nin N$ there exists an $min M$ such that $phi(m)=n$)

Can you take it from here?

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edited Mar 9 at 17:12

answered Mar 9 at 16:58

cansomeonehelpmeoutcansomeonehelpmeout

7,1273935

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Fix m,m′∈M, such that ϕ(m)=ϕ(m′)(1). We got: $psi(phi(m))=m$ and $psi(phi(m'))=m'.$ From (1): $psi(phi(m'))=psi(phi(m))=m=m'.$ So $phi$ is injective. Fix $nin N. phi(psi(n))=n,$ so it is enough to take $m:=psi(n). $ Therefore, $phi$ is surjective. Is it correct?
  $endgroup$
  – Toidi
  Mar 9 at 17:56
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

$$phi:Mrightarrow N\psi:Nrightarrow M$$

• $(Rightarrow)$ Define $psi$ in this way: Take $nin N$, then since $phi$ is surjective there is some $min M$ such that $phi(m)=n$. Therefore, let $psi(n)=m$. This $m$ is unique since $phi$ is injective, because if there were two different values $$psi(n)=mneq m'=psi(n')$$ then $phi(m)neq phi(m')$ and therefore $psi$ is well-defined.
  You can now check that indeed $phicircpsi=text{id}_N$, $psicircphi=text{id}_M$ and that $psi$ is an $R$-module homomorphism.




• $(Leftarrow)$ for this part you need to show that $phi$ is both injective (that is $phi(m)=phi(m')Rightarrow m=m'$) and surjective (that is given $nin N$ there exists an $min M$ such that $phi(m)=n$)

Can you take it from here?

share|cite|improve this answer

edited Mar 9 at 17:12

answered Mar 9 at 16:58

cansomeonehelpmeoutcansomeonehelpmeout

7,1273935

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Fix m,m′∈M, such that ϕ(m)=ϕ(m′)(1). We got: $psi(phi(m))=m$ and $psi(phi(m'))=m'.$ From (1): $psi(phi(m'))=psi(phi(m))=m=m'.$ So $phi$ is injective. Fix $nin N. phi(psi(n))=n,$ so it is enough to take $m:=psi(n). $ Therefore, $phi$ is surjective. Is it correct?
  $endgroup$
  – Toidi
  Mar 9 at 17:56
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

$$phi:Mrightarrow N\psi:Nrightarrow M$$

• $(Rightarrow)$ Define $psi$ in this way: Take $nin N$, then since $phi$ is surjective there is some $min M$ such that $phi(m)=n$. Therefore, let $psi(n)=m$. This $m$ is unique since $phi$ is injective, because if there were two different values $$psi(n)=mneq m'=psi(n')$$ then $phi(m)neq phi(m')$ and therefore $psi$ is well-defined.
  You can now check that indeed $phicircpsi=text{id}_N$, $psicircphi=text{id}_M$ and that $psi$ is an $R$-module homomorphism.




• $(Leftarrow)$ for this part you need to show that $phi$ is both injective (that is $phi(m)=phi(m')Rightarrow m=m'$) and surjective (that is given $nin N$ there exists an $min M$ such that $phi(m)=n$)

Can you take it from here?

share|cite|improve this answer

edited Mar 9 at 17:12

answered Mar 9 at 16:58

cansomeonehelpmeoutcansomeonehelpmeout

7,1273935

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Fix m,m′∈M, such that ϕ(m)=ϕ(m′)(1). We got: $psi(phi(m))=m$ and $psi(phi(m'))=m'.$ From (1): $psi(phi(m'))=psi(phi(m))=m=m'.$ So $phi$ is injective. Fix $nin N. phi(psi(n))=n,$ so it is enough to take $m:=psi(n). $ Therefore, $phi$ is surjective. Is it correct?
  $endgroup$
  – Toidi
  Mar 9 at 17:56
  
  
  

  
  
  
  

add a comment | 

$begingroup$

$$phi:Mrightarrow N\psi:Nrightarrow M$$

• $(Rightarrow)$ Define $psi$ in this way: Take $nin N$, then since $phi$ is surjective there is some $min M$ such that $phi(m)=n$. Therefore, let $psi(n)=m$. This $m$ is unique since $phi$ is injective, because if there were two different values $$psi(n)=mneq m'=psi(n')$$ then $phi(m)neq phi(m')$ and therefore $psi$ is well-defined.
  You can now check that indeed $phicircpsi=text{id}_N$, $psicircphi=text{id}_M$ and that $psi$ is an $R$-module homomorphism.




• $(Leftarrow)$ for this part you need to show that $phi$ is both injective (that is $phi(m)=phi(m')Rightarrow m=m'$) and surjective (that is given $nin N$ there exists an $min M$ such that $phi(m)=n$)

Can you take it from here?

share|cite|improve this answer

edited Mar 9 at 17:12

answered Mar 9 at 16:58

cansomeonehelpmeoutcansomeonehelpmeout

7,1273935

$endgroup$

share|cite|improve this answer

edited Mar 9 at 17:12

answered Mar 9 at 16:58

cansomeonehelpmeoutcansomeonehelpmeout

7,1273935

answered Mar 9 at 16:58

cansomeonehelpmeoutcansomeonehelpmeout

7,1273935

answered Mar 9 at 16:58

cansomeonehelpmeoutcansomeonehelpmeout

7,1273935