Equivalence in Injective Modules












0












$begingroup$


Can u help me prove the following statement



Statement about injective Modules



Proving the fact that if it is injective the sequence splits seems easy and i did it. Now the other way around ive tried using that fact that if the sequence splits then there exists inverse homomorphisms and i tried working with them but i didnt have any success. So any help is appreciated , thanks.






abstract-algebra modules







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  • $begingroup$
    Welcome to the site! It is usually better to include the statement of your question in the body of the text, rather than linking to it: the easier it is to see your question, the more likely it is that someone will answer it. (You can modify your question by clicking "edit").
    $endgroup$
    – Pierre-Guy Plamondon
    Dec 27 '18 at 12:29










  • $begingroup$
    Pehaps this question contains the answer you are looking for.
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    – Pierre-Guy Plamondon
    Dec 27 '18 at 12:32








  • 1




    $begingroup$
    @Pierre-GuyPlamondon: If you type [edit] in a comment, then it links to the edit tools of the post at hand, whether that be a question or an answer. Look: edit.
    $endgroup$
    – Shaun
    Dec 27 '18 at 13:01






  • 1




    $begingroup$
    @Shaun Thanks! I did not know of this very useful feature.
    $endgroup$
    – Pierre-Guy Plamondon
    Dec 27 '18 at 13:03
















0












$begingroup$


Can u help me prove the following statement



Statement about injective Modules



Proving the fact that if it is injective the sequence splits seems easy and i did it. Now the other way around ive tried using that fact that if the sequence splits then there exists inverse homomorphisms and i tried working with them but i didnt have any success. So any help is appreciated , thanks.






abstract-algebra modules







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to the site! It is usually better to include the statement of your question in the body of the text, rather than linking to it: the easier it is to see your question, the more likely it is that someone will answer it. (You can modify your question by clicking "edit").
    $endgroup$
    – Pierre-Guy Plamondon
    Dec 27 '18 at 12:29










  • $begingroup$
    Pehaps this question contains the answer you are looking for.
    $endgroup$
    – Pierre-Guy Plamondon
    Dec 27 '18 at 12:32








  • 1




    $begingroup$
    @Pierre-GuyPlamondon: If you type [edit] in a comment, then it links to the edit tools of the post at hand, whether that be a question or an answer. Look: edit.
    $endgroup$
    – Shaun
    Dec 27 '18 at 13:01






  • 1




    $begingroup$
    @Shaun Thanks! I did not know of this very useful feature.
    $endgroup$
    – Pierre-Guy Plamondon
    Dec 27 '18 at 13:03














0












0








0





$begingroup$


Can u help me prove the following statement



Statement about injective Modules



Proving the fact that if it is injective the sequence splits seems easy and i did it. Now the other way around ive tried using that fact that if the sequence splits then there exists inverse homomorphisms and i tried working with them but i didnt have any success. So any help is appreciated , thanks.






abstract-algebra modules







share|cite|improve this question











$endgroup$




Can u help me prove the following statement



Statement about injective Modules



Proving the fact that if it is injective the sequence splits seems easy and i did it. Now the other way around ive tried using that fact that if the sequence splits then there exists inverse homomorphisms and i tried working with them but i didnt have any success. So any help is appreciated , thanks.






abstract-algebra modules




abstract-algebra modules






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 13:01







Pedro Santos

















asked Dec 27 '18 at 12:23









Pedro SantosPedro Santos

1539




1539












  • $begingroup$
    Welcome to the site! It is usually better to include the statement of your question in the body of the text, rather than linking to it: the easier it is to see your question, the more likely it is that someone will answer it. (You can modify your question by clicking "edit").
    $endgroup$
    – Pierre-Guy Plamondon
    Dec 27 '18 at 12:29










  • $begingroup$
    Pehaps this question contains the answer you are looking for.
    $endgroup$
    – Pierre-Guy Plamondon
    Dec 27 '18 at 12:32








  • 1




    $begingroup$
    @Pierre-GuyPlamondon: If you type [edit] in a comment, then it links to the edit tools of the post at hand, whether that be a question or an answer. Look: edit.
    $endgroup$
    – Shaun
    Dec 27 '18 at 13:01






  • 1




    $begingroup$
    @Shaun Thanks! I did not know of this very useful feature.
    $endgroup$
    – Pierre-Guy Plamondon
    Dec 27 '18 at 13:03


















  • $begingroup$
    Welcome to the site! It is usually better to include the statement of your question in the body of the text, rather than linking to it: the easier it is to see your question, the more likely it is that someone will answer it. (You can modify your question by clicking "edit").
    $endgroup$
    – Pierre-Guy Plamondon
    Dec 27 '18 at 12:29










  • $begingroup$
    Pehaps this question contains the answer you are looking for.
    $endgroup$
    – Pierre-Guy Plamondon
    Dec 27 '18 at 12:32








  • 1




    $begingroup$
    @Pierre-GuyPlamondon: If you type [edit] in a comment, then it links to the edit tools of the post at hand, whether that be a question or an answer. Look: edit.
    $endgroup$
    – Shaun
    Dec 27 '18 at 13:01






  • 1




    $begingroup$
    @Shaun Thanks! I did not know of this very useful feature.
    $endgroup$
    – Pierre-Guy Plamondon
    Dec 27 '18 at 13:03
















$begingroup$
Welcome to the site! It is usually better to include the statement of your question in the body of the text, rather than linking to it: the easier it is to see your question, the more likely it is that someone will answer it. (You can modify your question by clicking "edit").
$endgroup$
– Pierre-Guy Plamondon
Dec 27 '18 at 12:29




$begingroup$
Welcome to the site! It is usually better to include the statement of your question in the body of the text, rather than linking to it: the easier it is to see your question, the more likely it is that someone will answer it. (You can modify your question by clicking "edit").
$endgroup$
– Pierre-Guy Plamondon
Dec 27 '18 at 12:29












$begingroup$
Pehaps this question contains the answer you are looking for.
$endgroup$
– Pierre-Guy Plamondon
Dec 27 '18 at 12:32






$begingroup$
Pehaps this question contains the answer you are looking for.
$endgroup$
– Pierre-Guy Plamondon
Dec 27 '18 at 12:32






1




1




$begingroup$
@Pierre-GuyPlamondon: If you type [edit] in a comment, then it links to the edit tools of the post at hand, whether that be a question or an answer. Look: edit.
$endgroup$
– Shaun
Dec 27 '18 at 13:01




$begingroup$
@Pierre-GuyPlamondon: If you type [edit] in a comment, then it links to the edit tools of the post at hand, whether that be a question or an answer. Look: edit.
$endgroup$
– Shaun
Dec 27 '18 at 13:01




1




1




$begingroup$
@Shaun Thanks! I did not know of this very useful feature.
$endgroup$
– Pierre-Guy Plamondon
Dec 27 '18 at 13:03




$begingroup$
@Shaun Thanks! I did not know of this very useful feature.
$endgroup$
– Pierre-Guy Plamondon
Dec 27 '18 at 13:03










1 Answer
1






active

oldest

votes


















1












$begingroup$

For the converse, if you have a diagram
begin{alignat}{3}
0longrightarrow& Mlongrightarrow N\
&downarrow \
&:I
end{alignat}
consider the amalgamated sum $;Icoprodlimits_M N$ and the canonical injection
$$0longrightarrow Ilongrightarrow Icoprodlimits_M N,$$
which has a retraction by the hypothesis on $I$. Compose it with the canonical injection $Nlongrightarrow Icoprodlimits_M N$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Im not familiar with the concept of an amalgamated sum and i didnt define it in my algebra course so i dont know :( , but thanks for the answer anyway.
    $endgroup$
    – Pedro Santos
    Dec 27 '18 at 17:17










  • $begingroup$
    Did you see fiber products? It's the dual notion.
    $endgroup$
    – Bernard
    Dec 27 '18 at 17:38










  • $begingroup$
    No sorry, im afraid i didnt go that far :/
    $endgroup$
    – Pedro Santos
    Dec 27 '18 at 17:39










  • $begingroup$
    The amalgamated sum isnt't very complex: with the present notations, you consider the direct sum $Ioplus N$, and quotient out by the submodule consisting of $bigl(f(m), -u(m)bigr)$ for all $min M$, where $f:Mto I$ and $u:Mto N$.
    $endgroup$
    – Bernard
    Dec 27 '18 at 17:53












  • $begingroup$
    Alright it doesnt seem very complicated , ill take a look into it , thx
    $endgroup$
    – Pedro Santos
    Dec 27 '18 at 17:54











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1 Answer
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1 Answer
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active

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active

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active

oldest

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1












$begingroup$

For the converse, if you have a diagram
begin{alignat}{3}
0longrightarrow& Mlongrightarrow N\
&downarrow \
&:I
end{alignat}
consider the amalgamated sum $;Icoprodlimits_M N$ and the canonical injection
$$0longrightarrow Ilongrightarrow Icoprodlimits_M N,$$
which has a retraction by the hypothesis on $I$. Compose it with the canonical injection $Nlongrightarrow Icoprodlimits_M N$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Im not familiar with the concept of an amalgamated sum and i didnt define it in my algebra course so i dont know :( , but thanks for the answer anyway.
    $endgroup$
    – Pedro Santos
    Dec 27 '18 at 17:17










  • $begingroup$
    Did you see fiber products? It's the dual notion.
    $endgroup$
    – Bernard
    Dec 27 '18 at 17:38










  • $begingroup$
    No sorry, im afraid i didnt go that far :/
    $endgroup$
    – Pedro Santos
    Dec 27 '18 at 17:39










  • $begingroup$
    The amalgamated sum isnt't very complex: with the present notations, you consider the direct sum $Ioplus N$, and quotient out by the submodule consisting of $bigl(f(m), -u(m)bigr)$ for all $min M$, where $f:Mto I$ and $u:Mto N$.
    $endgroup$
    – Bernard
    Dec 27 '18 at 17:53












  • $begingroup$
    Alright it doesnt seem very complicated , ill take a look into it , thx
    $endgroup$
    – Pedro Santos
    Dec 27 '18 at 17:54
















1












$begingroup$

For the converse, if you have a diagram
begin{alignat}{3}
0longrightarrow& Mlongrightarrow N\
&downarrow \
&:I
end{alignat}
consider the amalgamated sum $;Icoprodlimits_M N$ and the canonical injection
$$0longrightarrow Ilongrightarrow Icoprodlimits_M N,$$
which has a retraction by the hypothesis on $I$. Compose it with the canonical injection $Nlongrightarrow Icoprodlimits_M N$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Im not familiar with the concept of an amalgamated sum and i didnt define it in my algebra course so i dont know :( , but thanks for the answer anyway.
    $endgroup$
    – Pedro Santos
    Dec 27 '18 at 17:17










  • $begingroup$
    Did you see fiber products? It's the dual notion.
    $endgroup$
    – Bernard
    Dec 27 '18 at 17:38










  • $begingroup$
    No sorry, im afraid i didnt go that far :/
    $endgroup$
    – Pedro Santos
    Dec 27 '18 at 17:39










  • $begingroup$
    The amalgamated sum isnt't very complex: with the present notations, you consider the direct sum $Ioplus N$, and quotient out by the submodule consisting of $bigl(f(m), -u(m)bigr)$ for all $min M$, where $f:Mto I$ and $u:Mto N$.
    $endgroup$
    – Bernard
    Dec 27 '18 at 17:53












  • $begingroup$
    Alright it doesnt seem very complicated , ill take a look into it , thx
    $endgroup$
    – Pedro Santos
    Dec 27 '18 at 17:54














1












1








1





$begingroup$

For the converse, if you have a diagram
begin{alignat}{3}
0longrightarrow& Mlongrightarrow N\
&downarrow \
&:I
end{alignat}
consider the amalgamated sum $;Icoprodlimits_M N$ and the canonical injection
$$0longrightarrow Ilongrightarrow Icoprodlimits_M N,$$
which has a retraction by the hypothesis on $I$. Compose it with the canonical injection $Nlongrightarrow Icoprodlimits_M N$.






share|cite|improve this answer









$endgroup$



For the converse, if you have a diagram
begin{alignat}{3}
0longrightarrow& Mlongrightarrow N\
&downarrow \
&:I
end{alignat}
consider the amalgamated sum $;Icoprodlimits_M N$ and the canonical injection
$$0longrightarrow Ilongrightarrow Icoprodlimits_M N,$$
which has a retraction by the hypothesis on $I$. Compose it with the canonical injection $Nlongrightarrow Icoprodlimits_M N$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 27 '18 at 13:03









BernardBernard

123k741116




123k741116












  • $begingroup$
    Im not familiar with the concept of an amalgamated sum and i didnt define it in my algebra course so i dont know :( , but thanks for the answer anyway.
    $endgroup$
    – Pedro Santos
    Dec 27 '18 at 17:17










  • $begingroup$
    Did you see fiber products? It's the dual notion.
    $endgroup$
    – Bernard
    Dec 27 '18 at 17:38










  • $begingroup$
    No sorry, im afraid i didnt go that far :/
    $endgroup$
    – Pedro Santos
    Dec 27 '18 at 17:39










  • $begingroup$
    The amalgamated sum isnt't very complex: with the present notations, you consider the direct sum $Ioplus N$, and quotient out by the submodule consisting of $bigl(f(m), -u(m)bigr)$ for all $min M$, where $f:Mto I$ and $u:Mto N$.
    $endgroup$
    – Bernard
    Dec 27 '18 at 17:53












  • $begingroup$
    Alright it doesnt seem very complicated , ill take a look into it , thx
    $endgroup$
    – Pedro Santos
    Dec 27 '18 at 17:54


















  • $begingroup$
    Im not familiar with the concept of an amalgamated sum and i didnt define it in my algebra course so i dont know :( , but thanks for the answer anyway.
    $endgroup$
    – Pedro Santos
    Dec 27 '18 at 17:17










  • $begingroup$
    Did you see fiber products? It's the dual notion.
    $endgroup$
    – Bernard
    Dec 27 '18 at 17:38










  • $begingroup$
    No sorry, im afraid i didnt go that far :/
    $endgroup$
    – Pedro Santos
    Dec 27 '18 at 17:39










  • $begingroup$
    The amalgamated sum isnt't very complex: with the present notations, you consider the direct sum $Ioplus N$, and quotient out by the submodule consisting of $bigl(f(m), -u(m)bigr)$ for all $min M$, where $f:Mto I$ and $u:Mto N$.
    $endgroup$
    – Bernard
    Dec 27 '18 at 17:53












  • $begingroup$
    Alright it doesnt seem very complicated , ill take a look into it , thx
    $endgroup$
    – Pedro Santos
    Dec 27 '18 at 17:54
















$begingroup$
Im not familiar with the concept of an amalgamated sum and i didnt define it in my algebra course so i dont know :( , but thanks for the answer anyway.
$endgroup$
– Pedro Santos
Dec 27 '18 at 17:17




$begingroup$
Im not familiar with the concept of an amalgamated sum and i didnt define it in my algebra course so i dont know :( , but thanks for the answer anyway.
$endgroup$
– Pedro Santos
Dec 27 '18 at 17:17












$begingroup$
Did you see fiber products? It's the dual notion.
$endgroup$
– Bernard
Dec 27 '18 at 17:38




$begingroup$
Did you see fiber products? It's the dual notion.
$endgroup$
– Bernard
Dec 27 '18 at 17:38












$begingroup$
No sorry, im afraid i didnt go that far :/
$endgroup$
– Pedro Santos
Dec 27 '18 at 17:39




$begingroup$
No sorry, im afraid i didnt go that far :/
$endgroup$
– Pedro Santos
Dec 27 '18 at 17:39












$begingroup$
The amalgamated sum isnt't very complex: with the present notations, you consider the direct sum $Ioplus N$, and quotient out by the submodule consisting of $bigl(f(m), -u(m)bigr)$ for all $min M$, where $f:Mto I$ and $u:Mto N$.
$endgroup$
– Bernard
Dec 27 '18 at 17:53






$begingroup$
The amalgamated sum isnt't very complex: with the present notations, you consider the direct sum $Ioplus N$, and quotient out by the submodule consisting of $bigl(f(m), -u(m)bigr)$ for all $min M$, where $f:Mto I$ and $u:Mto N$.
$endgroup$
– Bernard
Dec 27 '18 at 17:53














$begingroup$
Alright it doesnt seem very complicated , ill take a look into it , thx
$endgroup$
– Pedro Santos
Dec 27 '18 at 17:54




$begingroup$
Alright it doesnt seem very complicated , ill take a look into it , thx
$endgroup$
– Pedro Santos
Dec 27 '18 at 17:54


















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