Can we deduce that $M_0$ is a submodule of the limit of the following diagram?












2














Let $M_0$ be an R-module, and suppose $M_{n+1}$ is the pushout of the diagram below as shown, for all $n in mathbb{N}$:



$$begin{array}{ccc}M_n&to& M_{n+1}\uparrow &&uparrow\A&to& Bend{array}$$



Define $M:= colim(M_0 rightarrow M_1 rightarrow M_2 rightarrow.....)$



Can we deduce that $M_0$ is a submodule of $M$? What about if the left arrow is surjective and the bottom arrow is injective? Would this help?



I am asking this in order to complete my understanding of the proof that $R-mod$ has enough injectives. In this proof, A and B are given, however there is no justification given for the claim which I am inquiring about, it is just stated to be obvious. Thanks.






category-theory homological-algebra limits-colimits diagram-chasing







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    2














    Let $M_0$ be an R-module, and suppose $M_{n+1}$ is the pushout of the diagram below as shown, for all $n in mathbb{N}$:



    $$begin{array}{ccc}M_n&to& M_{n+1}\uparrow &&uparrow\A&to& Bend{array}$$



    Define $M:= colim(M_0 rightarrow M_1 rightarrow M_2 rightarrow.....)$



    Can we deduce that $M_0$ is a submodule of $M$? What about if the left arrow is surjective and the bottom arrow is injective? Would this help?



    I am asking this in order to complete my understanding of the proof that $R-mod$ has enough injectives. In this proof, A and B are given, however there is no justification given for the claim which I am inquiring about, it is just stated to be obvious. Thanks.






    category-theory homological-algebra limits-colimits diagram-chasing







    share|cite|improve this question

























      2












      2








      2







      Let $M_0$ be an R-module, and suppose $M_{n+1}$ is the pushout of the diagram below as shown, for all $n in mathbb{N}$:



      $$begin{array}{ccc}M_n&to& M_{n+1}\uparrow &&uparrow\A&to& Bend{array}$$



      Define $M:= colim(M_0 rightarrow M_1 rightarrow M_2 rightarrow.....)$



      Can we deduce that $M_0$ is a submodule of $M$? What about if the left arrow is surjective and the bottom arrow is injective? Would this help?



      I am asking this in order to complete my understanding of the proof that $R-mod$ has enough injectives. In this proof, A and B are given, however there is no justification given for the claim which I am inquiring about, it is just stated to be obvious. Thanks.






      category-theory homological-algebra limits-colimits diagram-chasing







      share|cite|improve this question













      Let $M_0$ be an R-module, and suppose $M_{n+1}$ is the pushout of the diagram below as shown, for all $n in mathbb{N}$:



      $$begin{array}{ccc}M_n&to& M_{n+1}\uparrow &&uparrow\A&to& Bend{array}$$



      Define $M:= colim(M_0 rightarrow M_1 rightarrow M_2 rightarrow.....)$



      Can we deduce that $M_0$ is a submodule of $M$? What about if the left arrow is surjective and the bottom arrow is injective? Would this help?



      I am asking this in order to complete my understanding of the proof that $R-mod$ has enough injectives. In this proof, A and B are given, however there is no justification given for the claim which I am inquiring about, it is just stated to be obvious. Thanks.






      category-theory homological-algebra limits-colimits diagram-chasing




      category-theory homological-algebra limits-colimits diagram-chasing






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      asked yesterday









      Daven

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      33319






















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          If not yet done, convince yourself (by element chasing) that the pushout of an injective module homomorphism is injective.



          Then, if $Ato B$ is injective, then $M_i$ embeds to $M_{i+1}$, and it's easy to see that $M':=bigcup_iM_i$ satisfies the universal property of the colimit, hence $Mcong M'$ and the natural arrow $M_ito M$ is injective for all $i$.






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          • Thanks, this is exactly what I was looking for!
            – Daven
            yesterday











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          1 Answer
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          active

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          If not yet done, convince yourself (by element chasing) that the pushout of an injective module homomorphism is injective.



          Then, if $Ato B$ is injective, then $M_i$ embeds to $M_{i+1}$, and it's easy to see that $M':=bigcup_iM_i$ satisfies the universal property of the colimit, hence $Mcong M'$ and the natural arrow $M_ito M$ is injective for all $i$.






          share|cite|improve this answer





















          • Thanks, this is exactly what I was looking for!
            – Daven
            yesterday
















          3














          If not yet done, convince yourself (by element chasing) that the pushout of an injective module homomorphism is injective.



          Then, if $Ato B$ is injective, then $M_i$ embeds to $M_{i+1}$, and it's easy to see that $M':=bigcup_iM_i$ satisfies the universal property of the colimit, hence $Mcong M'$ and the natural arrow $M_ito M$ is injective for all $i$.






          share|cite|improve this answer





















          • Thanks, this is exactly what I was looking for!
            – Daven
            yesterday














          3












          3








          3






          If not yet done, convince yourself (by element chasing) that the pushout of an injective module homomorphism is injective.



          Then, if $Ato B$ is injective, then $M_i$ embeds to $M_{i+1}$, and it's easy to see that $M':=bigcup_iM_i$ satisfies the universal property of the colimit, hence $Mcong M'$ and the natural arrow $M_ito M$ is injective for all $i$.






          share|cite|improve this answer












          If not yet done, convince yourself (by element chasing) that the pushout of an injective module homomorphism is injective.



          Then, if $Ato B$ is injective, then $M_i$ embeds to $M_{i+1}$, and it's easy to see that $M':=bigcup_iM_i$ satisfies the universal property of the colimit, hence $Mcong M'$ and the natural arrow $M_ito M$ is injective for all $i$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Berci

          59.7k23672




          59.7k23672












          • Thanks, this is exactly what I was looking for!
            – Daven
            yesterday


















          • Thanks, this is exactly what I was looking for!
            – Daven
            yesterday
















          Thanks, this is exactly what I was looking for!
          – Daven
          yesterday




          Thanks, this is exactly what I was looking for!
          – Daven
          yesterday


















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