Can we deduce that $M_0$ is a submodule of the limit of the following diagram?
$begingroup$
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
asked Jan 3 at 15:02
DavenDaven
41229
$endgroup$
add a comment |
$begingroup$
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
asked Jan 3 at 15:02
DavenDaven
41229
$endgroup$
add a comment |
$begingroup$
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
asked Jan 3 at 15:02
DavenDaven
41229
$endgroup$
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
asked Jan 3 at 15:02
DavenDaven
41229
asked Jan 3 at 15:02
DavenDaven
41229
asked Jan 3 at 15:02
DavenDaven
41229
asked Jan 3 at 15:02
DavenDaven
41229
asked Jan 3 at 15:02
DavenDaven
41229
41229
add a comment |
add a comment |
1 Answer
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$begingroup$
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 $i$.
answered Jan 3 at 15:19
BerciBerci
61.8k23675
$endgroup$
$begingroup$
Thanks, this is exactly what I was looking for!
$endgroup$
– Daven
Jan 3 at 15:58
add a comment |
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1 Answer
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active
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1 Answer
1
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oldest
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$begingroup$
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 $i$.
answered Jan 3 at 15:19
BerciBerci
61.8k23675
$endgroup$
$begingroup$
Thanks, this is exactly what I was looking for!
$endgroup$
– Daven
Jan 3 at 15:58
add a comment |
$begingroup$
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 $i$.
answered Jan 3 at 15:19
BerciBerci
61.8k23675
$endgroup$
$begingroup$
Thanks, this is exactly what I was looking for!
$endgroup$
– Daven
Jan 3 at 15:58
add a comment |
$begingroup$
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 $i$.
answered Jan 3 at 15:19
BerciBerci
61.8k23675
$endgroup$
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 $i$.
answered Jan 3 at 15:19
BerciBerci
61.8k23675
answered Jan 3 at 15:19
BerciBerci
61.8k23675
answered Jan 3 at 15:19
BerciBerci
61.8k23675
answered Jan 3 at 15:19
BerciBerci
61.8k23675
61.8k23675
$begingroup$
Thanks, this is exactly what I was looking for!
$endgroup$
– Daven
Jan 3 at 15:58
add a comment |
$begingroup$
Thanks, this is exactly what I was looking for!
$endgroup$
– Daven
Jan 3 at 15:58
$begingroup$
Thanks, this is exactly what I was looking for!
$endgroup$
– Daven
Jan 3 at 15:58
$begingroup$
Thanks, this is exactly what I was looking for!
$endgroup$
– Daven
Jan 3 at 15:58
add a comment |
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