Mixing
$g circ f = id_A$ implies $ker(g) cong coker(f) $
$begingroup$
Let $R$ be a ring (for example $mathbb{Z}$) and $A,B$ two $R$-modules.
Futhermore we have two morphisms $A xrightarrow{f}B$ and $B xrightarrow{g}A$ with property $g circ f = id_A$.
My question is how to see that $ker(g) cong coker(f) $?
My considerations:
We have $coker(f)= B/im(f)$ by definition and since $g circ f = id_A$ $f$ is in injection so we can identify $A$ with $im(f)$.
So $coker(f) cong B/A$. But I don't see how to settle $ker(g) cong B/A$. Could anybody help?
Can the argument be generalized to any abelian category?
abstract-algebra modules
edited Jan 25 at 21:40
user26857
39.4k124183
asked Jan 25 at 17:55
KarlPeterKarlPeter
4641315
$endgroup$
|
show 3 more comments
$begingroup$
Let $R$ be a ring (for example $mathbb{Z}$) and $A,B$ two $R$-modules.
Futhermore we have two morphisms $A xrightarrow{f}B$ and $B xrightarrow{g}A$ with property $g circ f = id_A$.
My question is how to see that $ker(g) cong coker(f) $?
My considerations:
We have $coker(f)= B/im(f)$ by definition and since $g circ f = id_A$ $f$ is in injection so we can identify $A$ with $im(f)$.
So $coker(f) cong B/A$. But I don't see how to settle $ker(g) cong B/A$. Could anybody help?
Can the argument be generalized to any abelian category?
abstract-algebra modules
edited Jan 25 at 21:40
user26857
39.4k124183
asked Jan 25 at 17:55
KarlPeterKarlPeter
4641315
$endgroup$
1
$begingroup$
The ahort exact sequence $0to Ato Bto B/Ato0$ splits...
$endgroup$
– Lord Shark the Unknown
Jan 25 at 17:59
$begingroup$
@LordSharktheUnknown: yes, so we have $B cong A oplus B/A$. The question is if $g$ coincides in this case with the canonical projection $pr:A oplus B/A to A$?
$endgroup$
– KarlPeter
Jan 25 at 18:06
1
$begingroup$
When one has a split s.e.s. then there's a split s.e.s. the other way round.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 18:07
$begingroup$
@LordSharktheUnknown: Does it provide that there just exist some split $0 to B/A to B to A to 0$ or does it mean that the "new" spling conserves the previous maps in sense that the split is given exactly by $$0 to B/A to B xrightarrow{g} A to 0$$?
$endgroup$
– KarlPeter
Jan 25 at 18:12
1
$begingroup$
Yes the diagram above is commutative. When we have a split exact sequence as above, the isomorphism $Bto Aoplus B/A$ is exactly given by $(g,p)$ (where $p:Bto B/A$ is the canonical projection).
$endgroup$
– Roland
Jan 25 at 19:19
|
show 3 more comments
$begingroup$
Let $R$ be a ring (for example $mathbb{Z}$) and $A,B$ two $R$-modules.
Futhermore we have two morphisms $A xrightarrow{f}B$ and $B xrightarrow{g}A$ with property $g circ f = id_A$.
My question is how to see that $ker(g) cong coker(f) $?
My considerations:
We have $coker(f)= B/im(f)$ by definition and since $g circ f = id_A$ $f$ is in injection so we can identify $A$ with $im(f)$.
So $coker(f) cong B/A$. But I don't see how to settle $ker(g) cong B/A$. Could anybody help?
Can the argument be generalized to any abelian category?
abstract-algebra modules
edited Jan 25 at 21:40
user26857
39.4k124183
asked Jan 25 at 17:55
KarlPeterKarlPeter
4641315
$endgroup$
Let $R$ be a ring (for example $mathbb{Z}$) and $A,B$ two $R$-modules.
Futhermore we have two morphisms $A xrightarrow{f}B$ and $B xrightarrow{g}A$ with property $g circ f = id_A$.
My question is how to see that $ker(g) cong coker(f) $?
My considerations:
We have $coker(f)= B/im(f)$ by definition and since $g circ f = id_A$ $f$ is in injection so we can identify $A$ with $im(f)$.
So $coker(f) cong B/A$. But I don't see how to settle $ker(g) cong B/A$. Could anybody help?
Can the argument be generalized to any abelian category?
abstract-algebra modules
abstract-algebra modules
edited Jan 25 at 21:40
user26857
39.4k124183
asked Jan 25 at 17:55
KarlPeterKarlPeter
4641315
edited Jan 25 at 21:40
user26857
39.4k124183
asked Jan 25 at 17:55
KarlPeterKarlPeter
4641315
edited Jan 25 at 21:40
user26857
39.4k124183
edited Jan 25 at 21:40
user26857
39.4k124183
edited Jan 25 at 21:40
user26857
39.4k124183
39.4k124183
asked Jan 25 at 17:55
KarlPeterKarlPeter
4641315
asked Jan 25 at 17:55
KarlPeterKarlPeter
4641315
asked Jan 25 at 17:55
KarlPeterKarlPeter
4641315
4641315
1
$begingroup$
The ahort exact sequence $0to Ato Bto B/Ato0$ splits...
$endgroup$
– Lord Shark the Unknown
Jan 25 at 17:59
$begingroup$
@LordSharktheUnknown: yes, so we have $B cong A oplus B/A$. The question is if $g$ coincides in this case with the canonical projection $pr:A oplus B/A to A$?
$endgroup$
– KarlPeter
Jan 25 at 18:06
1
$begingroup$
When one has a split s.e.s. then there's a split s.e.s. the other way round.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 18:07
$begingroup$
@LordSharktheUnknown: Does it provide that there just exist some split $0 to B/A to B to A to 0$ or does it mean that the "new" spling conserves the previous maps in sense that the split is given exactly by $$0 to B/A to B xrightarrow{g} A to 0$$?
$endgroup$
– KarlPeter
Jan 25 at 18:12
1
$begingroup$
Yes the diagram above is commutative. When we have a split exact sequence as above, the isomorphism $Bto Aoplus B/A$ is exactly given by $(g,p)$ (where $p:Bto B/A$ is the canonical projection).
$endgroup$
– Roland
Jan 25 at 19:19
|
show 3 more comments
1
$begingroup$
The ahort exact sequence $0to Ato Bto B/Ato0$ splits...
$endgroup$
– Lord Shark the Unknown
Jan 25 at 17:59
$begingroup$
@LordSharktheUnknown: yes, so we have $B cong A oplus B/A$. The question is if $g$ coincides in this case with the canonical projection $pr:A oplus B/A to A$?
$endgroup$
– KarlPeter
Jan 25 at 18:06
1
$begingroup$
When one has a split s.e.s. then there's a split s.e.s. the other way round.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 18:07
$begingroup$
@LordSharktheUnknown: Does it provide that there just exist some split $0 to B/A to B to A to 0$ or does it mean that the "new" spling conserves the previous maps in sense that the split is given exactly by $$0 to B/A to B xrightarrow{g} A to 0$$?
$endgroup$
– KarlPeter
Jan 25 at 18:12
1
$begingroup$
Yes the diagram above is commutative. When we have a split exact sequence as above, the isomorphism $Bto Aoplus B/A$ is exactly given by $(g,p)$ (where $p:Bto B/A$ is the canonical projection).
$endgroup$
– Roland
Jan 25 at 19:19
1
1
$begingroup$
The ahort exact sequence $0to Ato Bto B/Ato0$ splits...
$endgroup$
– Lord Shark the Unknown
Jan 25 at 17:59
$begingroup$
The ahort exact sequence $0to Ato Bto B/Ato0$ splits...
$endgroup$
– Lord Shark the Unknown
Jan 25 at 17:59
$begingroup$
@LordSharktheUnknown: yes, so we have $B cong A oplus B/A$. The question is if $g$ coincides in this case with the canonical projection $pr:A oplus B/A to A$?
$endgroup$
– KarlPeter
Jan 25 at 18:06
$begingroup$
@LordSharktheUnknown: yes, so we have $B cong A oplus B/A$. The question is if $g$ coincides in this case with the canonical projection $pr:A oplus B/A to A$?
$endgroup$
– KarlPeter
Jan 25 at 18:06
1
1
$begingroup$
When one has a split s.e.s. then there's a split s.e.s. the other way round.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 18:07
$begingroup$
When one has a split s.e.s. then there's a split s.e.s. the other way round.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 18:07
$begingroup$
@LordSharktheUnknown: Does it provide that there just exist some split $0 to B/A to B to A to 0$ or does it mean that the "new" spling conserves the previous maps in sense that the split is given exactly by $$0 to B/A to B xrightarrow{g} A to 0$$?
$endgroup$
– KarlPeter
Jan 25 at 18:12
$begingroup$
@LordSharktheUnknown: Does it provide that there just exist some split $0 to B/A to B to A to 0$ or does it mean that the "new" spling conserves the previous maps in sense that the split is given exactly by $$0 to B/A to B xrightarrow{g} A to 0$$?
$endgroup$
– KarlPeter
Jan 25 at 18:12
1
1
$begingroup$
Yes the diagram above is commutative. When we have a split exact sequence as above, the isomorphism $Bto Aoplus B/A$ is exactly given by $(g,p)$ (where $p:Bto B/A$ is the canonical projection).
$endgroup$
– Roland
Jan 25 at 19:19
$begingroup$
Yes the diagram above is commutative. When we have a split exact sequence as above, the isomorphism $Bto Aoplus B/A$ is exactly given by $(g,p)$ (where $p:Bto B/A$ is the canonical projection).
$endgroup$
– Roland
Jan 25 at 19:19
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The trick is to prove the following equality (not merely an isomorphism): $$B=f(A)oplus {rm ker}(g)$$ and then the needed isomorphism is trivial, by factoring out by $f(A)$.
If you need help with this equality, just let me know.
answered Jan 25 at 21:58
Pace NielsenPace Nielsen
25526
$endgroup$
add a comment |
$begingroup$
Maybe it's not true. The kernel of $g : B -> A$ is a subobject of B. It's also the fiber $g^{-1}(0)=f(0)$. But subobjects are not quotients unless some very special situation occurs.
answered Jan 25 at 18:26
Esa PulkkinenEsa Pulkkinen
1
$endgroup$
4
$begingroup$
This is a special situation.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 18:33
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The trick is to prove the following equality (not merely an isomorphism): $$B=f(A)oplus {rm ker}(g)$$ and then the needed isomorphism is trivial, by factoring out by $f(A)$.
If you need help with this equality, just let me know.
answered Jan 25 at 21:58
Pace NielsenPace Nielsen
25526
$endgroup$
add a comment |
$begingroup$
The trick is to prove the following equality (not merely an isomorphism): $$B=f(A)oplus {rm ker}(g)$$ and then the needed isomorphism is trivial, by factoring out by $f(A)$.
If you need help with this equality, just let me know.
answered Jan 25 at 21:58
Pace NielsenPace Nielsen
25526
$endgroup$
add a comment |
$begingroup$
The trick is to prove the following equality (not merely an isomorphism): $$B=f(A)oplus {rm ker}(g)$$ and then the needed isomorphism is trivial, by factoring out by $f(A)$.
If you need help with this equality, just let me know.
answered Jan 25 at 21:58
Pace NielsenPace Nielsen
25526
$endgroup$
The trick is to prove the following equality (not merely an isomorphism): $$B=f(A)oplus {rm ker}(g)$$ and then the needed isomorphism is trivial, by factoring out by $f(A)$.
If you need help with this equality, just let me know.
answered Jan 25 at 21:58
Pace NielsenPace Nielsen
25526
answered Jan 25 at 21:58
Pace NielsenPace Nielsen
25526
answered Jan 25 at 21:58
Pace NielsenPace Nielsen
25526
answered Jan 25 at 21:58
Pace NielsenPace Nielsen
25526
25526
add a comment |
add a comment |
$begingroup$
Maybe it's not true. The kernel of $g : B -> A$ is a subobject of B. It's also the fiber $g^{-1}(0)=f(0)$. But subobjects are not quotients unless some very special situation occurs.
answered Jan 25 at 18:26
Esa PulkkinenEsa Pulkkinen
1
$endgroup$
4
$begingroup$
This is a special situation.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 18:33
add a comment |
$begingroup$
Maybe it's not true. The kernel of $g : B -> A$ is a subobject of B. It's also the fiber $g^{-1}(0)=f(0)$. But subobjects are not quotients unless some very special situation occurs.
answered Jan 25 at 18:26
Esa PulkkinenEsa Pulkkinen
1
$endgroup$
4
$begingroup$
This is a special situation.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 18:33
add a comment |
$begingroup$
Maybe it's not true. The kernel of $g : B -> A$ is a subobject of B. It's also the fiber $g^{-1}(0)=f(0)$. But subobjects are not quotients unless some very special situation occurs.
answered Jan 25 at 18:26
Esa PulkkinenEsa Pulkkinen
1
$endgroup$
Maybe it's not true. The kernel of $g : B -> A$ is a subobject of B. It's also the fiber $g^{-1}(0)=f(0)$. But subobjects are not quotients unless some very special situation occurs.
answered Jan 25 at 18:26
Esa PulkkinenEsa Pulkkinen
1
answered Jan 25 at 18:26
Esa PulkkinenEsa Pulkkinen
1
answered Jan 25 at 18:26
Esa PulkkinenEsa Pulkkinen
1
answered Jan 25 at 18:26
Esa PulkkinenEsa Pulkkinen
1
1
4
$begingroup$
This is a special situation.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 18:33
add a comment |
4
$begingroup$
This is a special situation.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 18:33
4
4
$begingroup$
This is a special situation.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 18:33
$begingroup$
This is a special situation.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 18:33
add a comment |
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1
$begingroup$
The ahort exact sequence $0to Ato Bto B/Ato0$ splits...
$endgroup$
– Lord Shark the Unknown
Jan 25 at 17:59
$begingroup$
@LordSharktheUnknown: yes, so we have $B cong A oplus B/A$. The question is if $g$ coincides in this case with the canonical projection $pr:A oplus B/A to A$?
$endgroup$
– KarlPeter
Jan 25 at 18:06
1
$begingroup$
When one has a split s.e.s. then there's a split s.e.s. the other way round.
$endgroup$
– Lord Shark the Unknown
Jan 25 at 18:07
$begingroup$
@LordSharktheUnknown: Does it provide that there just exist some split $0 to B/A to B to A to 0$ or does it mean that the "new" spling conserves the previous maps in sense that the split is given exactly by $$0 to B/A to B xrightarrow{g} A to 0$$?
$endgroup$
– KarlPeter
Jan 25 at 18:12
1
$begingroup$
Yes the diagram above is commutative. When we have a split exact sequence as above, the isomorphism $Bto Aoplus B/A$ is exactly given by $(g,p)$ (where $p:Bto B/A$ is the canonical projection).
$endgroup$
– Roland
Jan 25 at 19:19