Does an embedding necessarily make a short exact sequence split?
$begingroup$
Suppose we have a short exact sequence.
$0 to A to B to C to 0$.
Suppose also that we know there is some embedding $e : C hookrightarrow B$.
Can we conclude that the sequence splits? It seems that we should be able to, but it is not required that $e$ be a section of the surjection $B to C$.
Any help is appreciated.
abstract-algebra group-theory exact-sequence
asked Dec 12 '18 at 10:35
CuriousKid7CuriousKid7
1,676717
$endgroup$
add a comment |
$begingroup$
Suppose we have a short exact sequence.
$0 to A to B to C to 0$.
Suppose also that we know there is some embedding $e : C hookrightarrow B$.
Can we conclude that the sequence splits? It seems that we should be able to, but it is not required that $e$ be a section of the surjection $B to C$.
Any help is appreciated.
abstract-algebra group-theory exact-sequence
asked Dec 12 '18 at 10:35
CuriousKid7CuriousKid7
1,676717
$endgroup$
add a comment |
$begingroup$
Suppose we have a short exact sequence.
$0 to A to B to C to 0$.
Suppose also that we know there is some embedding $e : C hookrightarrow B$.
Can we conclude that the sequence splits? It seems that we should be able to, but it is not required that $e$ be a section of the surjection $B to C$.
Any help is appreciated.
abstract-algebra group-theory exact-sequence
asked Dec 12 '18 at 10:35
CuriousKid7CuriousKid7
1,676717
$endgroup$
Suppose we have a short exact sequence.
$0 to A to B to C to 0$.
Suppose also that we know there is some embedding $e : C hookrightarrow B$.
Can we conclude that the sequence splits? It seems that we should be able to, but it is not required that $e$ be a section of the surjection $B to C$.
Any help is appreciated.
abstract-algebra group-theory exact-sequence
abstract-algebra group-theory exact-sequence
asked Dec 12 '18 at 10:35
CuriousKid7CuriousKid7
1,676717
asked Dec 12 '18 at 10:35
CuriousKid7CuriousKid7
1,676717
asked Dec 12 '18 at 10:35
CuriousKid7CuriousKid7
1,676717
asked Dec 12 '18 at 10:35
CuriousKid7CuriousKid7
1,676717
asked Dec 12 '18 at 10:35
CuriousKid7CuriousKid7
1,676717
1,676717
add a comment |
add a comment |
1 Answer
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$begingroup$
There is no reason for this to hold. For example, there is a short exact sequence
$$0rightarrow mathbb{Z}/2mathbb{Z}rightarrow mathbb{Z}/4mathbb{Z}rightarrowmathbb{Z}/2mathbb{Z}rightarrow 0$$
but it doesn't split since $mathbb{Z}/4mathbb{Z}$ is not isomorphic to $mathbb{Z}/2mathbb{Z}oplus mathbb{Z}/2mathbb{Z}$. And even a sequence with $Bcong Aoplus C$ need not split in general, as can be seen in this question.
answered Dec 12 '18 at 10:49
Arnaud D.Arnaud D.
15.9k52443
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is no reason for this to hold. For example, there is a short exact sequence
$$0rightarrow mathbb{Z}/2mathbb{Z}rightarrow mathbb{Z}/4mathbb{Z}rightarrowmathbb{Z}/2mathbb{Z}rightarrow 0$$
but it doesn't split since $mathbb{Z}/4mathbb{Z}$ is not isomorphic to $mathbb{Z}/2mathbb{Z}oplus mathbb{Z}/2mathbb{Z}$. And even a sequence with $Bcong Aoplus C$ need not split in general, as can be seen in this question.
answered Dec 12 '18 at 10:49
Arnaud D.Arnaud D.
15.9k52443
$endgroup$
add a comment |
$begingroup$
There is no reason for this to hold. For example, there is a short exact sequence
$$0rightarrow mathbb{Z}/2mathbb{Z}rightarrow mathbb{Z}/4mathbb{Z}rightarrowmathbb{Z}/2mathbb{Z}rightarrow 0$$
but it doesn't split since $mathbb{Z}/4mathbb{Z}$ is not isomorphic to $mathbb{Z}/2mathbb{Z}oplus mathbb{Z}/2mathbb{Z}$. And even a sequence with $Bcong Aoplus C$ need not split in general, as can be seen in this question.
answered Dec 12 '18 at 10:49
Arnaud D.Arnaud D.
15.9k52443
$endgroup$
add a comment |
$begingroup$
There is no reason for this to hold. For example, there is a short exact sequence
$$0rightarrow mathbb{Z}/2mathbb{Z}rightarrow mathbb{Z}/4mathbb{Z}rightarrowmathbb{Z}/2mathbb{Z}rightarrow 0$$
but it doesn't split since $mathbb{Z}/4mathbb{Z}$ is not isomorphic to $mathbb{Z}/2mathbb{Z}oplus mathbb{Z}/2mathbb{Z}$. And even a sequence with $Bcong Aoplus C$ need not split in general, as can be seen in this question.
answered Dec 12 '18 at 10:49
Arnaud D.Arnaud D.
15.9k52443
$endgroup$
There is no reason for this to hold. For example, there is a short exact sequence
$$0rightarrow mathbb{Z}/2mathbb{Z}rightarrow mathbb{Z}/4mathbb{Z}rightarrowmathbb{Z}/2mathbb{Z}rightarrow 0$$
but it doesn't split since $mathbb{Z}/4mathbb{Z}$ is not isomorphic to $mathbb{Z}/2mathbb{Z}oplus mathbb{Z}/2mathbb{Z}$. And even a sequence with $Bcong Aoplus C$ need not split in general, as can be seen in this question.
answered Dec 12 '18 at 10:49
Arnaud D.Arnaud D.
15.9k52443
answered Dec 12 '18 at 10:49
Arnaud D.Arnaud D.
15.9k52443
answered Dec 12 '18 at 10:49
Arnaud D.Arnaud D.
15.9k52443
answered Dec 12 '18 at 10:49
Arnaud D.Arnaud D.
15.9k52443
15.9k52443
add a comment |
add a comment |
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