A question about split short exact sequence

In summary, the free abelian group defined by the correspondence between generators and pre-images is not the same as the group generated on a basis.
  • #1
lichen1983312
85
2
I am looking at a statement that, for a short exact sequence of Abelian groups

##0 \to A\mathop \to \limits^f B\mathop \to \limits^g C \to 0##

if ##C## is a free abelian group then this short exact sequence is split

I cannot figured out why, can anybody help?
 
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  • #2
So can you start by telling us what your definitions are of a split sequence and of a free Abelian group?
 
  • #3
Are there any theorems already, e.g. about projective modules?
 
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  • #4

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  • #5
fresh_42 said:
Are there any theorems already, e.g. about projective modules?
Please see the attached picture in the post above. Since I know nothing about modules, can we restrict our case to groups?
 
  • #6
the splitting would follow if the map g has a right inverse whiuch is a group homomorphism. free abelian groups are, almost by definition, those abelian groups on which it is very easy to define homomorphisms.
 
  • #7
And how is a free Abelian group defined? Is there a mapping property you can use to construct a commutative diagram for
$$ \begin{matrix} &&C&& \\ & & \downarrow & & \\ B & \longrightarrow & C & \longrightarrow & 0 \end{matrix} $$
 
  • #8
lichen1983312 said:
you mean if ##C## is free abelian, there must exist a ##g'## such that ##gg' = i{d_C}##? Can you please show me how?

What is your definition of a free abelian group?
 
  • #9
you mean if ##C## is free abelian, there must exist a ##g'## such that ##gg' = i{d_C}##? Can you please show me how?
 
  • #10
micromass said:
What is your definition of a free abelian group?
An abelian group that is generated on a basis? I guess you ask me this question trying to lead me to somewhere?
 
  • #11
lichen1983312 said:
An abelian group that is generated on a basis? I guess you ask me this question trying to lead me to somewhere?
So ##C= \oplus_{\iota \in I} \mathbb{Z}_\iota## for the basis ##I##?
 
  • #12
fresh_42 said:
So ##C= \oplus_{\iota \in I} \mathbb{Z}_\iota## for the basis ##I##?
Yes....?
 
  • #13
Now ##g## is surjective and for each ##1_{\iota}## in this component of ##C## there is a pre-image ##b_\iota \in B##, i.e. ##g(b_\iota)= id_C (1_\iota)##. Now how could you define your ##g' \, : \, C \rightarrow B## with ##gg'(c_\iota)=\underbrace{1_\iota + \ldots + 1_\iota}_{c_\iota \text{ times }}\;##?
 
  • #14
Here is an exact sequence of abelian groups. Why is it not split?

## 0→Z_2→Z_4→Z_2→0##
 
  • #15
lavinia said:
Here is an exact sequence of abelian groups. Why is it not split?

## 0→Z_2→Z_4→Z_2→0##
I don't see that ##Z_4## is that same as ##{Z_2} \oplus {Z_2}##, but you have another point to make, rihgt?
fresh_42 said:
Now ##g## is surjective and for each ##1_{\iota}## in this component of ##C## there is a pre-image ##b_\iota \in B##, i.e. ##g(b_\iota)= id_C (1_\iota)##. Now how could you define your ##g' \, : \, C \rightarrow B## with ##gg'(c_\iota)=\underbrace{1_\iota + \ldots + 1_\iota}_{c_\iota \text{ times }}\;##?
Ok guys, stop teasing, I already feel that I am in a zoo.
Is this understanding right?

Say ##C## is a free abelien group built on a basis ##\{ {a_i}\} ## . Since ##g## is surjective, for each ##{a_i}## there is a ##{b_i} \in B## such that ##g({b_i}) = {a_i}##. It is also true that ##g({n_i}{b_i}) = {n_i}{a_i}##. This correspondence on the other hand defines an isomorphism (is this right?) between ##\{ {n_i}{b_i}\} ## and ##\{ {n_i}{a_i}\} ##. So we find a ##g':C \to B## such that ##gg' = i{d_C}##.
 
  • #16
lichen1983312 said:
Say ##C## is a free abelien group built on a basis ##\{ {a_i}\} ## . Since ##g## is surjective, for each ##{a_i}## there is a ##{b_i} \in B## such that ##g({b_i}) = {a_i}##. It is also true that ##g({n_i}{b_i}) = {n_i}{a_i}##. This correspondence on the other hand defines an isomorphism (is this right?) between ##\{ {n_i}{b_i}\} ## and ##\{ {n_i}{a_i}\} ##. So we find a ##g':C \to B## such that ##gg' = i{d_C}##.
Well, I've already almost written the proof. Whether you call the generators ##a_i## or like me ##1_\iota## doesn't make a difference. Considering your proof, I cannot see this isomorphism. What happens in a situation like
$$ 0 \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z} \oplus \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow 0$$
and how exactly do you define ##g'(c)\;##?
 
  • #17
For the isomorphism, I mean I can choose a subset ##B'## of ##B## in this way
1. 0 is in ##B'## such that ##g(0) = 0##
2. For each generator ##{c_i} \in C##, find a ##{b_i} \in B## such that ##g({b_i}) \in {c_i}##
3. Take ## - {c_i}## in ##B'##
then if ##g## is a homomorphism and surjective, then ##B'## should also be a free abelian group and isomorphic to ##C##.
is this right ?
 
  • #18
lichen1983312 said:
For the isomorphism, I mean I can choose a subset ##B'## of ##B## in this way
1. 0 is in ##B'## such that ##g(0) = 0##
What for? And which isomorphism?
2. For each generator ##{c_i} \in C##, find a ##{b_i} \in B## such that ##g({b_i}) \in {c_i}##
Yes, because ##g## is surjective.
3. Take ## - {c_i}## in ##B'##
If the ##c_i## are still in ##C##, how can they be in ##B' \subseteq B \, ##?
then if ##g## is a homomorphism and surjective, then ##B'## should also be a free abelian group and isomorphic to ##C##.
is this right ?
To me this looks as if you proved ##C=\mathbb{F}_{B'}=C## but the main point is, that the ##b_i## aren't in ##C\,##.

What about what I've written in post #13? Simply define ##g'(c_\iota) := b_\iota## where the ##c_\iota## are generators of ##C## with pre-images ##b_\iota \in B## under ##g##. I only took ##\iota## as index instead of ##i## because usually ##i## denotes something countable and ##\iota## any index set ##I\,##. And we don't know anything about ##I\,##. You don't have to bother, whether ##B## is isomorphic to ##C## or not. All we can say for sure is, that ##g## is surjective and ##C## is freely generated over ##\{c_\iota \,\vert \, \iota \in I\}\,##, i.e. an arbitrary ##c \in C## looks like ##c= \Sigma_{\iota \in I} \,n_\iota \cdot c_\iota##. With that you can calculate ##gg'(c)\,##.
 
  • #19
fresh_42 said:
What for? And which isomorphism?

Yes, because ##g## is surjective.

If the ##c_i## are still in ##C##, how can they be in ##B' \subseteq B \, ##?

To me this looks as if you proved ##C=\mathbb{F}_{B'}=C## but the main point is, that the ##b_i## aren't in ##C\,##.

What about what I've written in post #13? Simply define ##g'(c_\iota) := b_\iota## where the ##c_\iota## are generators of ##C## with pre-images ##b_\iota \in B## under ##g##. I only took ##\iota## as index instead of ##i## because usually ##i## denotes something countable and ##\iota## any index set ##I\,##. And we don't know anything about ##I\,##. You don't have to bother, whether ##B## is isomorphic to ##C## or not. All we can say for sure is, that ##g## is surjective and ##C## is freely generated over ##\{c_\iota \,\vert \, \iota \in I\}\,##, i.e. an arbitrary ##c \in C## looks like ##c= \Sigma_{\iota \in I} \,n_\iota \cdot c_\iota##. With that you can calculate ##gg'(c)\,##.

My bad, I was in a hurry this afternoon, there are big typos in 2 and 3
lichen1983312 said:
For the isomorphism, I mean I can choose a subset ##B'## of ##B## in this way
1. 0 is in ##B'## such that ##g(0) = 0##
2. For each generator ##{c_i} \in C##, find a ##{b_i} \in B## such that ##g({b_i}) \in {c_i}##
3. Take ## - {c_i}## in ##B'##
then if ##g## is a homomorphism and surjective, then ##B'## should also be a free abelian group and isomorphic to ##C##.
is this right ?

I was trying to say, by doing the following 3 things, can I find a subset ##B'## in ##B## that is isomorphic to C?
1. 0 is in ##B'## such that ##g(0) = 0##
2. For each generator ##{c_i} \in C##, take a ##{b_i} \in B## in ##B'## such that ##g({b_i}) = c_i##
3. Take ## - {b_i}## in ##B'##
 
  • #20
What about ##\{0\} \longrightarrow \mathbb{Z}_2 \longrightarrow \mathbb{Z}_2 \oplus \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow \{0\}\; ##?

You still need to actually define ##g'\, : \, C \longrightarrow B## with ##gg' = id_C\;##. What you don't need is an isomorphism with ##B## (see my example) and homomorphy isn't the problem. If you define all ##g'(c_\iota)## you can extend it to an homomorphism. There is no need to introduce a set ##B'## or to consider the group it generates, only the elements ##b_\iota## with ##g(b_\iota) = c_\iota \,## are needed. Of course there is still the equivalence of the three properties in your definition (or theorem?) of a split exact sequence. (Just as a remark: In my book it says "direct exact sequence", which is somehow telling. However, "split" is the usual term.)
 
  • #21
lichen1983312 said:
I was trying to say, by doing the following 3 things, can I find a subset ##B'## in ##B## that is isomorphic to C?
1. 0 is in ##B'## such that ##g(0) = 0##
2. For each generator ##{c_i} \in C##, take a ##{b_i} \in B## in ##B'## such that ##g({b_i}) = c_i##
3. Take ## - {b_i}## in ##B'##

This is essentially correct. You still need to write down the splitting homomorphism from ##C## into ##B## and to show that it is actually a homomorphism.

To illustrate why, consider a couple examples.

For the sequence ##0→Z_4→Z_8→Z_2→0##, a generator ##c## of ##Z_2## has a lift ##b## to ##Z_8## so ##gg'(c) = c## but the map ##0→0## ##c→b## is not a homomorphism.

If ##C = Z⊕Z## and ##B## is the free group on two generators, one has an exact sequence

##1 →A →F(b_1,b_2) →Z⊕Z→0## where ##A## is the commutator subgroup of ##F(b_1,b_2)##. The set ##b_1## and ##b_2## and ##1## are your set ##B'## but the map ##c_1→b_1## ##c_2→b_2## ##0→1## does not determine a homomorphism.

Another example, similar to the free group example, is perhaps a little more clear. Let ##B## be a group with 3 generators ##a## ##b_1## and ##b_2## each of which generates a free abelian group and with the relations ##[b_1,b_2] = a## and ##[a,b_{i}] = 1## where ##[,]## denotes the commutator.

##B## is not abelian and it fits the exact sequence, ##0→ Z →B→Z⊕Z→0## This sequence is not split.

- It might be helpful to consider whether or not every sequence ##1→A→B→Z→0## is split whether or not ##B## is abelian.
 
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  • #22
fresh_42 said:
What about ##\{0\} \longrightarrow \mathbb{Z}_2 \longrightarrow \mathbb{Z}_2 \oplus \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow \{0\}\; ##?

You still need to actually define ##g'\, : \, C \longrightarrow B## with ##gg' = id_C\;##. What you don't need is an isomorphism with ##B## (see my example) and homomorphy isn't the problem. If you define all ##g'(c_\iota)## you can extend it to an homomorphism. There is no need to introduce a set ##B'## or to consider the group it generates, only the elements ##b_\iota## with ##g(b_\iota) = c_\iota \,## are needed. Of course there is still the equivalence of the three properties in your definition (or theorem?) of a split exact sequence. (Just as a remark: In my book it says "direct exact sequence", which is somehow telling. However, "split" is the usual term.)
Thanks very much, I got the point.
 

1. What is a split short exact sequence?

A split short exact sequence is a sequence of mathematical objects (usually groups or vector spaces) connected by homomorphisms in a specific way. It is called "short" because the sequence only has three terms, and "exact" because each homomorphism in the sequence is both injective and surjective.

2. How is a split short exact sequence different from a regular short exact sequence?

A split short exact sequence is different from a regular short exact sequence in that it has an additional property: it is split. This means that there exists a homomorphism in the opposite direction of each homomorphism in the sequence, making the sequence "split" into two sub-sequences.

3. What is the significance of a split short exact sequence?

A split short exact sequence is significant because it allows us to break down and understand more complicated mathematical objects by studying simpler ones. It also provides a way to connect different areas of mathematics and make connections between seemingly unrelated concepts.

4. Can you give an example of a split short exact sequence?

One example of a split short exact sequence is the sequence of vector spaces: 0 → V → V ⊕ W → W → 0, where V and W are vector spaces and the arrows represent the inclusion and projection maps. This sequence is split because there exists a homomorphism from V ⊕ W back to V that maps (v,w) to v, and a homomorphism from V to V ⊕ W that maps v to (v,0).

5. How are split short exact sequences used in mathematics?

Split short exact sequences are used in many areas of mathematics, such as algebra, topology, and differential equations. They are often used to prove theorems, make connections between different mathematical concepts, and simplify complex problems by breaking them down into smaller parts. They also have applications in physics and engineering, such as in the study of electromagnetic fields and control theory.

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