Graduate A question about split short exact sequence

[lichen1983312]

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?

 

[micromass]

So can you start by telling us what your definitions are of a split sequence and of a free Abelian group?

 

[fresh_42]

Are there any theorems already, e.g. about projective modules?

 

[lichen1983312]

So can you start by telling us what your definitions are of a split sequence and of a free Abelian group?

Actually it is from the online material
http://www.math.colostate.edu/~renzo/teaching/Topology09/UniversalCoefficient.pdf
here is the screen shot .
my question is the problem 2. Actually I need it to understand another proof, but I don't know how to proof this one.

 

[lichen1983312]

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?

 

[mathwonk]

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.

 

[fresh_42]

And how is a free Abelian group defined? Is there a mapping property you can use to construct a commutative diagram for
$$ \begin{matrix} &&C&& \\ & & \downarrow & & \\ B & \longrightarrow & C & \longrightarrow & 0 \end{matrix} $$

 

[micromass]

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?

 

[lichen1983312]

you mean if $C$ is free abelian, there must exist a $g'$ such that $gg' = i{d_C}$? Can you please show me how?

 

[lichen1983312]

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?

 

[fresh_42]

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 $I$?

 

[lichen1983312]

So $C= \oplus_{\iota \in I} \mathbb{Z}_\iota$ for the basis $I$?

Yes....?

 

[fresh_42]

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 $gg'(c_\iota)=\underbrace{1_\iota + \ldots + 1_\iota}_{c_\iota \text{ times }}\;$?

 

[lavinia]

Here is an exact sequence of abelian groups. Why is it not split?

$0→Z_2→Z_4→Z_2→0$

 

[lichen1983312]

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?

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 $gg' = i{d_C}$.

 

[fresh_42]

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 $g'(c)\;$?

 

[lichen1983312]

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 ?

 

[fresh_42]

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 $gg'(c)\,$.

 

[lichen1983312]

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

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 $B'$

 

[fresh_42]

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.)

 

[lavinia]

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.

 

[lichen1983312]

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.