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A A question about split short exact sequence

  1. Jan 13, 2017 #1
    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?
     
  2. jcsd
  3. Jan 13, 2017 #2

    micromass

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    So can you start by telling us what your definitions are of a split sequence and of a free Abelian group?
     
  4. Jan 13, 2017 #3

    fresh_42

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    Are there any theorems already, e.g. about projective modules?
     
    Last edited: Jan 13, 2017
  5. Jan 13, 2017 #4

    Attached Files:

  6. Jan 13, 2017 #5
    Please see the attached picture in the post above. Since I know nothing about modules, can we restrict our case to groups?
     
  7. Jan 13, 2017 #6

    mathwonk

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    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.
     
  8. Jan 13, 2017 #7

    fresh_42

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    And how is a free Abelian group defined? Is there a mapping property you can use to construct a commutative diagram for
    $$ \begin{matrix} &&C&& \\ & & \downarrow & & \\ B & \longrightarrow & C & \longrightarrow & 0 \end{matrix} $$
     
  9. Jan 13, 2017 #8

    micromass

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    What is your definition of a free abelian group?
     
  10. Jan 13, 2017 #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?
     
  11. Jan 13, 2017 #10
    An abelian group that is generated on a basis? I guess you ask me this question trying to lead me to somewhere?
     
  12. Jan 13, 2017 #11

    fresh_42

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    So ##C= \oplus_{\iota \in I} \mathbb{Z}_\iota## for the basis ##I##?
     
  13. Jan 13, 2017 #12
    Yes.............?
     
  14. Jan 13, 2017 #13

    fresh_42

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    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 }}\;##?
     
  15. Jan 13, 2017 #14

    lavinia

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

    ## 0→Z_2→Z_4→Z_2→0##
     
  16. Jan 13, 2017 #15
    I don't see that ##Z_4## is that same as ##{Z_2} \oplus {Z_2}##, but you have another point to make, rihgt?


    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}##.
     
  17. Jan 14, 2017 #16

    fresh_42

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    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)\;##?
     
  18. Jan 14, 2017 #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 ?
     
  19. Jan 14, 2017 #18

    fresh_42

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    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)\,##.
     
  20. Jan 14, 2017 #19
    My bad, I was in a hurry this afternoon, there are big typos in 2 and 3
    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'##
     
  21. Jan 14, 2017 #20

    fresh_42

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    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.)
     
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