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Given an exact sequence of vector spaces: $$0\longrightarrow U \longrightarrow V \longrightarrow W\longrightarrow 0$$ with $f:U\rightarrow V$ and $g: V \rightarrow W$

I want to prove the that following are equivalent:

$\bullet$ The sequence splits on the right ($\exists s:W\rightarrow V$ such that $g\circ s =1_W$)

$\bullet$ The sequence splits on the left ($\exists t: V\rightarrow U$ such that $t\circ f=1_U$)

$\bullet$ $\exists\gamma : V\rightarrow U\oplus W$ an isomorphism satisfying $\gamma \circ f=i_1$ and $p_2\circ \gamma=g$. Where $i_1$ and $p_2$ are the usual inclusion and projection into the first and second summand respectively.

So exactness gives us that $f$ is 1:1 and $g$ is onto, so clearly there exists a function with the property of the second bullet, but I can't quite figure out how to know it's linear on all of $V$ (or how this uses any assumptions from the first bullet.) I'm more confused about the latter two implications. Don't really know where to start with those unfortunately...

Edit: It says explicitly that I'm supposed to avoid using anything about bases here! For some reason...

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en.wikipedia.org/wiki/Splitting_lemma#Proof –  user17794 Oct 5 '12 at 4:45
    
Hint: pick basis vectors for $U$ and seeing where they go. Then do the same for $W$ and see where they come from. –  KReiser Oct 5 '12 at 4:46
    
@KReiser Allegedly, I'm supposed to do this without knowing that any bases exist...for some reason.. –  AsinglePANCAKE Oct 5 '12 at 4:47
    
You can avoid using bases if you try to prove the statement in a more general category, i.e. merely assume a concrete abelian category, i.e. objects and morhisms are essentially sets and maps and morphisms are an abelian group (i.e. can be added). This allows fun with diagram chasing... –  Hagen von Eitzen Oct 5 '12 at 5:41
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You don't even need concreteness or elements here – a completely morphism-theoretic proof is possible and isn't awkward in the least. –  Zhen Lin Oct 5 '12 at 6:43
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1 Answer 1

up vote 2 down vote accepted

(1) The sequence splits on the right ($\exists s:W\rightarrow V$ such that $g\circ s =1_W$)

(2) The sequence splits on the left ($\exists t: V\rightarrow U$ such that $t\circ f=1_U$)

(3) $\exists\gamma : V\rightarrow U\oplus W$ an isomorphism satisfying $\gamma \circ f=i_1$ and $p_2\circ \gamma=g$. Where $i_1$ and $p_2$ are the usual inclusion and projection into the first and second summand respectively.

(1) $\Rightarrow$ (3):

Let $x \in V$. $g(x - sg(x)) = g(x) - g(x) = 0$. Hence there exists $y \in U$ such that $x - sg(x) = f(y)$. Hence $V = f(U) + s(W)$. Let $a \in f(U) \cap s(W)$. There exist $u \in U, w \in W$ such that $a = f(u) = s(w)$. $g(a) = g(f(u)) = gs(w) = w$. Hence $w = 0$. Hence $a = s(w) = 0$. Therefore $V = f(U) \oplus s(W)$.

(3) $\Rightarrow$ (1): Clear.

(2) $\Rightarrow$ (3):

Let $K = Ker(t)$. Let $x \in V$. $t(x - ft(x)) = t(x) - t(x) = 0$. Hence $x - ft(x) \in K$. Hence $V = f(U) + K$. Let $a \in f(U) \cap K$. There exist $u \in U, k \in K$ such that $a = f(u) = k$. $t(a) = tf(u) = u = t(k) = 0$. Hence $a = f(u) = 0$. Hence $V = f(U) \oplus K$. It remains to prove that $g|K\colon K \rightarrow W$ is an isomorphism. Suppose $g(k) = 0$, where $k \in K$. There exists $u \in U$ such that $k = f(u)$. Since $0 = t(k) = tf(u) = u$, $k = 0$. Hence $g|K$ is injective. Let $w \in W$. There exists $x \in V$ such that $w = g(x)$. Then $x - ft(x) \in K$ and $g(x - ft(x)) = g(x) = w$. Hence $g|K$ is surjective.

(3) $\Rightarrow$ (2): Clear.

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