4
$\begingroup$

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...

  • 0
    @ZhenLin: Oh, I see you are right. It's because sum=coproduct and product behave so nicely here.2012-10-05

1 Answers 1

2

(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.