Splitting Exact Sequences 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...
 A: (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.
