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8:42 PM
let $0 \to A \to B \to C \to 0$ be exact
$C \to B$ be section
the maps are f,g,s
let b in B. then s(g(b)) in B
consider b-s(g(b))
g(b-s(g(b))) = g(b) - g(s(g(b))) = g(b) - g(b) = 0
so it is in the kernel
so b=f(a)+s(g(b)) for some a in A
now let f(a)+s(c) = f(a')+s(c')
f(a-a') = s(c'-c)
g(f(a-a')) = g(s(c'-c))
0 = c'-c
f(a-a') = 0
a-a' = 0
therefore B is isomorphic to A oplus C
(this is the splitting lemma)
 

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