Proof
First, to show that (3) implies both (1) and (2), we assume (3) and take as t the natural projection of the direct sum onto A, and take as u the natural injection of C into the direct sum.
To prove that (1) implies (3), first note that any member of B is in the set (ker t + im q). This follows since for all b in B, b = (b - qt(b)) + qt(b); qt(b) is obviously in im q, and (b - qt(b)) is in ker t, since
- t(b - qt(b)) = t(b) - tqt(b) = t(b) - (tq)t(b) = t(b) - t(b) = 0.
Next, the intersection of im q and ker t is 0, since if there exists a in A such that q(a) = b, and t(b) = 0, then 0 = tq(a) = a; and therefore, b = 0.
This proves that B is the direct sum of im q and ker t. So, for all b in B, b can be uniquely identified by some a in A, k in ker t, such that b = q(a) + k.
By exactness ker r = im q. The subsequence B → C → 0 implies that r is onto; therefore for any c in C there exists some b = q(a) + k such that c = r(b) = r(q(a) + k) = r(k). Therefore, for any c in C, exists k in ker t such that c = r(k), and r(ker t) = C.
If r(k) = 0, then k is in im q; since the intersection of im q and ker t = 0, then k = 0. Therefore the restriction of the morphism r : ker t → C is an isomorphism; and ker t is isomorphic to C.
Finally, im q is isomorphic to A due to the exactness of 0 → A → B; so B is isomorphic to the direct sum of A and C, which proves (3).
To show that (2) implies (3), we follow a similar argument. Any member of B is in the set ker r + im u; since for all b in B, b = (b - ur(b)) + ur(b), which is in ker r + im u. The intersection of ker r and im u is 0, since if r(b) = 0 and u(c) = b, then 0 = ru(c) = c.
By exactness, im q = ker r, and since q is an injection, im q is isomorphic to A, so A is isomorphic to ker r. Since ru is a bijection, u is an injection, and thus im u is isomorphic to C. So B is again the direct sum of A and C.
Read more about this topic: Splitting Lemma
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