Concept
In mathematics, and more specifically in homological algebra, the splitting lemma states that in any , the following statements are equivalent for a short exact sequence If any of these statements holds, the sequence is called a split exact sequence, and the sequence is said to split. In the above short exact sequence, where the sequence splits, it allows one to refine the first isomorphism theorem, which states that: C ≅ B/ker r ≅ B/q(A) (i.e., C isomorphic to the of r or cokernel of q) to: B = q(A) ⊕ u(C) ≅ A ⊕ C where the first isomorphism theorem is then just the projection onto C. It is a generalization of the rank–nullity theorem (in the form V ≅ ker T ⊕ im T) in linear algebra. 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 + q). This follows since for all b in B, b = (b − qt(b)) + qt(b); qt(b) is 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 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.