Given a short exact sequence with maps q and r:
one writes the additional arrows t and u for maps that may not exist:
Then the following statements are equivalent:
- 1. left split
- there exists a map t: B → A such that tq is the identity on A,
- 2. right split
- there exists a map u: C → B such that ru is the identity on C,
- 3. direct sum
- B is isomorphic to the direct sum of A and C, with q corresponding to the natural injection of A and r corresponding to the natural projection onto C.
The short exact sequence is called split if any of the above statements hold.
(The word "map" refers to morphisms in the abelian category we are working in, not mappings between sets.)
It allows one to refine the first isomorphism theorem:
- the first isomorphism theorem states that in the above short exact sequence,
- if the sequence splits, then , and the first isomorphism theorem is just the projection onto C.
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.
In the form stated here, the splitting lemma does not hold in the full category of groups, which is not an abelian category.
It is partially true: if a short exact sequence of groups is left split or a direct sum (conditions 1 or 3), then all of the conditions hold. For a direct sum this is clear, as one can inject from or project to the summands. For a left split sequence, the map gives an isomorphism, so B is a direct sum (condition 3), and thus inverting the isomorphism and composing with the natural injection gives an injection splitting r (condition 2).
However, if a short exact sequence of groups is right split (condition 2), then it need not be left split or a direct sum (neither condition 1 nor 3 follows): the problem is that the image of the right splitting need not be normal. What is true in this case is that B is a semidirect product, though not in general a direct product.
To form a counterexample, take the smallest non-abelian group , the symmetric group on three letters. Let A denote the alternating subgroup, and let . Let q and r denote the inclusion map and the sign map respectively, so that
is a short exact sequence. Condition (3) fails, because is not abelian. But condition (2) holds: we may define u: C → B by mapping the generator to any two-cycle. Note for completeness that condition (1) fails: any map t: B → A must map every two-cycle to the identity because the map has to be a group homomorphism, while the order of a two-cycle is 2 which can not be divided by the order of the elements in A other than the identity element, which is 3 as A is the alternating subgroup of , or namely the cyclic group of order 3. But every permutation is a product of two-cycles, so t is the trivial map, whence tq: A → A is the trivial map, not the identity.