Splitting lemma

From Wikipedia, the free encyclopedia

In mathematics, and more specifically in homological algebra, the splitting lemma states that the following statements regarding the below short exact sequence in any abelian category are equivalent:

$0 \rightarrow A {{q \atop \longrightarrow} \atop {\longleftarrow \atop t}} B {{r \atop \longrightarrow} \atop {\longleftarrow \atop u}} C \rightarrow 0 \,$

there exists a map t: B → A such that tq is the identity on A,
there exists a map u: C → B such that ru is the identity on C,
B is isomorphic to the direct sum of A and C, with q being the natural injection of A and r being the natural projection onto C.

The short exact sequence is called split if any the above statements hold.

[edit] 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 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 rq = A, and so 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.

[edit] Non-abelian groups

This article or section is in need of attention from an expert on the subject.

WikiProject Mathematics or the Mathematics Portal may be able to help recruit one.

If a more appropriate WikiProject or portal exists, please adjust this template accordingly.

In the form stated here, the splitting lemma does not hold in the full category of groups, which is not an abelian category.

To form a counterexample, take the smallest non-abelian group $B\cong S_3$ , the symmetric group on three letters. Let A denote the alternating subgroup, and let $C=B/A\cong\{\pm 1\}$ . Let q and r denote the inclusion map and the sign map respectively, so that

$0 \rightarrow A \stackrel{q}{\longrightarrow} B \stackrel{r}{\longrightarrow} C \rightarrow 0 \,$

is a short exact sequence. Condition (3) fails, because $S 3$ 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, by Lagrange's theorem. 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.