Splitting lemma - Citizendia

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. Mathematics is the body of Knowledge and Academic discipline that studies such concepts as Quantity, Structure, Space and Homological algebra is the branch of Mathematics which studies homology in a general algebraic setting In Mathematics, especially in Homological algebra and other applications of Abelian category theory as well as in Differential geometry and Group In Mathematics, an abelian category is a category in which Morphisms and objects can be added and in which kernels and Cokernels exist In Mathematics, an identity function, also called identity map or identity transformation, is a function that always returns the same value that The symbol \oplus \! denotes direct sum it is also the astrological and astronomical symbol for Earth, and a symbol for the Exclusive disjunction

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

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). In the various branches of Mathematics that fall under the heading of Abstract algebra, the kernel of a Homomorphism measures the degree to which the homomorphism An image (from Latin imago) or picture is an artifact usually two-dimensional that has a similar appearance to some subject &mdashusually 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.

Non-abelian groups

In the form stated here, the splitting lemma does not hold in the full category of groups, which is not an abelian category. In Mathematics, the category Grp has the class of all groups for objects and Group homomorphisms for Morphisms As such

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. Lagrange's theorem, in the Mathematics of Group theory, states that for any Finite group G, the order (number of elements of 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.

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