Splitting lemma

related topics
{math, number, function}
{group, member, jewish}

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

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 are equivalent:

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

It allows one to refine the first isomorphism theorem:

  • the first isomorphism theorem states that in the above short exact sequence, C \cong B/q(A)
  • if the sequence splits, then B \cong q(A) \oplus u(C) \cong A \oplus C, and the first isomorphism theorem is just the projection onto C.

It is a categorical generalization of the rank–nullity theorem (in the form V \approx \ker T \oplus \operatorname{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 + 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

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 BC → 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 tC is an isomorphism; and ker t is isomorphic to C.

Finally, im q is isomorphic to A due to the exactness of 0 → AB; 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.

Full article ▸

related documents
Extended real number line
Richard's paradox
Haar measure
Unicity distance
Axiom of pairing
Legendre symbol
Ring (mathematics)
Functional analysis
Assignment problem
Meromorphic function
Elementary group theory
Presburger arithmetic
Examples of groups
Queue (data structure)
Mathematical model
Chain rule
Lagrange inversion theorem
Extended Backus–Naur Form
Oracle machine
Statistical independence
XSL Transformations
Referential transparency (computer science)
Boolean ring
ML (programming language)