# Splitting lemma

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

Then the following statements are equivalent:

- Right split
- There exists a morphism
*u*:*C*→*B*such that*ru*is the identity on*C*, id_{C},

- Direct sum
- There is an isomorphism h from
*B*to the direct sum of*A*and*C*, such that*hq*is the natural monomorphism of*A*in the direct sum, and is the natural projection of the direct sum onto*C*.

If these statements hold, 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:

to:

*B*=*q*(*A*) ⊕*u*(*C*) ≅*A*⊕*C*

where the first isomorphism theorem is then just the projection onto *C*.

It is a categorical generalization of the rank–nullity theorem (in the form V ≅ ker *T* ⊕ im *T*) in linear algebra.

## Proof for the category of abelian groups

### 3. ⇒ 1. and 3. ⇒ 2.

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.

### 1. ⇒ 3.

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).

### 2. ⇒ 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*.

An alternative "abstract nonsense" proof of the splitting lemma may be formulated entirely in category theoretic terms.

## 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.

### Partially true

It is partially true: if a short exact sequence of groups is left split or a direct sum (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 t × r: *B* → *A* × *C* gives an isomorphism, so *B* is a direct sum (3.), and thus inverting the isomorphism and composing with the natural injection *C* → *A* × *C* gives an injection *C* → *B* splitting *r* (2.).

However, if a short exact sequence of groups is right split (2.), then it need not be left split or a direct sum (neither 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.

### Counterexample

To form a counterexample, take the smallest non-abelian group *B* ≅ *S*_{3}, the symmetric group on three letters. Let *A* denote the alternating subgroup, and let *C* = *B*/*A* ≅ {±1}. Let *q* and *r* denote the inclusion map and the sign map respectively, so that

is a short exact sequence. 3. fails, because *S*_{3} is not abelian. But 2. holds: we may define *u*: *C* → *B* by mapping the generator to any two-cycle. Note for completeness that 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 *S*_{3}, 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.

## References

- Saunders Mac Lane:
*Homology*. Reprint of the 1975 edition, Springer Classics in Mathematics, ISBN 3-540-58662-8, p. 16 - Allen Hatcher:
*Algebraic Topology*. 2002, Cambridge University Press, ISBN 0-521-79540-0, p. 147