# nLab split exact sequence

### Context

#### Homological algebra

homological algebra

and

nonabelian homological algebra

diagram chasing

# Contents

## Definition

### In an abelian category

Let $\mathcal{A}$ be an abelian category.

###### Definition

A short exact sequence $0\to A \stackrel{i}{\to} B \stackrel{p}{\to} C\to 0$ in $\mathcal{A}$ is called split if either of the following equivalent conditions hold

1. There exists a section of $p$, hence a morphism $s \colon C\to B$ such that $p \circ s = id_C$.

2. There exists a retract of $i$, hence a morphism $r \colon B\to A$ such that $r \circ i = id_A$.

3. There exists an isomorphism of sequences with the sequence

$0\to A\to A\oplus C\to C\to 0$

given by the direct sum and its canonical injection/projection morphisms.

###### Lemma

(splitting lemma)

The three conditions in def. 1 are indeed equivalent.

###### Proof

It is clear that the third condition implies the first two: take the section/retract to be given by the canonical injection/projection maps that come with a direct sum.

Conversely, suppose we have a retract $r \colon B \to A$ of $i \colon A \to B$. Write $P \colon B \stackrel{r}{\to} A \stackrel{i}{\to} B$ for the corresponding idempotent.

Then every element $b \in B$ can be decomposed as $b = (b - P(b)) + P(b)$ hence with $b - P(b) \in ker(r)$ and $P(b) \in im(i)$. Moreover this decomposition is unique since if $b = i(a)$ while at the same time $r(b) = 0$ then $0 = r(i(a)) = a$. This shows that $B \simeq im(i) \oplus ker(r)$ is a direct sum and that $i \colon A \to B$ is the canonical inclusion of $im(i)$. By exactness it then follows that $ker(r) \simeq im(p)$ and hence that $B \simeq A \oplus C$ with the canonical inclusion and projection.

The implication that the second condition also implies the third is formally dual to this argument.

### In a semi-abelian category

There is a nonabelian analog of split exact sequences in semiabelian categories. See there.

## Properties

### Relation to chain homotopy

###### Proposition

A long exact sequence $C_\bullet$ is split exact precisely if the weak homotopy equivalence from the 0-chain complex, namely the quasi-isomorphism $0 \to C_\bullet$ is actually a chain homotopy equivalence, in that the identity on $C_\bullet$ has a null homotopy.

### Of free modules and vector spaces

###### Proposition

Every exact sequence of finitely generated free abelian groups is split.

###### Proposition

Every exact sequence of free modules which is bounded below is split.

Let $k$ be a field and denote by $\mathcal{A} \coloneqq k$Vect the category of vector spaces over $k$.

###### Corollary

Every short exact sequence of vector spaces is split.

### Involving injective/projective objects

###### Lemma

If in a short exact sequence $0 \to A \to B \to C \to 0$ in an abelian category the first object $A$ is an injective object or the last object is a projective object then the sequence is split exact.

###### Proof

Consider the first case. The other is formally dual.

By the properties of a short exact sequence the morphism $A \to B$ here is a monomorphism. By definition of injective object, if $A$ is injective then it has the right lifting property against monomorphisms and so there is a morphism $q : B \to A$ that makes the following diagram commute:

$\array{ A &\stackrel{id_A}{\to}& A \\ \downarrow & \nearrow_{q} \\ B } \,.$

Hence $q$ is a retract as in def. 1.

## References

For instance section 1.4 of

Revised on December 20, 2015 08:24:24 by Tim Porter (2.27.40.9)