nLab split exact sequence

Contents

Context

Homological algebra

homological algebra

Introduction

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. 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 free abelian groups is split, assuming the axiom of choice.

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. .

References

For instance section 1.4 of

Last revised on September 25, 2019 at 21:49:08. See the history of this page for a list of all contributions to it.