nLab split exact sequence

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.

(e.g. Hatcher (2002), p. 147)

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

Assuming the axiom of choice:

Proposition

Every exact sequence of 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.

(Essentially by the basis theorem, for exposition see for instance here.)

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

• Charles Weibel, Section 1.4 of: An Introduction to Homological Algebra (1994)

• Allen Hatcher, pp. 147 of: Algebraic Topology, Cambridge University Press (2002) [ISBN:9780521795401, webpage]