Split short exact sequence

0 \to 𝐴 𝑓 ↪ 𝐵 𝑔 ↠ 𝐶 \to 0

A split short exact sequence¹ is a short exact sequence (depicted above) in an Abelian category that is equivalent to

0 \to 𝐴 𝑖 \to 𝐴 \oplus 𝐶 𝑝 \to 𝐶 \to 0

which is always exact.

Equivalent characterizations

The following characterisations are equivalent:² homology

the sequence splits;
$𝑔$ is a split epimorphism;
$𝑓$ is a split monomorphism.

Proof

We prove for a sequence in $𝑅 𝖬 𝗈 𝖽$ and thus for any Abelian category via Freyd-Mitchell theorem.

Consider a split sequence, i.e. the following diagram commutes.

$𝑝$ has a right-inverse, namely $𝑝 𝑖' = i d 𝐶$ for $𝑖' : 𝐶 ↣ 𝐴 \oplus 𝐶$ . Then $𝑔 𝛽 - 1 𝑖' = 𝑝 𝑖' = i d 𝐶$ , so $𝛽 - 1 𝑖'$ is a right-inverse of $𝑔$ . Therefore 1. implies 2.

Now take a short sequence such that $𝑔$ has a right-inverse $𝑟$ with $𝑔 𝑟 = i d 𝐶$ . Since $𝑓 : 𝐴 ↣ 𝐵$ is injective, there exists an inverse on its range $𝑓' : k e r 𝑔 ↠ 𝐴$ . $𝑏 - 𝑟 𝑔 (𝑏) \in k e r 𝑔$ for all $𝑏 \in 𝐵$ since
$𝑔 (𝑏 - 𝑟 𝑔 (𝑏)) = 𝑔 (𝑏) - 𝑔 𝑟 𝑔 (𝑏) = 𝑔 (𝑏) - 𝑔 (𝑏) = 0$
Thus we may define $𝑞 (𝑏) = 𝑓' (𝑏 - 𝑟 𝑔 (𝑏))$ , which is a left-inverse of $𝑓$ since
$𝑞 𝑓 (𝑎) = 𝑓' (𝑓 (𝑎) - 𝑟 𝑔 𝑓 (𝑎)) = 𝑓' (𝑓 (𝑎) - 𝑟 (0)) = 𝑎$
for all $𝑎 \in 𝐴$ . Therefore 2. implies 3..

Finally take a short sequence such that $𝑓$ has a left-inverse $𝑞$ with $𝑞 𝑓 = i d 𝐴$ . Let $𝛽 = 𝑞 \oplus 𝑔$ . Then $(i d 𝐴, 𝛽, i d 𝐶)$ is a morphism of short exact sequences, since
$𝛽 𝑓 = 𝑞 𝑓 \oplus 𝑔 𝑓 = i d 𝐴 \oplus 0 = 𝑖$
and
$𝑝 𝛽 = 𝑝 (𝑞 \oplus 𝑔) = 𝑔$
Hence by the Five lemma $𝛽$ is an isomorphism, whence $(i d 𝐴, 𝛽, i d 𝐶)$ is an isomorphism of short sequences. Therefore 3. implies 1..

tidy | en | SemBr

German spaltete kurze exakte Sequenz ↩
2010, Algebraische Topologie, ¶3.1.11, pp. 132ff ↩

Table of Contents

Split short exact sequence

Equivalent characterizations

Graph View

Backlinks

Table of Contents

Split short exact sequence

Equivalent characterizations

Footnotes

Graph View

Backlinks