Splitting lemma

Remarks

From (3), the morphisms $t = p_A \circ h$ and $u = h^{-1} \circ i_C$ give (1) and (2). From (1), the morphism $h = (t, g) \colon B \to A \oplus C$ satisfies $h \circ f = i_A$ and $p_C \circ h = g$, and it is an isomorphism by the five lemma applied to the map from the given sequence to $0 \to A \to A \oplus C \to C \to 0$. From (2), one argues dually, using $B \cong A \oplus C$ through $f$ and $u$. Example: in $0 \to \Z/2\Z \to \Z/4\Z \to \Z/2\Z \to 0$, the quotient map has no right inverse. A right inverse would send the generator of $\Z/2\Z$ to an element of $\Z/4\Z$ of order at most $2$, that is, to $0$ or $2$, and both map to $0$. So the sequence does not split, in agreement with $\Z/4\Z \not\cong \Z/2\Z \oplus \Z/2\Z$. The lemma fails for non-abelian groups: for the symmetric group, $1 \to A_3 \to S_3 \to \Z/2\Z \to 1$ has a right inverse of the quotient map, but $S_3$ is not isomorphic to $A_3 \times \Z/2\Z$.

Going deeper

For a student

The lemma is the standard test for splitting. For example, every short exact sequence ending in a projective object splits, since the identity of $C$ lifts along $g$. Dually, every short exact sequence starting with an injective object splits. For vector spaces every short exact sequence splits, because a basis of $C$ can be lifted to $B$. In the theory of extensions, the split sequence is the zero element of $\mathrm{Ext}^1(C, A)$. For groups, condition (2) alone gives a semidirect product, and condition (1) gives a direct product.