Extensive category

Remarks

An equivalent form: $\mathcal{C}$ has pullbacks along the coproduct injections, and the following holds. For every commutative diagram whose bottom row is $A \to A + B \leftarrow B$ and whose top row is $X_1 \to X \leftarrow X_2$, the top row is a coproduct if and only if both squares are pullbacks. In an extensive category coproducts are disjoint, and the initial object is strict.

Example: the category of sets is extensive. A map $h \colon X \to A + B$ is, up to isomorphism, the sum of its restrictions $h^{-1}(A) \to A$ and $h^{-1}(B) \to B$, and this decomposition is unique.

Non-example: the category of real vector spaces, where the coproduct is the direct sum. Let $\Delta \colon \R \to \R \oplus \R$ be the diagonal $x \mapsto (x, x)$. The pullbacks of $\Delta$ along both injections are the zero space, since $(x, x) = (a, 0)$ forces $x = 0$. But $0 \oplus 0 = 0$ is not $\R$, so in the equivalent form the squares are pullbacks while the top row is not a coproduct.