Defining "distributive category" by requiring the existence of an isomorphism instead of requiring that the canonical morphism is an iso A category $\mathfrak C$ is distributive if it has binary products $\times$ and binary coproducts $\sqcup$, and the canonical morphism
$$X\times Y\sqcup X\times Z\longrightarrow X\times(Y\sqcup Z)$$
is an iso.
Why does one require that this particular morphism is an isomorphism? Why doesn't one require that there exists an isomorphism between $X\times Y\sqcup X\times Z$ and $X\times(Y\sqcup Z)$?
In particular: Is a category distributive if and only if there's a natural isomorphism between $X\times Y\sqcup X\times Z$ and $X\times(Y\sqcup Z)$ (natural in $X,Y,Z$)?
 A: Distributivity can be phrased abstractly as follows. There are 2-monads $\mathrm{Cart}$ and $\mathrm{Cocart}$ for cartesian categories and cocartesian categories respectively, between which there exists a distributive law establishing $\mathrm{Cocart} \circ \mathrm{Cart}$ as a composite 2-monad. The algebras for this 2-monad are exactly distributive categories. An algebra for a distributive law is an algebra for each of the two monads forming the composite, together with the requirement that the latter structure is preserved by the former structure: this is precisely the condition Qiaochu Yuan mentions in the comments about the finite product functors preserving finite coproducts. This shows that the distributivity condition is very natural, arising from distributive laws of 2-monads.
However, it happens (surprisingly) that the distributivity condition can be replaced by the requirement for any such natural transformation to be invertible. This is proven in Lack's Non-canonical isomorphisms. One should not expect this in general, but under certain circumstances holds more generally: for this, see Nunes's Pseudoalgebras and non-canonical isomorphisms.
