Isomorphism of intervals of a distributive lattice Let $P$ be a distributive lattice and let for $a,b \in P$ by $a \land b$ and $a \lor b$ denote the usual notion of infimum and supremum.
How can I show the following isomorphism?
$$[a \land b , a \lor b] \cong [a \land b,a] \times [a \land b,b]$$
The following is obviously true if for example $a \land b = b$ and $a \lor b = a$ but I don't know how to prove it in general.
 A: The first step is to find a reasonable candidate for explicit isomorphism.
Here the most natural idea is $f(x)=(x\wedge a, x\wedge b)$. And for the reciprocal, $g(x_a,x_b)=x_a\vee x_b$.
We can show that they are reciprocal, using the fact that the lattice is distributive: $g(f(x))= (x\wedge a)\vee(x\wedge b)= x\vee (a\wedge b)=x$.
The fact that is is a lattice morphism is also a consequence of distributivity, left as exercise...
A: To get an intuitive idea of the isomorphism think for example of $P$ as being the set of subsets of some fixed set $S$, partially ordered by inclusion. If $A,B \subseteq S$ are fixed subsets, then in order to specify a set $X \subseteq S$ contained in $A \cup B$ and containing $A \cap B$ you just need to say which elements from $A$ and which elements from $B$ it contains. Thus $X$ is uniquely determined by the pair $(X \cap A, X \cap B)$. Conversely every pair $(Y,Z)$ with $A \cap B \subseteq Y \subseteq A$ and $A \cap B \subseteq Z \subseteq B$ determines such an $X$ via $X = Y \cup Z$.
Thus, in the general distributive lattice $P$, you suspect that the isomorphism $[a \wedge b, a \vee b] \cong [a\wedge b,a] \times [a\wedge b, b]$ is given by $x \mapsto (x \wedge a, x \wedge b)$ in one direction, and by $(y,z) \mapsto y \vee z$ in the other. It is now easily checked using the lattice axioms and the distributivity that these two maps are lattice homomorphisms and each other's inverse.
