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.
On
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.
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...