Dedekind Completion is Unique up to Isomorphism

Theorem
Let $S$ be an ordered set.

Let $\left({X, f}\right)$ and $\left({Y, g}\right)$ be Dedekind completions of $S$.

Then there exists a unique order isomorphism $\phi: X \to Y$ such that $\phi \circ f = g$.

Proof
By assumption, $\left({X, f}\right)$ is a Dedekind completion of $S$.

Also, $g: S \to Y$ is an order embedding and $Y$ is Dedekind complete.

Hence by definition of Dedekind completion, there exists a unique $\phi: X \to Y$ such that:


 * $\phi \circ f = g$

It only remains to show that $\psi$ is an order isomorphism.

By reversing the roles of $X$ and $Y$ in the above, we obtain a unique $\psi: Y \to X$ subject to:


 * $\psi \circ g = f$

Therefore:


 * $\psi \circ \phi \circ f = f$
 * $\phi \circ \psi \circ g = g$

By the uniqueness clause of the definition of Dedekind completion, it follows that:


 * $\psi \circ \phi = \operatorname{id}_X$
 * $\phi \circ \psi = \operatorname{id}_Y$

Hence $\psi$ is an order isomorphism.

Also see

 * Existence of Dedekind Completion