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 $\psi: X \to Y$ such that $\psi \circ f = g$.

Proof
The uniqueness of $\psi$ follows directly from the definition of a Dedekind completion.

By definition, there exist unique increasing mappings $\tilde{g}: X \to Y$ and $\tilde{f}: Y \to X$ such that:
 * $\tilde{g} \circ f = g$
 * $\tilde{f} \circ g = f$

It follows that:
 * $\bigl({\tilde{g} \circ \tilde{f}}\bigr) \circ g = {\operatorname{id}_Y} \circ g$
 * $\bigl({\tilde{f} \circ \tilde{g}}\bigr) \circ f = {\operatorname{id}_X} \circ f$

By definition, it follows that:
 * $\tilde{g} \circ \tilde{f} = \operatorname{id}_Y$
 * $\tilde{f} \circ \tilde{g} = {\operatorname{id}_X}$

Hence, $\tilde{f} = \tilde{g}^{-1}$, and thus $\tilde{g}$ is an order isomorphism.