Although the assignment $X\mapsto \mathcal{P}\webleft (X\webright )^{\mathsf{op}}$ is called the free completion of $X$, it is not an idempotent operation, i.e. we have $\mathcal{P}\webleft (\mathcal{P}\webleft (X\webright )^{\mathsf{op}}\webright )^{\mathsf{op}}\neq \mathcal{P}\webleft (X\webright )^{\mathsf{op}}$.


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