ScalingStacks

Proof. [02KM]

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Proof.

Using the H-representation of polyhedra, one verifies that, if Λ1\Lambda_{1} and Λ2\Lambda_{2} are polyhedra with non-empty intersection, then any face of Λ1∩Λ2\Lambda_{1}\cap\Lambda_{2} is the intersection of a face of Λ1\Lambda_{1} with a face of Λ2\Lambda_{2}. This implies that Π1⋅Π2\Pi_{1}\cdot\Pi_{2} is a polyhedral complex.

Now suppose that Π1\Pi_{1} and Π2\Pi_{2} are complete. Let σ∈rec⁡(Π1⋅Π2)\sigma\in\operatorname{rec}(\Pi_{1}\cdot\Pi_{2}). This means that σ=rec⁡(Λ)\sigma=\operatorname{rec}(\Lambda) and Λ=Λ1∩Λ2\Lambda=\Lambda_{1}\cap\Lambda_{2} with Λi∈Πi\Lambda_{i}\in\Pi_{i}. It is easy to verify that Λ≠∅\Lambda\not=\emptyset implies rec⁡(Λ)=rec⁡(Λ1)∩rec⁡(Λ2)\operatorname{rec}(\Lambda)=\operatorname{rec}(\Lambda_{1})\cap\operatorname{rec}(\Lambda_{2}). Therefore σ∈rec⁡(Π1)⋅rec⁡(Π2)\sigma\in\operatorname{rec}(\Pi_{1})\cdot\operatorname{rec}(\Pi_{2}). This shows

rec⁡(Π1⋅Π2)⊂rec⁡(Π1)⋅rec⁡(Π2).\operatorname{rec}(\Pi_{1}\cdot\Pi_{2})\subset\operatorname{rec}(\Pi_{1})\cdot\operatorname{rec}(\Pi_{2}).

Since both complexes are complete, they agree. ∎

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