TITLE: Penrose tiling substitution is bijective
QUESTION [7 upvotes]: Let $\mathcal{P}$ a Penrose tiling built by a substitution $\omega$ with two triangles.
It is claimed, for instance, in the article of Anderson and Putnam "Topological invariants for substitution tilings and their $C^{\star}$-algebras" that $\omega$ is bijective.
Why is it true?

REPLY [4 votes]: This was proved in much greater generality in the paper by B. Solomyak "Nonperiodicity implies unique composition for self-similar translationally finite Tilings", Disc. Comp. Geom. 20 (1998), 265–279.
The main theorem says that every non-periodic translationally-finite self-similar tiling (which includes Penrose tilings) has the unique composition property (which implies that your $\omega$ is a bijection).