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Unitary in the corner algebra! #4

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@maximilianweinhold

Instantly obtain an "intertwining/commutation-on-the-support relation"! I don't think we need persistent bounds to be unitary themselves, only to always give back the same! If that breaks you're on your own.

Much rejoicing in the square.

Or maybe just a normal operator with a prescribed sandwich F F* (+ isometry).~~~ Depends on the existence of a projection.

It also might not be setwise, though sets are very pretty. Operator inequalities and yard-side tacs. Don't think I can play Goedel if its not setwise.

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