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Mar
15
comment Proving $(f^{-1}(U))^0 = f^*(U^0)$
I think ${(U^0)}^0 \cong U$ only holds in the finite dimensional case. Also – sorry for that – I edited the question removing the requirement that V and W be Euclidean, since I really want to prove the general case (it just happens that the only application I have for it right now deals with Euclidean spaces). I think, though, that the proof might still work with minor modifications, let me check…
Mar
15
comment Proving $(f^{-1}(U))^0 = f^*(U^0)$
@ArturoMagidin, Benjamin Lim: Clarified the question.