(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

For my proof, it tells me that f:X->Y is bijective.

I understand that it is one-to-one and onto, but I just want to be clear about this from a neighborhood (open subset by our def) standpoint.

Just to be clear: if f is bijective continuous, then that means the for all open neighborhoods "V" in Y, there is a unique open neighborhood "U" in X s.t. f^{-1}(V) = U (which is open in X).

Is that correct?

I want to affirm the uniqueness part.

Thank you!

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Questions about a f:X->Y being a bijective continuous function.

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