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

  • #1
Hodgey8806
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Homework Statement


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!


Homework Equations





The Attempt at a Solution

 
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