- #1
Hodgey8806
- 145
- 3
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!