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Mapping and inverse mapping of open sets and their complements

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Assume that \(\displaystyle f: E \to Y \,\,\, , E \subset X\) then can we say that \(\displaystyle f(E^c)=f(E)^c\) what about the inverse mapping \(\displaystyle f^{-1}: V \to X \,\,\, , V\subset Y\) do we have to have some restrictions on f and its inverse ? My immediate answer is that we have to have a bijection in order to conclude that but I am not sure.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908
Assume that \(\displaystyle f: E \to Y \,\,\, , E \subset X\) then can we say that \(\displaystyle f(E^c)=f(E)^c\) what about the inverse mapping \(\displaystyle f^{-1}: V \to X \,\,\, , V\subset Y\) do we have to have some restrictions on f and its inverse ? My immediate answer is that we have to have a bijection in order to conclude that but I am not sure.
If we have \(\displaystyle f: E \to Y,\ E \subset X\) then can we say that \(\displaystyle f(E^c)=\varnothing\), since f is not defined for any element that is not in E, while $f(E)^c = Y \backslash f(E)$, which is not necessarily empty.

The inverse as you define it, is only defined if f is injective.
That is since each element in the domain of $f^{-1}$ must have exactly 1 image.
Or put otherwise, the mapping between E and V must be bijective.