# Mapping and inverse mapping of open sets and their complements

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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
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.