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- #1

- Jan 17, 2013

- 1,667

- Thread starter ZaidAlyafey
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- Thread starter
- #1

- Jan 17, 2013

- 1,667

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- #2

- Mar 5, 2012

- 8,779

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.