Why does f(f^-1(E)) $\subset$ E?

[IniquiTrance]

Hey all,

How is it that $f(f^{-1}(E))\subset E$ for some induced metric $E\subset Y$, can be a proper subset of E, rather than E itself?

Thanks!

 

[A. Bahat]

The function f might not be surjective. If it were not, there would be an x in E such that f^-1({x})=∅. Then f(f^-1({x}))=∅, which is a proper subset of {x}.

 

[IniquiTrance]

The function f might not be surjective. If it were not, there would be an x in E such that f^-1({x})=∅. Then f(f^-1({x}))=∅, which is a proper subset of {x}.

Thank you.

 

[IniquiTrance]

One more question, say $E\subset X$ s.t. $f^{-1}(f(E))\supset E$. How can this occur? Is this the case when a function is non-bijective, such as where f(a)=f(b) for $a \neq b,a \in E, b\notin E$

 

[HallsofIvy]

Let's take a specific example: $f(x)= x^2$ which is neither surjective ("onto" the real numbers) nor injective ("one to one").
Let E= [-1, 4]. Then $f^{-1}(E)= [-2, 2]$ and $f(f^{-1}(E))= [0, 4]$ which is a proper subset of E.

Let E= [0, 2]. Then $f(E)= [0, 4]$ and $f^{-1}(f(E))= [-2, 2]$ which contains E as a proper subsert.

 

[IniquiTrance]

Thank you.