# Preimage proof?

## Homework Statement

Define f:S→T, where B$\subseteq$T. Let f-1(B)={x$\in$S:f(x)$\in$B} be the preimage of B.

Demonstrate that for any such map f, f(f-1(B))=B.

My main question is, would I prove this using set inclusion both ways?

I was going to begin by letting an element be in the preimage of B, and explain what that means, then mapping that element to B. Would this be correct?

I just need a push in the right direction.

Thank you.

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SammyS
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## Homework Statement

Define f:S→T, where B$\subseteq$T. Let f-1(B)={x$\in$S:f(x)$\in$B} be the preimage of B.

Demonstrate that for any such map f, f(f-1(B))=B.

My main question is, would I prove this using set inclusion both ways?

I was going to begin by letting an element be in the preimage of B, and explain what that means, then mapping that element to B. Would this be correct?

I just need a push in the right direction.

Thank you.
Generally, yes. To show set equality you need to show inclusion in both directions.

Define f:S→T, where B$\subseteq$T. Let f-1(B)={x$\in$S:f(x)$\in$B} be the preimage of B.

Demonstrate that for any such map f, f(f-1(B))=B.

Just to let you know, but that equality is false in general. You need extra assumptions on f.

SammyS- Thank you

Micromass- Yes, the function needs to be surjective, right? I'm not sure why there were no assumptions...

SammyS
Staff Emeritus
Homework Helper
Gold Member
Just to let you know, but that equality is false in general. You need extra assumptions on f.
I'm very reluctant to question you, micromass.

But isn't the extra assumption of f needed for $f^{-1}\left(f(A)\right)=A$ rather than for $f\left(f^{-1}(B)\right)=B\ ?$

I'm very reluctant to question you, micromass.

But isn't the extra assumption of f needed for $f^{-1}\left(f(A)\right)=A$ rather than for $f\left(f^{-1}(B)\right)=B\ ?$
An extra assumption is needed for both. Consider $f(x)=x^2$ and B=[-1,0]. Then $f^{-1}(B)=\{0\}$ and thus $f(f^{-1}(B))=\{0\}$. The problem is that f is not surjective.

SammyS
Staff Emeritus
An extra assumption is needed for both. Consider $f(x)=x^2$ and B=[-1,0]. Then $f^{-1}(B)=\{0\}$ and thus $f(f^{-1}(B))=\{0\}$. The problem is that f is not surjective.