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Question:

Let $f:( X,d )\rightarrow ( Y,e )$ be a map between metric spaces, let $U \subseteq X$ and$ V\subseteq Y$.

Show that $V=f\left ( f^{-1}\left ( V \right ) \right )$.

Recall by definition the pre-image of a set V:

$$f^{-1} ( V )=\{ x \in X \mid f ( x ) \in V \}$$

If it helps: $f\left ( x \right )=x^{2}$

$$f^{-1}(\{x\})= \left\{\begin{matrix} \sqrt x, -\sqrt x &x>0 \\ 0 & x=0\\ \emptyset & x<0 \end{matrix}\right.$$

Any help is appreciated.

Thanks in advance.

Zev Chonoles
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Mathematicing
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1 Answers1

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The statement is false in general (more specifically, whenever $f$ is not surjective).

For example, if $f:\mathbb{R}\to\mathbb{R}$ is defined by $f(x)=x^2$, and $V$ is the set $(-\infty,0)$ of negative real numbers, then $$f^{-1}(V)=\{x\in\mathbb{R}:f(x)\in V\}=\{x\in\mathbb{R}:x^2<0\}=\varnothing$$ and therefore $f(f^{-1}(V))=\varnothing$, which is not $V$.

Zev Chonoles
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  • Strictly speaking, we don't need $f$ to be surjective, we just need $V\subset f(X)$. – florence Aug 15 '16 at 06:58
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    @florence: I see what you mean, though it just depends on whether you want to impose a condition on $f$ for the statement to be true for arbitrary $V$, or whether you want to impose a condition on $V$ for the statement to be true for a specific $f$. – Zev Chonoles Aug 15 '16 at 07:00