0
$\begingroup$

Let $X$ and be $Y$ be two metric spaces and $f:X\to Y$ a continuous function. We know that if $A$ is a closed set in $Y$ then $f^{-1}(A)=\{x\in X, \ \ f(x)\in A \}$ is a closed set in $X$.

Now if we have $B$ is a closed set in $X$, does $B$ have the form $B=f^{-1}(A)$ where $A$ is a closed set in $Y$ ?

$\endgroup$
3
  • 2
    $\begingroup$ No. Give $X=\Bbb R$ the discrete metric and $Y=\Bbb R$ the usual metric and consider the identity mapping from $X$ to $Y$. $\endgroup$ – David Mitra Jul 4 '14 at 0:35
  • $\begingroup$ Or, perhaps more interesting, take $f(x)=\sin x$ on $[0,\pi]$ with the usual topology and $B=[0,\pi/3]$ (here, there is no $A$ with $f^{-1}(A)=B$). $\endgroup$ – David Mitra Jul 4 '14 at 0:43
  • $\begingroup$ Another counterexample: take $f$ to be any constant function, and let $X$ be any metric space with a proper, non-empty, closed subset. $\endgroup$ – Ben Grossmann Jul 4 '14 at 0:53
1
$\begingroup$

Multiple examples of when this fails were already given. Let's look at the general question: how could this property hold? First, the only candidate for a set $A$ such that $f^{-1}(A)=B$ is $f(A)$. So, we need $$f^{-1}(f(A))=A$$ for every closed set. Since single points are closed, it follows that $f$ is injective. Let $g$ be the inverse of $f$, which is defined on $f(X)$.

Also, we need $f(A)$ to be closed whenever $A$ is. This is equivalent to $g$ being continuous.

Conclusion: the property you described holds if and only if $f$ is a homeomorphism onto its image. Such a map is called a (topological) embedding of $X$ into $Y$.

$\endgroup$
1
  • $\begingroup$ Ah, thank you, it is even better to know when it holds ! $\endgroup$ – user50618 Jul 4 '14 at 13:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.