1
$\begingroup$

I have a problem with the construction of a homeomorphism. This is the task:

Suppose $f : X \to Y$ is a homeomorphism and $A$ is a subset of $X$. Prove that $X\setminus A$ is homeomorphic to $Y \setminus f(A)$ (each being given the subspace topology).

I know that I need to construct a bijective continuous map from $X \setminus A$ to $Y\setminus f(A)$ whose inverse is also continuous but I don't know in which manner. I would be grateful for your help.

$\endgroup$
2
$\begingroup$

First, show that $f^*: X \setminus A \to Y \setminus f(A)$, $f^*(x) = f(x)$, is still a bijection.

Then, show that $f|_{A^c}: X \setminus A \to Y$, $f|_{A^c}(x) = f(x)$ is continuous. This might already be a theorem in your book/material; this is just the restriction of $f$ to $A^c$ (the complement of $A$).

Similarly, show that $f^{-1}|_{f(A)^c}$ is continuous.

Combine these to show that $f^*: X \setminus A \to Y \setminus f(A)$ is a homeomorphism.

$\endgroup$
1
  • $\begingroup$ Thank you both very much. I will accept this answer as it has more details but both are helpful. $\endgroup$
    – Jim
    Feb 8 '15 at 11:25
2
$\begingroup$

Indeed. Show that $g: X\setminus A \rightarrow Y \setminus f[A]$ defined by $g(x) = f(x)$ (so just $f$ restricted) works.

Main fact: the restriction of a continuous function is continuous.

$\endgroup$
0
$\begingroup$

For any sets $B, C$ with topologies $T_B ,T_C,$ a function $f:B\to C$ is a homeomorphism iff (i) $f:B\to C$ is a bijection, and (ii) $\{f(U):U\in T_B\}=\{V:V\in T_C\}=T_C.$

For your Q: Let $B= X$ \ $A$ and $C=Y$ \ $f(A).$

Let $f_B$ be the restriction of $f$ to the domain $B.$ So $f_B(B)=f(B).$

Now $f_B:B\to f_B(B)$ is a bijection because $f$ is a bijection.

And since $f$ is a bijection and $A\subset X,$ we have $$f_B(B)=f(B)=f(X \setminus A)=f(X) \setminus f(A)=Y \setminus f(A)=C.$$

$\bullet$ . Therefore $f_B:B\to C$ is a bijection.

Let $T_X$ be the topology on $X$ and let $T_Y$ be the topology on $Y.$ The topology on $B$ as a sub-space of $X$ is $T_B=\{s\cap B: s\in T_X\}.$ The topology on $C$ as a sub-space of $Y$ is $T_C=\{t\cap C: t\in T_Y\}.$

Since every $b\in T_B$ is a subset of the domain of $f_B,$ and since $f$ is a homeomorphism, therefore $$\bullet \quad \{f_B(U):U\in T_B\}=$$ $$=\{f_B(s\cap B):s\in T_X\}=$$ $$=\{f(s\cap B):s\in T_X\}=$$ $$=\{f(s)\cap f(B):s\in T_X\}=$$ $$=\{t\cap f(B):t\in T_Y\}=$$ $$=\{t\cap C:t\in T_Y\}=T_C.$$

So $f_B:B\to C$ is a homeomorphism.

Remark: The short version is that $f$ maps the open sets of $X$ to the open sets of $Y$ so $f$ maps the intersections of $B$ with the open sets of $X$ to the intersections of $f(B)$ with the open sets of $Y.$

$\endgroup$

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.