If $X$ and $Y$ are homeomorphic, then for every $A\subset X$, $X-A$ and $Y-f(A)$ are homeomorphic.

Question

Let $X$ and $Y$ be metric spaces and $X$ and $Y$ are homeomorphic under $f:X\to Y$, then for every $A\subset X$, $X-A$ and $Y-f(A)$ are homeomorphic.

It is quite intuitive but how can we write the proof rigorously? How can we construct the new homeomorphism $g:A-X\to Y-f(A)$?

Could anyone please give some hints? I could not find any such theorem in the internet.

Thanks!

score 2 · Accepted Answer · answered Jul 07 '16 at 13:19

2

$g$ is defined the easiest way it can be. Take an $x \in X-A$ and let $g(x) = f(x)$. Since $f$ is injective, and $x \notin A$, we have $g(x) \notin f(A)$, and therefore $g$ is a function from $X-A$ to $Y-f(A)$.

answered Jul 07 '16 at 13:19

Arthur

199,419

@user338393: Or in other words, $g=f\upharpoonright X\setminus A$. – Brian M. Scott Jul 07 '16 at 20:17
Oh yes, it is a simple construction. Thanks! – user338393 Jul 09 '16 at 14:16

If $X$ and $Y$ are homeomorphic, then for every $A\subset X$, $X-A$ and $Y-f(A)$ are homeomorphic.

1 Answers1