4
$\begingroup$

Let $X$ and $Y$ be two topological spaces. Let $f : X \to Y$ and $g: Y \to X$, such that

  1. $f$ and $g$ are surjective;
  2. $f$ and $g$ are continous.

Does this imply that $X$ and $Y$ are homeomorphic?

It seems similar to Bernstein's theorem in set theory, and many of the topological properties like compactness, connectedness, etc are getting preserved.

Any help would be appreciated.

$\endgroup$
3
  • $\begingroup$ You should ask a different question. Editing an already answered question changing the question altoghether disrupts the question system and disregards the work which was already put in the answers below. $\endgroup$ – Jackozee Hakkiuz May 13 at 20:54
  • $\begingroup$ @JackozeeHakkiuz Apologies. I'll do that. $\endgroup$ – user499096 May 13 at 20:55
  • $\begingroup$ @JackozeeHakkiuz math.stackexchange.com/q/4138074/499096 $\endgroup$ – user499096 May 13 at 20:59
2
$\begingroup$

No. There exists a continuous surjection $f$ between $\mathbb{R}$ and $\mathbb{R}^2$ (see here), and a continuous surjection $g$ between $\mathbb{R}^2$ and $\mathbb{R}$ (obvious), but they are not homeomorphic.

$\endgroup$
1
  • $\begingroup$ Thanks :). A follow up question, what would happen if f and g were taken to be bijective ? $\endgroup$ – user499096 May 13 at 20:43
3
$\begingroup$
  • For a continuous surjection $S^1 \mapsto [-1,+1]$ take the first coordinate projection map $(x,y) \mapsto x$.
  • For a continuous surjection $[-1,+1] \mapsto S^1$ take the map $x \mapsto (\cos(2 \pi x),\sin(2\pi x))$.
$\endgroup$
1
$\begingroup$

No it does not.

Consider $[0,1]$ and $[0,1]^2$. The projection on any component certainly constitutes a continuous surjection $[0,1]^2 \rightarrow [0,1]$.

Conversely, there are space filling curves, giving continuous surjections $[0,1]\rightarrow [0,1]^2$.

But $[0,1]$ and $[0,1]^2$ are not homeomorphic. Deleting a point in the interior of $[0,1]$ results in a disconnected space, which is not true, when deleting a point in the interior of $[0,1]^2$.

$\endgroup$
2
  • $\begingroup$ Thanks for replying. A follow up question, what would happen if f and g were taken to be bijective ? $\endgroup$ – user499096 May 13 at 20:45
  • $\begingroup$ That’s a good one. $\endgroup$ – PrudiiArca May 13 at 20:48

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.