Homeomorphic Spaces in Topology

Question

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.

Answer 1

• 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 $x \mapsto (\cos(2 \pi x),\sin(2\pi x))$.

Answer 2

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.

Answer 3

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 $[0,1]^2$.