I want to find a topological embedding $f : X \rightarrow Y$ and $g: Y \rightarrow X$, yet $X$ is not homeomorphic to $Y$.

Question

This is related to a problem of showing that none of the intervals $(0,1], (0,1), [0,1]$ are homeomorphic to another, since it's in the same problem block. I've tried using $f(x) = 1 - x \ , f(x) = 1/(x-1)$ between varying pairs of those intervals to no avail. Do I need a function as complicated as the topologist's sine to show this?

You can embedd each interval in the other one by $f(x)=\frac13+\frac13 x$ — Stefan Hamcke, Sep 24 '13 at 23:19
Ah, that makes sense now. I guess I kept looking for a surjection as well. — Daniel Donnelly, Sep 24 '13 at 23:23
With your additional condition (that it is also surjection), it seems to be identical to this question — Martin Sleziak, Sep 26 '13 at 09:02
This questin is about the other problem you mentioned: Is there exist a homemoorphism between either pair of $(0,1),(0,1],[0,1]$; http://math.stackexchange.com/questions/240414/is-there-exist-a-homeorphism-between-either-pair-of-0-1-0-1-0-1 — Martin Sleziak, Sep 26 '13 at 09:03

score 2 · Accepted Answer · answered Dec 10 '18 at 00:40

This community wiki solution is intended to clear the question from the unanswered queue.

Define $f : (0, 1) \to [0, 1)$ to be the inclusion. This is an embedding.

Define $g : [0, 1) \to (0, 1)$ by $\displaystyle t \mapsto \frac12t + \frac12$. This embeds into $[1/2, 1) \subseteq (0, 1)$.

But $(0, 1)$ and $[0, 1)$ are not homeomorphic. If you remove any point from $(0, 1)$ you get two disconnected components. However, if you remove $0$ from $[0, 1)$ you still have only one component.

I want to find a topological embedding $f : X \rightarrow Y$ and $g: Y \rightarrow X$, yet $X$ is not homeomorphic to $Y$.

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