Does there exist a continuous injection $f:X \to Y$ that is not an embedding but $X$ is homeomorphic with $f(X)$?

Asked Nov 07 '23 at 14:55

Active Nov 07 '23 at 14:55

Viewed 49 times

Suppose $f :X \to Y$ is continuous and injective. If $f(X)$ is not homeomorphic to $X$ then $f$ is obviously not an embedding. Every standard example (that I've seen, anyways) of a continuous injection that is not an embedding is of this variety.

I am curious about the converse - if $f$ is not an embedding is $f$ necessarily not homeomorphic with its image? Said differently, can there exist a continuous injection $f:X \to Y$ and a homeomorphism $X \to f(X)$ such that $f$ is not an embedding?

asked Nov 07 '23 at 14:55

Math

Hint: try to find an example. – Moishe Kohan Nov 07 '23 at 14:58
Hint: without loss of generality, you can take $Y = X$ and the homeomorphism between $X$ and $f(X)$ to be the identity. – Mees de Vries Nov 07 '23 at 15:00
This has been answered here: https://math.stackexchange.com/q/20913/473276, https://math.stackexchange.com/q/1702979/473276, https://math.stackexchange.com/q/1353578/473276 – Izaak van Dongen Nov 07 '23 at 15:29
Thanks. In the case that $X$ is a manifold i guess this is impossible by invariance of domain? – Math Nov 07 '23 at 15:38

Does there exist a continuous injection $f:X \to Y$ that is not an embedding but $X$ is homeomorphic with $f(X)$?

0 Answers0