Timeline for Lebesgue dimension of images
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 21, 2010 at 18:26 | comment | added | Gerald Edgar | I guess what I was remembering has $X$ and $Y$ compact. | |
| Jul 21, 2010 at 15:48 | comment | added | Henno Brandsma | Indeed Henrik. We need f to be closed and surjective, and X,Y to be normal (which is pretty mild). The theorem then states that $\dim(Y) \le \dim(X) + (k-1)$ whenever all fibres have size at most $k$. | |
| Jul 21, 2010 at 11:56 | comment | added | HenrikRüping | Maybe there is a condition like $f$ is open missing. For any space $Y$, one can consider $id_Y:Y\rightarrow Y$, where the first $Y$ is equipped with the discrete topology. Then it is bijective, the first space is zero dimensional and $Y$ could be of any dimension. Am I missing something ? | |
| Jul 21, 2010 at 11:37 | history | answered | Gerald Edgar | CC BY-SA 2.5 |