- #1
LAHLH
- 409
- 1
Hi,
When we define a manifold, and in particular define what a chart is, one of the conditions we specify is that the image [tex] \phi(U) [/tex] is open in [tex]R^n[/tex]. Why do we specify this?
For example if we didn't specify this and allowed closed balls, then we could cover [tex] S^1 [/tex] by [tex] \theta [/tex] where [tex]\theta \in [0,2\pi) [/tex], and wouldn't need two charts.
I know that open balls and continuity go hand in hand, so I understand if we take [tex] U \subset M [/tex] as an open interval and want to define a continuous map [tex] \phi [/tex] then it must be that [tex] \phi(U) \subset R^n [/tex] is open. But what if our [tex] U \subset M [/tex] is not open, then why do we care if the image is open or not?
Basically why is this part of the definition?
When we define a manifold, and in particular define what a chart is, one of the conditions we specify is that the image [tex] \phi(U) [/tex] is open in [tex]R^n[/tex]. Why do we specify this?
For example if we didn't specify this and allowed closed balls, then we could cover [tex] S^1 [/tex] by [tex] \theta [/tex] where [tex]\theta \in [0,2\pi) [/tex], and wouldn't need two charts.
I know that open balls and continuity go hand in hand, so I understand if we take [tex] U \subset M [/tex] as an open interval and want to define a continuous map [tex] \phi [/tex] then it must be that [tex] \phi(U) \subset R^n [/tex] is open. But what if our [tex] U \subset M [/tex] is not open, then why do we care if the image is open or not?
Basically why is this part of the definition?