Why do charts map to open sets?

1. Mar 5, 2013

dEdt

My textbook says that "a chart or coordinate system consists of a subset $U$ of a set $M$, along with a one-to-one map $\phi :U\rightarrow\mathbf{R}^n$, such that the image $\phi(U)$ is open in $\mathbf{R}^n$."

What's the motivation for demanding that the image of $U$ under $\phi$ be open?

2. Mar 5, 2013

WannabeNewton

We do calculus on the coordinate domains by doing calculus on the coordinate representations of the coordinate domains under the coordinate map so a primary motivation for making them open is the same reason we make domains of regions in euclidean space open when doing calculus. From a purely topological standpoint, there are many advantages because we can show the characterization of locally euclidean in terms of open subsets of euclidean space is equivalent to the characterization of locally euclidean in terms of open balls which leads to nice properties of topological manifolds such as having a countable basis of pre - compact coordinate balls (and even smooth ones if you have an atlas).

3. Mar 5, 2013

Bacle2

AFAIK, the map $\phi$ is usually defined to be a diffeomorphism, which is a closed map (as well as an open map) . Doesn't your book define $\phi$ that way?

4. Mar 5, 2013

WannabeNewton

Bacle do you mean a homeomorphism? The notion of diffeomorphism between smooth manifolds is defined using the diffeomorphic nature of the transition maps between coordinate domains so the notion of coordinate maps being diffeomorphisms doesn't make sense a priori when first constructing a smooth atlas for a manifold. Also, you can start with just injective mappings for the charts making up the differentiable structure and there is a naturally induced topology that makes them homeomorphisms anyways. See for example Do Carmo Riemannian Geometry chapter 0.

5. Mar 5, 2013

Bacle2

Right, I misread your question, my bad. I know we pre- and post- compose with chart maps to determine properties of maps between manifolds.

6. Mar 5, 2013

micromass

Staff Emeritus
The map $\phi:U\rightarrow \phi(U)$ is a homeomorphism. But the map $\phi:U\rightarrow \mathbb{R}^n$ does not need to be a homeomorphism. So while $\phi(U)$ is certainly closed in itself, it needs not be open in $\mathbb{R}^n$. We do require it to be open in $\mathbb{R}^n$.

7. Mar 5, 2013