# Open submanifolds; why does $\mathcal{A}_U$ cover $U$?

I am currently reading Lee's Introduction to Smooth Manifolds, and have come across open submanifolds. Suppose that $$M$$ is a smooth manifold, and that $$U \subseteq M$$ is an open set, and define $$\mathcal{A}_U := \{\text{smooth charts } (V, \varphi) \text{ of } M \text{ such that } V \subseteq U\}$$.

Lee says that it is easy to verify that $$\mathcal{A}_U$$ is a smooth atlas for $$U$$. I am stuck on showing that the charts in $$\mathcal{A}_U$$ cover $$U$$.

If I take an element $$u \in U$$, how do I assert the existence of a chart $$(V, \varphi)$$ in $$\mathcal{A}_U$$ such that $$u \in V$$? I am not even sure how to see that $$\mathcal{A}_U$$ is non-empty when $$U$$ is non-empty.

There are other questions on this website about why $$\mathcal{A}_U$$ is a smooth atlas for $$U$$, but none that I have found address why $$\mathcal{A}_U$$ covers $$U$$. I feel that I am missing something simple here.

## 2 Answers

Your concern is valid since without additional assumptions, there is no reason for $$\mathcal A_U$$ to cover $$U$$. Consider the case of $$U < \mathbb R^n$$ a proper open subset where the atlas on $$\mathbb R^n$$ is $$\mathcal A_{\mathbb R^n} = \{(\mathbb R^n, id)\}$$. This is a completely valid atlas but its restriction to $$\mathcal A_U$$ is empty. What you're missing here is that different atlases can induce the same smooth strucure. Indeed, Lee discusses this on page 9 on his book where he gives two atlases on $$\mathbb R^n$$ - the one I listed above and $$\{(B_1(x), id) : x \in \mathbb R^n\}$$. These induce the same smooth structure. The definition of a smooth manifold (per Lee) is then a pair $$(M, \mathcal A)$$ where $$M$$ is a topological manifold and $$\mathcal A$$ is a maximal atlas. i.e. one not contained in any strictly larger smooth atlas. There is no loss of generality in working with maximal atlases, as Lee proves that every atlas in contained in a unique smooth atlas.

Now, given that $$\mathcal A_M$$ is maximal, we can prove that $$\mathcal A_U$$ covers $$U$$. Indeed, let $$u \in U$$. As $$\mathcal A_M$$ covers $$M$$ there is a $$(W, \phi) \in \mathcal A_M$$ such that $$u \in W$$. As $$\mathcal A_M$$ is maximal, the restriction $$(W \cap U, \phi|_{W \cap U}) \in \mathcal A_M$$ as well. Of course, this means that $$(W \cap U, \phi|_{W \cap U}) \in \mathcal A_U$$ so this atlas covers $$U$$.

For any atlas $$\{(U_i, \phi_i)\}$$ of $$M$$, the restriction $$\bigcap (U\cap U_i)$$ covers $$U$$. The restriction of the map provides the atlas for the submanifold.

• There is no guarantee that $U \cap U_i$ is a chart in $\mathcal A_U$ as defined by OP unless the atlas is taken to be maximal. Jun 16 '20 at 4:01