# Topology, atlas, smooth manifold

Let $$X$$ be a set, $$n\in\mathbb{N}$$ and $$((U_i,\phi_i))_i$$ a family of subsets $$U_i\subseteq X$$ with injective functions $$\phi_i: U_i\to\mathbb{R}^n$$, which hold the following conditions:

1. $$\bigcup_{i} U_i=X$$

2. for every $$i$$ is $$\phi_i(U_i)\subseteq\mathbb{R}^n$$ open

3. for all $$i,j$$ is $$\phi_i(U_i\cap U_j)\subseteq\mathbb{R}^n$$ open and $$\phi_j\circ\phi^{-1}_i:\phi_i(U_i\cap U_j)\to\mathbb{R}^n$$ is smooth.

1. Show, that $$X$$ has exactly one topology, in regards to the family $$((U_i,\phi_i))_i$$ is an atlas.

2. Conclude: Is the family $$(U_i.\phi_i)_i$$ countable, then is $$X$$ with this atlas a smooth manifold.

Hello,

I Have a question to this task. How can I characterize the open subsets of $$X$$ with $$(U_i,\phi_i)$$ conveniently?

To get an atlas, it needs to be $$\bigcup_{(V_i,\psi_i)} V_i=X$$ with charts $$(V_i,\psi_i)$$.

The problem is, that $$(U_i,\phi_i)$$ are not charts, since $$\phi_i$$ is not a homeomorphism.

So I need a topology, such that $$\phi_i$$ is a homeomorphism, or not? How can I give such a topology?

Since the functions $\phi_i$ are injective and the images of the $U_i$ are open you can let $\phi^{-1}(U_i)$ be a basis for the topology on X. The maps are now automatically homeomorphisms and you can check that this does indeed define a topology on X.
• Ok. So I observe to topology which is induced by $\mathcal{B}:=\{\phi_i^{-1}(U_i)|i\in\mathbb{N}\}$. I have to show, that this is a topology. Therefore includes $\emptyset, X$, the union of sets and countable cut of sets. I am not sure, how to start... Jul 5, 2016 at 19:31
• Yes, sure. So we declare $\mathcal{B}$ as the base. Therefore every open set has to be representable as union of sets which are in $\mathcal{B}$. Now I have to show, that I can represent $\emptyset$ and $X$ like that. $\emptyset$ is clear, since you can just take the "empty" union. How can I represent $X$? Jul 5, 2016 at 19:46
• I think it's rather taking $\{U_i\}$ as a subbase, unless the altlas under consideration is maximal. Jul 5, 2016 at 19:47
• You cannot that the primages of the U_i as a basis: you need to takr the preimages of all open subsets of the U_i. Otherwise the topology is not even T_1 in general (this happens, for example, if there is only one $i$) Jul 6, 2016 at 2:53