Topology, atlas, smooth manifold

Question

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.

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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?

Thanks in advance for your hints and comments.

Answer 1

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.