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.


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.


1 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.

  • $\begingroup$ 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... $\endgroup$
    – MrTopology
    Jul 5, 2016 at 19:31
  • $\begingroup$ Well, it doesn't directly include all of that. We declare them to be a basis of the topology. The actual topology will contain all of those things. $\endgroup$ Jul 5, 2016 at 19:37
  • $\begingroup$ 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 $X$? $\endgroup$
    – MrTopology
    Jul 5, 2016 at 19:46
  • $\begingroup$ I think it's rather taking $\{U_i\}$ as a subbase, unless the altlas under consideration is maximal. $\endgroup$
    – Qiyu Wen
    Jul 5, 2016 at 19:47
  • 1
    $\begingroup$ 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 $i$) $\endgroup$ Jul 6, 2016 at 2:53

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