I was reading this post, but unfortunately I don't understand the following statement made in the following answer given by Amitai Yuval:

"About the topology: Every $(U_\alpha,\varphi_\alpha)$ needs to be a chart, meaning that $\varphi_\alpha$ is a homeomorphism on its image, so for every subset $V\subset U_\alpha$, $V$ is open if and only if $\varphi_\alpha(V)\subset\mathbb{R}^n$ is open. Since the $U_\alpha$s cover $M$, it means that the collection of open subsets of elements in $\{\varphi_\alpha(U_\alpha)\}$ induces a basis of a topology on $M$, thus the topology is indeed unique."

Denote with $\mathcal{T}_\mathcal{B}$ the topology induced by $\mathcal{B}$ e we suppose that for all $\alpha$, the pair $(U_\alpha,\varphi_\alpha)$ are chart for $M$, then on $M$ exists a topology $\mathcal{T}$ such that $U_\alpha\in\mathcal{T}$ and $\varphi_\alpha\colon U_\alpha\to\varphi_\alpha(U_\alpha)$ is an homeomorphism.

Claim: $\mathcal{T}_\mathcal{B}=\mathcal{T}.$

First, for each charts we denote with $\mathcal{T}_{\mathcal{B}_{U\alpha}}$ the topology induced by $\mathcal{T}_\mathcal{B}$ on $U_\alpha$, and with $\mathcal{T}_{U_\alpha}$ the topology induce by $\mathcal{T}$ on $U_\alpha$.

Let $A\in\mathcal{T}_\mathcal{B}$, since $U_\alpha\in\mathcal{T}_B$ for all $\alpha$, $A\cap U_\alpha\in\mathcal{T}_\mathcal{B}$, then $A\cap\ U_\alpha=(A\cap U_\alpha)\cap U_\alpha\in\mathcal{T}_{\mathcal{B}_{U_\alpha}}.$

Since $\varphi_\alpha\colon (U_\alpha,\mathcal{T}_{\mathcal{B}_{U_\alpha}})\to\varphi_\alpha(U_\alpha)$ is an homeomorphism , $\varphi_\alpha(A\cap U_\alpha)\subseteq\varphi_\alpha(U_\alpha)$ is open. On the other hand, also

$\varphi_\alpha\colon (U_\alpha,\mathcal{T}_{U_\alpha})\to\varphi_\alpha(U_\alpha)$ is an homeomorphism, then we have that $$\varphi_\alpha^{-1}(\varphi_\alpha(A\cap U_\alpha))=A\cap U_\alpha\in\mathcal{T}_{U_\alpha}\subseteq\mathcal{T}.$$

Since $$A=\bigcup_\alpha (A\cap U_\alpha)\Rightarrow A\in\mathcal{T}.$$

The above shows that $\mathcal{T}_{\mathcal{B}}\subseteq\mathcal{T}.$ In the same way the vice versa can be shown.

It's correct?

Question. What is explicitly meant in this answer, in which we want to show the uniqueness of the topology that satisfies the hypothesis of the lemma?



What "each $(U_\alpha,\varphi_\alpha)$ needs to be a chart" means is "the topology on $M$ needs to be such that, for any $\alpha$, a subset $V \subseteq U_\alpha$ is open if and only if $\varphi_\alpha(V)$ is open in $\mathbb{R}^n$". The only point that Amitai Yuval was making is that "for any $\alpha$, a subset $V \subseteq U_\alpha$ is open if and only if $\varphi_\alpha(V)$ is open in $\mathbb{R}^n$" uniquely determines a topology on $M$, namely, the topology in which a subset $V$ of $M$ is open if and only if $\varphi_\alpha(V\cap U_\alpha)$ is open in $\mathbb{R}^n$ for each $\alpha$.

Side remark: you should have commented this question on his answer rather than made it a new question on its own.

  • $\begingroup$ @mathworker21Thanks for your answer. But I don't understand why the two topologies are the same. We have a first topology induced by the base $\mathcal{B}$, then we suppose that the pairs $(U_\alpha,\varphi_\alpha)$ are charts, then necessarily $U_\alpha$ is open in $M$ (compared to which topology?) and $\varphi_\alpha\colon U_\alpha\to \varphi_\alpha(U_\alpha)$ is an homeomorphism. How can we say that the two topologies are the same? $\endgroup$
    – Jack J.
    Oct 21 '19 at 7:30
  • $\begingroup$ @JackJ. what two topologies are there? The charts impose a necessary condition for a topology on 𝑀, namely that, for any $\alpha$, a subset $V \subseteq U_\alpha$ is open in $M$ iff $\varphi_\alpha(V)$ is open in $\mathbb{R}^n$. This is just a condition that needs to be satisfied. The point is that the topology induced by the base $\mathcal{B}$ is the only topology satisfying this condition. $\endgroup$ Oct 21 '19 at 9:17
  • $\begingroup$ @mothworker21I modified my question, could you check if i got it right? $\endgroup$
    – Jack J.
    Oct 21 '19 at 11:43
  • 1
    $\begingroup$ @JackJ. looks correct to me! $\endgroup$ Oct 21 '19 at 12:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.