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To remove ambiguity, I already gave related definitions at the end of this question.

  • A subset $X \subseteq \mathbb{R}^{N}$ is called a smooth $n$-dimensional manifold if $\forall x \in X$, $\exists$ a diffeomorphism $\varphi: U \to V$ such that $V$ is open in $X$, $U$ is open in $\mathbb{R}^{n}$, and $x \in V$. Then $\varphi$ is called a local parameterization of $V$. The inverse $\varphi^{-1}$ is called a local coordinate system, or chart, on $V$.

  • Lemma: Let $A$ be open in $\mathbb R^m$, $B$ open in $\mathbb R^n$, and $f:A \to B$ a diffeomorphism. Then $m = n$. [A proof can be found here]

Let $X \subseteq \mathbb R^N$ be a smooth $n$-dimensional manifold and $x \in X$. Let $\varphi: U \to V$ be a local parameterization around $x \in V$. Here $U \subseteq \mathbb R^n$ and $V \subseteq \mathbb R^N$. It follows that $\varphi$ is a diffeomorphism. By our Lemma, $n=N$.

Clearly, the statement $n=N$ is not correct, but I could not spot the error. Could you elaborate on where is my logical mistake?

Akira
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    The point is that unless $n=N$ an open nbhd $x \in V \subseteq X$ is not an open subset of $\mathbb{R}^N$ and so you are not in a position to invoke the stated Lemma. – Frieder Jäckel Nov 21 '21 at 10:06
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    @FriederJäckel Ahh It's that $V$ is open in $X$ and that $V$ is not necessarily open in $\mathbb R^N$. So we can not invoke our lemma. – Akira Nov 21 '21 at 10:09
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    Exactly. As an example take the following: Let $X=S^2$, $U=B_1(0) \subseteq \mathbb{R}^2$, $V\subseteq S^2$ the upper half sphere and $\varphi: U \to V$ given by $\varphi (x,y)=(x,y,\sqrt{1-x^2-y^2})$. This is clearly a parametrization and $V$ is open in $S^2$, but not open in $\mathbb{R}^3$. – Frieder Jäckel Nov 21 '21 at 10:30
  • You should define what you mean by a local parameterization of a manifold. I guess it is not that what one usually denotes as a chart around $x \in X$. – Paul Frost Nov 22 '21 at 09:57
  • Hi @PaulFrost question updated. – Akira Nov 22 '21 at 10:28
  • And "diffeomorphism" in the definition of local parameterization has to be understood in the sense of your question https://math.stackexchange.com/q/4312369 ? By the way, writing " Here $U \subseteq \mathbb R^n$ and $V \subseteq \mathbb R^N$ " is not consistent with the definition. You need $V \subseteq X$. – Paul Frost Nov 22 '21 at 11:07
  • @PaulFrost yess, diffeomorphism is understood in the sense given in that thread. That's why I posted that thread :) Writing $V \subseteq \mathbb R^N$ is not incorrect but it made me miss the point that $V$ is open in $X$, not in $\mathbb R^N$. This is where the collapse begins. – Akira Nov 22 '21 at 11:33
  • @PaulFrost my recent questions seem to be pedantic and repetitive of basic concepts. I'm sorry if they cause inconvenience to other people. The lecture note will be progressively harder with the introduction of new concepts and interaction of old ones. This approach to study is to make things clear from the beginning, i.e., just one time and for all. – Akira Nov 22 '21 at 11:42
  • @Akira Don't worry, in my opinion your questions do not cause inconvenience, but show that you are thinking about something. There are other questions in this forum which seem to have the character "here is my homework, solve the problem for me", but this does not apply here. But I suggest that you answer your own question to clear it from the unanswered queue. – Paul Frost Nov 22 '21 at 14:40
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    One nitpick: You say "writing $V \subseteq \mathbb R^N$ is not incorrect". Of course this inclusion holds true when $V \subseteq \mathbb X$, but imo it is a very infelicitous wording to say that a $V \subseteq \mathbb R^N$ is open in $X$ because that only makes sense for $V \subseteq \mathbb X$. And perhaps you find https://math.stackexchange.com/q/3511868 useful. – Paul Frost Nov 22 '21 at 14:51
  • @PaulFrost I actually did it before. When I post an answer like "@somebody's comment solves my question. I post it here as an answer to remove my question from the unanswered list", my answer has been deleted or voted to be deleted by some users. Those people feel that such my "answer" does not meet the standard of an answer. Then I feel that it's safer to let it be, i.e., not post such answers. – Akira Nov 22 '21 at 15:01

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