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I'm afraid this is a stupid question — I'm not a mathematician, so please correct me when I'll be saying something wrong — but I've been stuck at this point for so long that I thought it would be wise to ask for help.

Hereafter is an excerpt from Introduction to Smooth Manifold by John Lee. The last line is where I'm lost.

enter image description here page 503

Essentially, $D$ is an involutive (smooth) distribution of rank $k$ on a (smooth) manifold $\mathcal M$ of dimension $m$, $\mathcal N_1$ and $\mathcal N_2$ are integral manifolds for $D$, with $\emptyset\ne\mathcal N_1 \cap \mathcal N_2$, and we want to show that $\mathcal N_1 \cap \mathcal N_2$ is open with respect to, say, $\mathcal N_1$ (whose topology may be finer than the subspace topology induced by $\mathcal M$).

To this end, let $q\in \mathcal N_1 \cap \mathcal N_2$, and let us consider a chart $(W,\varphi)$ of $\mathcal M$ in $q$ that is flat for $D$ – this means just that $\varphi(W)$ is a cube in $\mathbb R^m$, and that $\partial/\partial x^1,\dots,\partial/\partial x^k$ is a (local) frame for $D$ over $W$. We call $V_i$ ($i=1,2$) the connected component (I guess with respect to the topology of $\mathcal N_i$) of $\mathcal N_i \cap W$ containing $q$. From previous results, we know that our $V_i$'s are open sets of the same slice $S=\{p\in W \,\vert\, \varphi^{k+1}(p)=c^{k+1},\dots,\varphi^{m}(p)=c^{m}\}$ of $W$, therefore $V_1\cap V_2$ is open in $S$ with respect to the subspace topology.

Here, as far as I understand, Lee is saying: since $V_1\cap V_2$ is open in $S$, then it is open in $\mathcal N_1$. What is the line of reasoning one should be following?

PS: my sketch argument would be along the following lines. Since $V_1$ is a connected component of $\mathcal N_1\cap W$, it should be open in $\mathcal N_1$. Also, for a subset of $V_1$, being open in $V_1$, or in $\mathcal N_1$ should be the same. From the fact that $V_1\cap V_2$ is open in $S$, we have $V_1\cap V_2= G\cap S$, for some open set $G$ of $\mathcal M$. Then $V_1\cap V_2= G\cap V_1$, hence is open in the subspace topology induced by $\mathcal M$, hence in $V_1$, hence in $\mathcal N_1$. But this seems to be too convoluted, I feel like I'm missing some very basic stuff.

atlantropa
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    "$\mathcal{N}_1\cup\mathcal{N}_2$ is open with respect to $\mathcal{N}_1$..." do you mean $\mathcal{N}_1\color{red}{\cap}\mathcal{N}_2$? – C Squared Mar 11 '24 at 01:34
  • Yes, sorry for the mess, I wrote a cup instead of a cap (btw, in the most crucial point of my post). I'll immediately edit it. – atlantropa Mar 11 '24 at 01:35
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    For what it's worth, I had the same reasoning as your sketch to convince myself that $V_1 \cap V_2$ is open in $N_1$. I can't really help further, but just saying I think it's right.

    Although one nitpick: $V_1$ is not a connected component of $N_1$. It is a connected component of $N_1 \cap W$.

     – Tob Ernack Mar 11 '24 at 01:49
  • Thanks a lot, I am completely self-taught on the subject, so this is reassuring as far as I'm concerned (maybe you should be worried a little bit?, just joking, somewhat) – atlantropa Mar 11 '24 at 02:02
  • And yes, you're right, I'm going to amend that point too… – atlantropa Mar 11 '24 at 02:11
  • I didn't read the whole arguments, but I think this all boils down to the definition of a topology. A topology is a family of open sets which is closed under arbitrary union and finite intersection, (and empty set and the whole space are open). So Lee is just checking that the intersection of two $N_1\cap N_2$ is open (hence any finite intersection is open by induction), so he can declare the family of such $N$ to define the topology. – Three aggies Mar 14 '24 at 03:56
  • Yes, I think I'm aware of what you are saying. The point, here, should be that there are different topologies at play. Each $N_\alpha$ has the subspace topology, induced by $M$, and its own topology, under which it is an immersed submanifold of $M$, which may be finer than the former one. In particular, when Lee says that $\alpha\cap _\beta$ is open in $\alpha$, he means that it is open with respect to the topology of $\alpha$ as a submanifold. The goal should be that the topology induced on each $N\alpha$ by the new topology defined on $\cup N_\alpha$ is its submanifold topology. – atlantropa Mar 15 '24 at 13:25
  • Let me add that, throughout the proof, connectedness (in particular, the notion of connected component) is now nontrivial. Under the above hypotheses, consider $\subset \cap$, where $$ is a single slice of $$, which is given the subspace topology induced by $$. Then you have two possibly different notions of connectedness of $$, one as a subset of $$, the other as a subset of $$. – atlantropa Mar 15 '24 at 13:28

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It's been a while since you have asked the question, but for the sake of an answer, note that the required result is mentioned in the referenced Proposition 19.16 of the book: "each slice is open in the integral integral submanifold containing it and is embedded in $M$." First couple of lines of the argument in that proposition is used here: Let $\iota:N_\alpha \to M$ be the embedding of $N_\alpha$ in $M$. Then $W$ is open in $M$ and so $\iota^{-1}(W)=W\cap N_\alpha$ is open in $N_\alpha$. And $S$ is a connected component of this subset $W\cap N_\alpha$ open in $N_\alpha$, so $S$ is itself open in $N_\alpha$.

So when here it is proved that $V_1 \cap V_2$ is open in $S$, as $S$ is also open in $N_\alpha$ (and in $N_\beta$), we have $V_1\cap V_2$ is open in $N_\alpha$ (and $N_\beta$).

Kooranifar
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