I have two submanifolds $M^n,N^n\subset W^m$. What are some sufficient condition for their union to be a submanifold too? (by submanifold I always mean smooth)
My guess is that if $M \cap N$ is an n-submanifold which is open and $\overline{M}\cap \overline{N}=\overline{M \cap N}$, then $M\cup N$ is a submanifold. Does anyone have a reference for a statement similar to this one? Or a proof of this? My idea to prove this would be to find for each point on $M \cup N$ a neighborhood completely contained in either $M$ or $N$.
According to the accepted answer in this mathoverflow question which is very similar to mine (https://mathoverflow.net/questions/78733/when-is-the-union-of-embedded-smooth-manifolds-a-smooth-manifold), a sufficient condition would be that $M\cap N$ is a n-submanifold and $\overline{M}\cap N=\overline{M\cap N}\cap N$. I think this is wrong: Consider $M=[-1,1] \subset \mathbb{R}^2$ and $N \subset \mathbb{R}^2$ to be the graph of the smooth function $f:[-1,1] \to \mathbb{R},t\mapsto \begin{cases} 0 \ \ \ \ \ \ \ \ \ if \ \ t \in [-1,0] \\ e^{-t^{-2}} \ \ \ if \ \ t>0\end{cases}$. Then the previous conditions hold, but obviously $M \cup N$ does not define a submanifold (removing the point (0,1) produces 3 components). But I am looking for a similar condition.
Thanks for any help
For the purpose of my answer, all manifolds are assumed to have empty boundary.
Consider the following curves in the Cartesian plane:
$C$ is the unit circle centered at $(0,1)$, $J$ is the interval in the $x$-axis given by the inequalities $-1<x\le 0$. I let $C_+$ denote the open semicircle in $C$ given by the inequality $0<x$.
I will define two smooth curves: $$ M= J\cup C_+, N= C \setminus \{(0,0)\}. $$
Remark. Note that $M$ is only $C^1$-smooth, but one can easily modify the example to achieve $C^\infty$-smoothness.
Clearly, both $M, N$ are diffeomorphic to the real line, hence, are 1-dimensional submanifolds in the plane. Their intersection is $C_+$, clearly also a 1-dimensional smooth submanifold in the plane. At the same time, $M\cup N$ is not a even a topological manifold (it is not locally Euclidean at the point $(0,0)$).
Now, let's check your condition: $$ \bar{M}\cap \bar{N}=\overline{M\cap N}. $$ Both sides are equal to $\overline{C_+}$, hence, your condition is satisfied. Thus, your conjecture is incorrect.
Let us test the same example agains the linked MO answer. There, there were two requirements (not one):
(a) $\bar{M}\cap N= \overline{M\cap N} \cap N$,
(b) $\bar{N}\cap M= \overline{N\cap M} \cap M$.
It is straightforward to see that both are satisfied: The intersection in (a) equals the half-closed arc $\overline{C_+}\setminus \{(0,0)\}$ and the intersection in (b) equals the half-closed arc $C_+ \cup \{(0,0)\}$. Hence, the linked answer is also wrong.
Edit. Here is a sufficient condition for $M\cup N$ to be a smooth submanifold, provided that $M$, $N$ are $n$-dimensional submanifolds in $W$ and $M\cap N$ is also an $n$-dimensional submanifold.
Assume that the pair $(M,N)$ satisfies the following property:
(a) For every sequence $(x_i)$ in $N$ converging to a point $x\in M$, for all sufficiently large $i$, $x_i\in M$.
(b) Same as (a) with the roles of $M, N$ swapped.
It is easy to see that under this condition, both inclusion maps $M\to M\cup N$, $N\to M\cup N$ are open. From this it is immediate that $M\cup N$ is a smooth submanifold of $W$. (The domain $U$ of an smooth chart $(U,\phi)$ in $M$ or $N$ is open in $M\cup N$, hence, $(U,\phi)$ is a smooth chart in $M\cup N$.)