I have two smoothly embedded orientable surfaces $S\_1,S\_2\subset S^3 \times [0,1]$ with boundary such that


$(i)$ $S\_1\cap S\_2$ is a smoothly embedded surface without boundary and


$(ii)$ $\overline{S\_1}\cap \overline{S\_2}=\overline{S\_1\cap S\_2}$


Now I want to prove that $S\_1 \cup S\_2$ is a smoothly embedded surface, by showing that each point in it has an open neighborhood $U \subset S\_1 \cup S\_2$ completely contained in either $S\_1$ or $S\_2$.


I tried to do this by doing a case distinction depending on where the point lies. The cases where I can't show this are when the point lies in $ int(S\_1) \cap \partial S\_2$, $\partial S\_2 \cap \partial S\_1$ or $\partial S\_1 \cap int(S\_2)$ (the last case is of course equivalent to the first).


Here are my questions:


1. How can one show that there is an open neighborhood in $S\_1 \cup S\_2$ completely contained in $S\_1$ or $S\_2$ in those last three cases? (Somehow I feel that one should make use of the tubular neighborhood theorem...)
2. Is $S\_1 \cup S\_2$ even guaranteed to be a smooth submanifold with boundary under my conditions?
3. What are alternative conditions which could guarantee that $S\_1 \cup S\_2$ is a submanifold when $S\_1 \cap S\_2 \neq \emptyset$?


According to the accepted answer in this mathoverflow question which is very similar to mine ([When is the union of embedded smooth manifolds a smooth manifold?](https://mathoverflow.net/questions/78733/when-is-the-union-of-embedded-smooth-manifolds-a-smooth-manifold)), a sufficient condition would be that $S\_1 \cap S\_2$ is a n-submanifold and $\overline{S\_1}\cap S\_2=\overline{S\_1\cap S\_2}\cap S\_2$.
I think this is not true (under the assumption that submanifolds with boundary are considered submanifolds, else there was another counterexample in the comments of the answer): Consider $S\_1=[-1,1] \subset \mathbb{R}^2$ and $S\_2 \subset \mathbb{R}^2$ 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 $S\_1 \cup S\_2$ does not define a submanifold (removing the point (0,1) produces 3 components). But correct me if I am wrong here, since this would provide me with an alternative sufficient condition I could use to show that $S\_1 \cup S\_2$ is a submanifold.


Most of the ideas I had so far, did not use the fact that I deal with surfaces inside $S^3 \times [0,1]$, and I believe that this should work in every dimension. But if someone knows how to make use of this assumption, I would be happy because that is the specific case I am interested in the most.


Thank you for your time.



Sufficient condition for the union of two submanifolds to be a submanifold
In the meantime, a very similar question of mine has been answered here <https://math.stackexchange.com/a/4642619/857154> , which answers these questions aswell. Moishe Kohan has provided a counterexample to my claim for 1-manifold which most likely carries over to surfaces. Therefore 1) cannot be shown, the answer to 2) is no, and an alternative condition would be the following:


(a) For every sequence $(x\_i)$ in $S\_2$ converging to a point $x\in S\_1$, for all sufficiently large $i$, $x\_i\in S\_1$.


(b) Same as (a) with the roles of $S\_1, S\_2$ swapped.


