Sufficient condition for the union of two submanifolds to be a submanifold

Question

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?), 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.

Answer 1

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.