Suppose we have k embeddings of one single smooth manifold into one other, such that the intersections are manifolds,too. What are sufficient conditions, such that the union of those embeddings is a smooth submanifold in the image manifold?
To be more precise: Suppose we have a manifold M and a manifold N (both smooth). Moreover we have k-many embeddings $f_1: M \rightarrow N$ , $\ldots$, $f_k: M \rightarrow N$ such that the intersections of $f_j(M) \cap f_i(M) \neq \emptyset$ are in general not empty but it is required that
$f_j(M) \cap f_i(M)$ is itself a submanifold of $N$. Is it possible at all that the union
$\cup f_j(M)$ for all $1 \leq j \leq k$ is a submanifold of N?
Of course it is if all $f_j$'s are equal, but suppose some of them are not. Then in general the union is not a submanifold as we can see for example if we embedded $\mathbb{R}^2$ into $\mathbb{R}^3$ by $f_1(x_1,x_2):=(0,x_1,x_2)$ and $f_2(x_1,x_2):=(x_1,x_2,0)$ and $f_3(x_1,x_2):=(x_1,0,x_3)$. Then the intersections are lines and hence are manifolds by themself, but the union of the images of the $f_j$'s is not a manifold.
The question is: Are there conditions under which the union is a submanifold or not?
Of course one sufficient condition is that there is a $i$ such that $f_i(M)=\cup f_j(M)$ $i \neq j$. So the more interesting situation is, when we have $f_i(M)\neq \cup f_j(M)$ for all $i \neq j$.
Best Answer
To expand on my comment, say for all $i$, $M_i$ is an $m$-dimensional submanifold of $N$. So in your question, $M_i = f_i(M)$. But the $M_i$'s need not be diffeomorphic for the answer to hold, below.
Further assume that $\overline{M_i} \cap M_j = \overline{M_i \cap M_j} \cap M_j$ for all pairs $i \neq j$, and $M_i \cap M_j$ is also an $m$-dimensional submanifold of $N$ for all $i \neq j$.
Then I claim $\cup_i M_i$ is an $m$-dimensional submanifold of $N$. The proof is fairly mechanical, basically the closure condition rules out the "topologist's sine curve" example Anton Lukayenko gives. It ensures that for any point in the union, a chart neighbourhood for an $M_i$ can be restricted to a chart neighbourhood of any overlapping $M_j$ because $M_j$ can't approach from a transverse direction. The condition that intersections all are $m$-dimensional rules out the figure-8 case, etc.