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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 $i \neq j$.

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  • $\begingroup$ Do you want to assume your manifolds are closed (compact with empty boundary)? $\endgroup$
    – Mark Grant
    Commented Oct 21, 2011 at 7:04
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    $\begingroup$ I think there are two possible reasons for the union of the embedded sumbanifolds $M_k\subset N$ not being a submanifold of $N$: i) some intersection $M_k\cap M_j$ has lower dimension; ii) some $M_i$ has a (topological) boundary in $N$ that meets another $M_k$. $\endgroup$
    – Pietro Majer
    Commented Oct 21, 2011 at 7:19
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    $\begingroup$ This question lacks motivation and could be homework. It seems not suitable for MO to me. $\endgroup$
    – Benoît Kloeckner
    Commented Oct 21, 2011 at 9:42
  • $\begingroup$ As a motivation I guess that the spirit of the question is, to provide a criterion to make a submanifold patching together smaller pieces (of course under this respect closed submanifolds are a trivial case) $\endgroup$
    – Pietro Majer
    Commented Oct 21, 2011 at 11:47
  • $\begingroup$ I had a typo in my question... The intersections are manifolds but NOT images of the preimage manifold. In the situations I have in mind, they are of lower dimension. Moreover nothing like being compact or closed is assumed. I can't find anything about it and I think this is not trivial or a homework exercise at all. $\endgroup$
    – Mirco
    Commented Oct 21, 2011 at 14:03

3 Answers 3

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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.

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  • $\begingroup$ Unfortunately the requirement $dim(M_i \cap M_j)=m$ for $dim(M_i)=m$ is not true in the simplicial setting $\endgroup$
    – Mirco
    Commented Oct 21, 2011 at 18:44
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    $\begingroup$ What do you mean? Certainly it can be true, unless you add some extra hypothesis that nullifies this possibility. For example, in $\mathbb R^2$ you can take the set $\{(x,y): x > -1, y=0 \}$ together with $\{(x,y): x < 1, y=0 \}$. $\endgroup$
    – Ryan Budney
    Commented Oct 21, 2011 at 21:09
  • $\begingroup$ Its worth noting that every closed manifold arises as such a union: just take a finite atlas. $\endgroup$
    – Mark Grant
    Commented Oct 22, 2011 at 7:30
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    $\begingroup$ @RyanBudney I am guessing that the submanifolds in your answer are intended to be embedded submanifolds -- so that the inclusion is a homeomorphism onto its image with the subspace topology. But then can we not construct a counterexample to your claim as follows? Let the ambient manifold be $M=\mathbb{R}^2$. Let $M_1$ be the interval $(-1,1)$ on the $x$-axis, and $M_2$ be the one-dimensional submanifold traced by going along the $x$-axis from $0$ (not included) to $1$, traveling up a little and then back again to the origin, approaching the latter in a direction transverse to the $x$-axis. $\endgroup$
    – Umberto Lupo
    Commented Oct 7, 2016 at 20:19
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    $\begingroup$ Then $M_1$ and $M_2$ are (I think) both embedded one-dimensional submanifolds, their intersection is the interval $(0,1)$ on the $x$-axis which is again a one-dimensional embedded submanifold, and drawing some figures one sees that the conditions on the closures as written in this answer are satisfied. But the union $M_1 \cup M_2$ is a "lasso" figure, which is an immersed but not an embedded submanifold of the plane. Have I gone wrong somewhere? $\endgroup$
    – Umberto Lupo
    Commented Oct 7, 2016 at 20:21
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The union will usually not be a manifold. Here are three examples:

1) x (two intersecting lines)

2) 8 (two tangent circles)

3) The topologist's sine curve, the union of two disjoint embeddings of R.

Based on these, I'm guessing there's no local condition that guarantees the union is a manifold. You probably can't do much better than requiring the manifolds to be closed and disjoint.

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  • $\begingroup$ The link leads to nowhere. $\endgroup$
    – Aaron Maroja
    Commented Oct 27, 2015 at 11:52
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For simplicity, I will assume that all manifolds are connected and closed.

An obvious sufficient condition is that the for each pair of embeddings, their images are either disjoint or equal.

To see that this is also necessary, suppose that the two embeddings $f\colon M \hookrightarrow N,\, f'\colon M' \hookrightarrow N$ of copies of $M$ intersect in a submanifold $Q$. Note that $Q = f^{-1}(M') \subseteq M$ can be viewed as a closed submanifold of $M$. This implies that either $Q=M=M'$, or $Q$ has dimension less than that of $M$. In this latter case, the union $f(M)\cup f'(M')$ will not be a submanifold (each point of $Q$ is a double-point singularity of the immersion $f\sqcup f'$).

The same argument extends to disconnected manifolds and shows that the images of each connected component must be pairwise equal or disjoint.

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  • $\begingroup$ The intersections are manifolds but NOT images of the preimage manifold. Sorry for that! $\endgroup$
    – Mirco
    Commented Oct 21, 2011 at 14:03

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