Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose I have $M$ and $N$, two $k$-manifolds in $\mathbb{R}^n$. Is it true that $M\cup N$ is also a manifold? What is a sufficient condition for positive answer?

share|cite|improve this question
I don't think there's really much to say besides that $M\cup N$ is a manifold iff near any $p\in M\cap N$, there's a neighborhood of $p$ in $M\cup N$ that looks like (an open subset of) $\mathbb{R}^k$. – Aaron Mazel-Gee Apr 29 '12 at 21:55
up vote 9 down vote accepted

No. Take two lines in $\mathbb{R}^n$ which intersect only at the origin. Disjointness is sufficient, although not necessary, for a positive answer.

share|cite|improve this answer
why two intersecting line is not a manifold? – Un Chien Andalou Apr 29 '12 at 20:06
a neighborhood of the point of intersection looks like $X$ which is not homeomorphic to $\mathbb{R^n}$ for any $n$. – user29743 Apr 29 '12 at 20:10
(i mean looks topologically like the letter x). To see that that's not so, note that you can remove a point from the letter X to leave it with four connected components, a property that isn't true of $\mathbb{R}^n$ for any $n$. – user29743 Apr 29 '12 at 20:10
@countinghaus : Topologically the union $\cup$ does not have a meaning unless you are talking about topological disjoint union which is defined even if the two spaces have elements in common since we can force them to be disjoint by $X\simeq X\times \{1\}$ where $1$ is some "tag" so saying that disjoint union of two lines intersecting at the origin is not a manifold is not true since $L_1\times \{1\}$ does not intersect $L_2\times \{2\}$.am i right? – palio Aug 27 '12 at 21:21
I don't think the OP is asking about disjoint unions since he is considering both manifolds as subsets of $\mathbb{R}^n$ (so the honest union makes sense), but if he is asking about disjoint unions then yes, the answer to his question is that it's always true. – user29743 Sep 3 '12 at 23:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.