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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?

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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

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

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why two intersecting line is not a manifold? – Taxi Driver Apr 29 '12 at 20:06
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a neighborhood of the point of intersection looks like $X$ which is not homeomorphic to $\mathbb{R^n}$ for any $n$. – countinghaus Apr 29 '12 at 20:10
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(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 $n$. – countinghaus 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. – countinghaus Sep 3 '12 at 23:46

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