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Suppose I have a manifold $\mathcal{M}$ and a closed submanifold $\mathcal{N} \subset \mathcal{M}$ of codimension 1. If I remove the closed submanifold $\mathcal{N}$ from $\mathcal{M}$ will I be left with a manifold?

I am not sure if it is true but it looks very plausible. However, I am pretty sure that if would only hold for codimension 1. For example in the manifold $\mathbb{R}^2$ I can take the submanifold complement to a figure "8".

Furthermore, would the statement also be true for smooth manifolds or symplectic manifolds?

Any help is welcome.

Thanks in advance.

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No, for example remove $(0,1) \times \{0\}$ from $\mathbb{R}^2$. – Chris Eagle Jun 3 '12 at 16:58
@ChrisEagle What about if you remove a closed submanifold? – JSchlather Jun 3 '12 at 16:59
Then the result is open, hence clearly a manifold. – Chris Eagle Jun 3 '12 at 17:00
But open subsets of manifolds are always manifolds, right? Just restrict the charts. Also, it's not obvious to me that the figure 8 is a manifold. – Dylan Moreland Jun 3 '12 at 17:01
@user29751: But it's not just that you can remove a closed manifold, you can even remove any closed set. So, if something like a Cantor set is removed, the resulting thing is still a manifold, being an open subset of a manifold. – Jason DeVito Jun 3 '12 at 17:32
up vote 0 down vote accepted

As pointed out in comments, removing any closed set from a manifold (topological, smooth, complex, symplectic...) leaves a manifold of the same kind, because the charts restrict to open subsets. One should note that the remaining manifold may be disconnected.

Removing a set which is not closed, generally, does not preserve the manifold structure, even if the removed set is itself a manifold. For example, remove an open line segment from a plane (example given by Chris Eagle).

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