I am reading Milnor's Topology from the differentiable viewpoint and I came across the following:
Let $f: M\rightarrow N$ be a smooth map between manifolds of the same dimension. Let $x \in M$ be a regular point.

It follows from inverse function theorem that $f$ maps a neighbourhood of $x$ in $M$ diffeomorphically onto an open set in $N$.

Here should we assume $x \in M$ and $y=f(x)\in N$ to not to be on the boundary to have an open neighbourhood around them in the first place?

  • $\begingroup$ The definition of a regular point ensure the existence of such a neighborhood. $\endgroup$ – CyclotomicField Jul 3 '18 at 13:02
  • $\begingroup$ Oh, so such $x$ can't be a boundary point right? $\endgroup$ – mathemather Jul 3 '18 at 13:24
  • $\begingroup$ Yes. Being a regular point ensures there exists some small open neighborhood of $x$ which makes it an interior point and therefore not a boundary point. $\endgroup$ – CyclotomicField Jul 3 '18 at 13:48

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