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In general, considering function $f: M_{1} \rightarrow M_{2}$ between 2 manifolds, how does one formalize the idea of that function being continuous? Specifically, I am asking this in the context of needing to prove that if a neighbourhood of a point $x$ in the first manifold is in its' interior, then $f(x)$ is in the interior of $M_{2}$

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Well both $M_1$ and $M_2$ are both topological spaces, so continuity of a function $f : M_1 \to M_2$ means the usual topological definition of continuity, i.e. $f$ is continuous if for every open set $U$ of $M_2$ we have $f^{-1}[U]$ to be an open subset of $M_1$.

Furthermore if $x$ is an interior point of $M_1$, then $x$ is contained in some chart $(U, \phi)$ where $U$ is an open set of $M_1$ and $\phi : U \to \phi[U] \subseteq \mathbb{R}^{2}$ is a homeomorphism and where $\phi[U]$ is an open subset of $\mathbb{R}^2$.

To show that $f(x)$ is an interior point of $M_2$ you need to show that $f(x)$ is contained in some chart $(V, \psi)$ where $V$ is an open set of $M_2$ and $\psi : V \to \psi[V] \subseteq \mathbb{R}^{2}$ is a homeomorphism for which $\psi[V]$ is an open subset of $\mathbb{R}^2$.

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  • $\begingroup$ Ahh yes, forgot to add that in, thanks @Arthur $\endgroup$ Jan 4, 2019 at 7:02
  • $\begingroup$ Thank you so much. Could you please also tell me whether the answer you have provided applies in general to $R^n$.?I'm sorry if that is a bit obvious, but I have only recently began studying topology, and I am thus still getting used to it. $\endgroup$ Jan 4, 2019 at 7:40
  • $\begingroup$ @AryamanGupta No problem, I'm glad to help. Since $\mathbb{R}^n$ is a topological space this answer also applies to $\mathbb{R}^n$, but since $\mathbb{R}^n$ is a metric space to show continuity of a function $f : \mathbb{R}^n \to \mathbb{R}^m$ we have another way (sometimes more useful) to show continuity of $f$ apart from using open sets, known as the $\epsilon-\delta$ formulation of continuity, in this case we say $f$ is continuous at $x \in \mathbb{R}^n$ if $\forall \epsilon > 0$ there exists a $\delta > 0$ such that $d(x, y) < \delta \implies d(f(x), f(y)) < \epsilon$ $\endgroup$ Jan 4, 2019 at 8:24

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