# Defining continuity for functions between surfaces

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 $$M_{2}$$

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 $$\mathbb{R}^2$$.
• 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. Jan 4, 2019 at 7:40
• @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 $d(x, y) < \delta \implies d(f(x), f(y)) < \epsilon$ Jan 4, 2019 at 8:24