# Manifold: induced topology on the graph of a function.

Let $$M$$ a set and $$\mathcal{A}=\{(U_\alpha,\varphi_\alpha)\}$$ an atlas we said that $$A\subseteq M$$ is open iif $$\varphi_\alpha(A\cap U_\alpha)$$ is open in $$\mathbb{R}^n$$ for all cha rt $$(U_\alpha,\varphi_\alpha).$$ Moreover this is the only topology on $$M$$ for which all $$U_\alpha$$ are opens and all $$\varphi_\alpha$$ are homeomorphisms with the image.

Let $$U\subseteq\mathbb{R}^n$$, and $$F\colon U\to\mathbb{R}^m$$ any application. Then the graph of $$F$$ $$\Gamma_F=\{(x,F(x))\in\mathbb{R}^{n+m}\ |\ x\in U\}\subset \mathbb{R}^{n+m}$$ is a $$n$$-dimensional smooth manifold with atlas $$(\Gamma_F,\varphi)$$, where $$\varphi\colon\Gamma_F\to U$$ defined as $$\varphi(x,F(x))=x$$.

Problem. The induced topology of the atlas $$(\Gamma_F,\varphi)$$ coincides with the topology of $$\Gamma_F$$ as a subspace of $$\mathbb{R}^{n+m}$$ if and only if $$F$$ is continuous.

Question. It is not ideas on how to solve the above problem, perhaps I don't remember any topological properties. Could anyone give me any suggestions?

Thanks!

Since $$\Gamma_F$$ has an atlas consisting of a single chart, the induced topology is simply the set $$\tau_{induced}=\{ O\subset\Gamma_F \ | \varphi(O)\in\tau_{\mathbb{R^n}}\}$$ whereas the subspace topology is $$\tau_{subspace}=\{ O\subset\Gamma_F \ | \ \exists V\subset\mathbb{R}^{n+m},\ V\text{ is open},\ O=V\cap\Gamma_F\}$$ You want to show that $$\tau_{induced}=\tau_{subspace}$$ if and only if $$F$$ is continuous. Do you know how to start the proof from here?