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?



1 Answer 1


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?

  • $\begingroup$ @PrototankThanks for your answer. This is exactly where I arrived, but then I don't know where to use the hypotheses. Could you give me a hand? $\endgroup$
    – Jack J.
    Oct 16, 2019 at 16:44

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