0
$\begingroup$

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!

$\endgroup$

1 Answer 1

1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.