Graph (topological manifold)
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Important to do
Contents
Definition
Let [ilmath]U\in\mathcal{P}(\mathbb{R}^n)[/ilmath] be an open set, with [ilmath]\mathbb{R}^n[/ilmath] denoting Euclidean [ilmath]n[/ilmath]-space. Let [ilmath]f:U\rightarrow\mathbb{R}^k[/ilmath] be a continuous map and recall the graph of [ilmath]f[/ilmath], [ilmath]\Gamma(f)[/ilmath] is defined as follows[1]:
- [ilmath]\Gamma(f):\eq\{(x,y)\in\mathbb{R}^n\times\mathbb{R}^k\ \big\vert\ x\in U\wedge f(x)\eq y\} [/ilmath][Note 1]
We claim that [ilmath]\Gamma(f)[/ilmath] is a topological [ilmath]n[/ilmath]-manifold (literally a topological manifold of dimension [ilmath]n[/ilmath])
Furthermore, it has a global chart (a chart whose domain is the entire of [ilmath]\Gamma(f)[/ilmath]):
- [ilmath](\Gamma(f),\varphi)[/ilmath] with [ilmath]\varphi:\Gamma(f)\rightarrow U\subseteq\mathbb{R}^n[/ilmath] by [ilmath]\varphi:(x,f(x))\mapsto x[/ilmath]
Proof of claims
- Hausdorff property of a topological manifold - a subspace of a Hausdorff space is a Hausdorff space so "inherited" from [ilmath]\mathbb{R}^{n+k} [/ilmath]
- Second countable topological space - a subspace of a second countable space is a second countable space so also inherited from [ilmath]\mathbb{R}^{n+k} [/ilmath]
- Locally Euclidean of dimension [ilmath]n[/ilmath] - which we will now show
- If we show that [ilmath](\Gamma(f),\varphi)[/ilmath] is a homemorphism we're done.
- Lemma 1: [ilmath]\varphi:\Gamma(f)\rightarrow U[/ilmath] is continuous
- Consider [ilmath]\pi:\mathbb{R}^n\times\mathbb{R}^k\rightarrow\mathbb{R}^n[/ilmath] - the fist canonical projection we get from the product topology
- By the characteristic property of the subspace topology (on [ilmath]U[/ilmath]) we see that [ilmath]\varphi[/ilmath], which is the restriction of [ilmath]\pi[/ilmath] to [ilmath]U[/ilmath], is continuous.
- This completes the proof
- Consider [ilmath]\pi:\mathbb{R}^n\times\mathbb{R}^k\rightarrow\mathbb{R}^n[/ilmath] - the fist canonical projection we get from the product topology
- Lemma 1: [ilmath]\varphi:\Gamma(f)\rightarrow U[/ilmath] is continuous
- If we show that [ilmath](\Gamma(f),\varphi)[/ilmath] is a homemorphism we're done.
Grade: A*
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Notes
- ↑ This could surely be written:
- [ilmath]\Gamma(f):\eq\{(x,y)\in U\times\mathbb{R}^k\ \big\vert y\eq f(x)\} [/ilmath]
TODO: Check this