TITLE: How to show that the parabola $\{(x, a x^2+b x+c) \mid x\in [0,1]\}$ is a manifold.
QUESTION [1 upvotes]: I have not had any geometry courses yet but have had one topology course where did not learn about manifolds. Unfortunately i am in a situation where i need to show that for $a,b,c\in\mathbb{Q}$ the set $\mathcal{M}=\{(x, a x^2+b x+c)\mid x\in[0,1]\}$ is a sub-manifold of $\mathbb{R}^n$.
I have read that the idea is to show that for any point $x\in\mathcal{M}$ we can find an open set $U_x$ in $\mathcal{M}$ (open w.r.t. $\mathcal{M}$ i believe) with $x\in U_x$, a function $\varphi_x :U_x\to \varphi_x(U_x)\subset\mathbb{R}^n$ where $\varphi_x(U_x)$ is open.
My guess is that i need to find a way to map open sets in $\mathcal{M}$ to open sets on the line $[0,1]$ since that probably is a manifold, but i can not seem to find anywhere where it is all explained in a way that i understand.

REPLY [1 votes]: This $\mathcal{M}$ is going to be a manifold with boundary, and $\partial\mathcal{M} = \{(0,c), (1, a+b+c)\}$. You have a global chart $\varphi: \mathcal{M} \to [0,1]$, which is the projection into the first factor. Namely, $\varphi = \pi|_{\mathcal{M}}$, where $\pi\colon \Bbb [0,1]\times \Bbb R \to \Bbb [0,1]$ is given by $\pi(x,y) = x$. Then $\varphi$ is continuous because it is the restriction of a continuous map. The inverse is $\varphi^{-1}\colon [0,1] \to \mathcal{M}$ given by $\varphi^{-1}(x) = (x,ax^2+bx+c)$.
Replacing $[0,1]$ with any open interval would give a manifold without boundary instead. The fact that $a$, $b$ and $c$ are rational is irrelevant. Same strategy works for showing that if $f\colon \Bbb R^n \to \Bbb R^k$ is continuous, then the graph of $f$ is homeomorphic to $\Bbb R^n$, and thus it is a topological manifold (if $f$ is smooth instead, then a smooth manifold).
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