Let $f_{1},f_{2}:\mathbb R^{3} \to \mathbb R$, defined as
$f_{1}(x,y,z)=x^2+xy-y-z$
$f_{2}(x,y,z)=2x^2+3xy-2y-3z$
and $M:=\{(x,y,z) \in \mathbb R^{3}:f_{1}(x,y,z)=f_{2}(x,y,z)=0\}$
I have shown that $M$ is a one-dimensional $C^{1}-$manifold. How do I go on to show that
the chart defined by: $\varphi: \mathbb R \to \mathbb R^{3}, \varphi(t)=(t,t^2,t^3)$ is a global chart.
My understanding of a global chart is that $\varphi$ would cover $M$ but how can I show this?
The map $f:\mathbb{R}\rightarrow\mathbb{R}^3$ defined by $f(t)=f(t,t^2,t^3)$ induces an homeomorphism between $\mathbb{R}$ and $M$.
The image of $f$ is contained in $M$.
$t^2+tt^2-t^2-t^3=f_1(t,t^2,t^3)=0$
$2t^2+3tt^2-2t^2-3t^3=f_2(t,t^2,t^3)=0$.
This implies that $f(\mathbb{R})\subset M$.
Define $g:\mathbb{R}^3\rightarrow\mathbb{R}$ by $g(x,y,z)=x$, its $h$ restriction to $M$ is the inverse of $f$.
$h\circ f(t)=g(t,t^2,t^3)=t$.
Let $(x,y,z)\in M$, $f\circ h(x,y,z)=f(x)=(x,x^2,x^3)$.
We have $x^2+xy-y-z=0, 2x^2+3xy-2y-3z=0$ implies that:
$-3(x^2+xy-y-z)+2x^2+3xy-2y-3z=-3x^2-3xy+3y+3z+2x^2+3xy-2y-3z$
$=-x^2+y=0$, we deduce that $y=x^2$ and $z=x^2+xy-y=x^2+xx^2-x^2=x^3$. This shows that if $(x,y,z)\in M, (x,y,z)=(x,x^2,x^3)$ and for every $(x,y,z)\in M$, $f\circ h(x,y,z)=(x,y,z)$.