Let $\phi : U\subset R^2 \to S$ be a local parametrization(local chart) of regular
surface $S$ and let $\sigma: V\subset R^2 \to U $ be a diffeomorphism. Then prove
that $\phi \circ \sigma : V \to S$ is also a local parametrization of
the surface $S$
I am given that the surface is regular and $\phi : U\subset R^2 \to S$ is one of the local parametrization. This means that $\phi$ is smooth, homeomorphism ($\phi$ is continuous, and has a continuous inverse) and the jacobian matrix of $Df : U \to S$ has rank $2$, i.e, the linear transformation $Df$ is one-one
I know that $\sigma$ being diffeomorphism means that it is smooth and has a smooth inverse.
Since I already know that $S$ is smooth, I need to show only the following:
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$\phi \circ \sigma$ is smooth, which is true because it is composed of two smooth functions
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The jacobian matrix of $D(\phi \circ \sigma) = D(\phi(\sigma))_{3\times 2}\times D(\sigma)_{2\times 2}$ has rank $2$. I am struggling in proving this.
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$\phi \circ \sigma$ is one-one. I am also struggling in providing an accurate argument why this is true.
Please help me filling the gaps.
Best Answer