Parametrization composed with diffeomorphism is again a parametrization.?

Question

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:

1. $\phi \circ \sigma$ is smooth, which is true because it is composed of two smooth functions

2. 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.

3. $\phi \circ \sigma$ is one-one. I am also struggling in providing an accurate argument why this is true.

Please help me filling the gaps.

Answer 1

1. Right.
2. Multiplying a matrix by a square non-singular matrix, by either side, preserves rank.
3. A composition of two one-to-one functions is also one-to-one.