Prove that $\varphi\circ X_1$ is also a parametrization if $\varphi$ is a diffeomorphism and $X_1$ is a parametrization

Question

Suppose that $X_1:U_1\subset R^2 \to S_1$ and $\varphi:S_1 \to S_2$ is a diffeomorphism, I want to prove that $\varphi \circ X_1:U_1 \to S_2$ is a parametrization of $S_2$

Here is my attempt:

Suppose that $X_2:U_2\subset R^2 \to S_2$ is a parametrization of $S_2$, by definition of diffeomorphism, we know that $X_2^{-1}\circ \varphi \circ X_1:U_1 \to U_2$ is differentiable.

From $\varphi \circ X_1=X_2\circ(X_2^{-1}\circ \varphi \circ X_1)$, we know that:

1. $\varphi \circ X_1$ is differentiable
2. $\varphi \circ X_1$ is one-to-one, because both $\varphi$ and $ X_1$ are all one-to-one

The rest I want to prove is $(\varphi \circ X_1)_{u}$ and $(\varphi \circ X_1)_{v}$ are independent, I am stuck with it and any hints will be appreciated!