Given the map $\phi: \mathbb{R} \rightarrow S_1\times S_1$ defined by $t\mapsto (e^{it}, e^{i\alpha t})$; $\alpha$ is an irrational number and $S_1$ is the complex unit circle.
I want to show that $d\phi|_t$ is injective (or called non-singular) at each $t\in \mathbb{R}$, then together with $\phi$ is $C^\infty$ and injective, this would imply $(\mathbb{R},\phi)$ is a submanifold of $S_1\times S_1$.
What I wanted to do is to write down the image vector in the basis of $T(S_1\times S_1)$ explicitly, $$d\phi|_t \left(\frac{\partial}{\partial r}\bigg|_t\right) = \sum_i \frac{\partial (x_i \circ \phi)}{\partial r}\bigg |_t \frac{\partial}{\partial x_i}\bigg|_{\phi_t} $$ Is this the correct approach, and is there a commonly used (global) coordinate system on $S_1\times S_1$ so I could calculate $\frac{\partial (x_i \circ \phi)}{\partial r}$?
As the comments suggested, we can look at the differential of $d\phi|_t$ as $$d\phi|_t = (2\pi i e^{2\pi i t} dr|_t, 2\pi i \alpha e^{2\pi i \alpha t} dr|_t),$$ only looking at the first component, we see that when acting on any nonzero vector of the form $c\frac{\partial}{\partial r}$, it will not be zero, thus it is injective.