Checking a map is an immersion Let $f: \mathbb{R} \rightarrow \mathbb{S}^{1} \times \mathbb{S}^{1}$ given by:
$f(t)=(\exp(it),\exp(qit))$.
I want to show that $f$ is an immersion. OK I know the definition: we compute its derivative and check it is injective.
My question is: can we view the map as $f(t)=(\cos(t),\sin(t),\cos(qt),\sin(qt))$ i.e $f: \mathbb{R} \rightarrow \mathbb{R}^{4}$ or do we have to introduce charts? I'm familiar with maps $g: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ but here we have the torus and the map in terms of complex numbers so I'm confused.
Can you please explain how to see $f$ is an immersion? What are the steps?
 A: The map you mention is an immersion for all values of $q\in \mathbb R $ , rational or not.
Indeed , consider the first projection $p:S^1\times S^1\to S^1: (z,w)\mapsto z$.
Clearly $p \circ f:\mathbb R\to S^1:t\mapsto e^{it}$ is an immersion and so $f$ is an immersion. 
Do you see why, quite generally,  $v\circ u  \quad immersion \Rightarrow u \quad immersion$ ?
A: Let us use the definition, with local charts, to compute the differential of $f$ at $0$. We can choose a chart of the torus $\mathbb{T}$ as follows:
$\phi: (-\pi,\pi)\times(-\pi,\pi) \to \mathbb{T}$ defined as $\phi(\alpha,\beta) = (e^{i\alpha},e^{i\beta})$.
The domain of this chart is the open set $\mathbb{T} \setminus \{\{-1\}\times\mathbb{S}^1,\mathbb{S}^1\times\{-1\}\}$, which contains the point $f(0) = (1,1)$. The local form of $f$ is $\hat{f}(t) = \phi^{-1} \circ f =(t,qt)$.
It's differential at $0$ is, seen as a linear map from $\mathbb{R}$ to $\mathbb{R}^2$ $df|_0 = (1,q)$. This is indeed an injective map (for any $q$). Actually, by using the same local charts, you can compute the differential 
You can similarly compute the differential at other points $t$.
You can also see $\mathbb{T}^2$ as an embedded submanifold of $\mathbb{R}^4$, then, you can simply compute the differential of the map as seen as a map from $\mathbb{R}$ to $\mathbb{R}^4$. This is injective. Than use the fact that the torus is embedded to conclude that the map f is injective.
