Chapter 0, Do Carmo differential geometry, section 3 examples. I'm reading through the examples in section 3.
The examples 3.1, 3.2 and 3.3. give the curves $\alpha : \mathbb{R} \to \mathbb{R}^2$ defined as
$$
\begin{array}{ll}
\alpha(t) : (t,|t|) & (1)\\
\alpha(t) = (t^3,t^2) & (2) \\
\alpha(t) = (t^3 - 4t, t^2 - 4) & (3)
\end{array}
$$
The three examples are supposed to clarify the concepts of embedding and immersion.
First of all the author gives the following definition:

Let $M^m$ and $N^n$ be differentiable manifolds. A differentiable mapping $\varphi : M \to N$ is said to be an immersion if $d\varphi_p : T_p M \to T_{\varphi(p)} N$ is injective for all $p \in M$. If, in addition, $\varphi$ is a homeomorphism onto $\varphi(M) \subset N$, where $\varphi(N)$ has the subspace topology induced from $N$, we say that $\varphi$ is an embedding.

I know now it may sound trivial but to me it's not clear in the definition of the $\alpha$ curves in the example which are the manifolds and what are the coordinate neighborhoods.
To me the two manifolds are clearly $\mathbb{R}$ and $\mathbb{R}^2$ and the coordinate neighborhoods are the very same set. The reason I'm specifying this because a map between two manifold is differentiable iff it's expression in terms of coordinate neighborhoods is differentiable (by definition).
Therefore to me each map $\alpha$ is actually is an expression for the actual $\varphi$. In other words it is my understanding that
$$
\alpha(t) = \left( y^{-1} \circ \varphi \circ x \right)(t),
$$
therefore the three examples should aim to prove that $\varphi$ is/isn't an embedding/immersion.
Now for case $(1)$ the expression isn't differentiable, therefore it cannot be either an immersion or an embedding. Case $(2)$ is slightly trickier to me.
First of all it's expression is differentiable therefore the map $\varphi$ can potentially be an immersion. To prove this is an immersion we need to prove that the linear map $d \varphi_p$ is injective, for each $p \in M = \mathbb{R}$. Here is where I get confused, because I've tried to use the definition of differential to understand.
Do Carmo defines
$$
d \varphi_p = \beta'(0) = \sum_{j=1}^n \frac{\partial y_1}{\partial x_j} x'_j(0) \frac{\partial}{\partial y_1} + \ldots + \sum_{j=1}^n \frac{\partial y_m}{\partial x_j} x'_j(0) \frac{\partial}{\partial y_m}
$$
Or equivalently he characterizes $d\varphi_p$ as the Jacobian of the coordinate transformation
$$
J = \frac{\partial(y_1,\ldots,y_m)}{\partial(x_1,\ldots,x_n)}
$$
However I don't know how to use this definition to reach the same conclusion the author does to prove that $(2)$ isn't an embedding.
Can you help?
 A: These curves $\alpha\colon \Bbb R\to \Bbb R^2$ stand for the map $\varphi$ in the general definition. You do not need to mix $\alpha$ and $\varphi$. Moreover, since $\Bbb R$ and $\Bbb R^2$ have their standard smooth structures and we have natural coordinates $t$ in $\Bbb R$ and $(x,y)$ in $\Bbb R^2$, you can compute $\alpha'(t)$ just like in calculus courses. More precisely, $\alpha'(t) = {\rm d}\alpha_t(1)$. 


*

*$\alpha(t) = (t,|t|)$ is not smooth because $|t|$ is not smooth at $0$.

*$\alpha(t) = (t^3,t^2)$ is not an immersion because $\alpha'(0) = 0$. This means that ${\rm d}\alpha_0$ is the zero map, and so it is not injective.

*$\alpha(t) = (t^3-4t, t^2-4)$ has derivative given by $\alpha'(t) = (3t^2-4, 2t)$, and this is never the zero vector, so ${\rm d}\alpha_t\colon T_t\Bbb R \to T_{\alpha(t)}\Bbb R^2$ is injective (by dimensional reasons), and $\alpha$ is an immersion. If $\alpha$ is also injective, then it will be an embedding if $\alpha$ is an open map, or a closed map, or a proper map (this is proposition 4.22 in Lee's Introduction to Smooth Manifolds, 2nd ed.).



Risking being overly pedantic, when we write that $\alpha\colon \Bbb R \to \Bbb R^2$ is given by $\alpha(t) = (x(t),y(t))$, we can see this as coordinate representation of an "abstract map", and when I wrote that $t$ is the natural coordinate in $\Bbb R$ and $(x,y)$ are the natural coordinates in $\Bbb R^2$, I mean that we are looking at the charts $(\Bbb R, {\rm Id}_{\Bbb R})$ and $(\Bbb R^2, {\rm Id}_{\Bbb R^2})$, and $\alpha'(t)$ is the local representation of $\alpha$ under these charts. What is really going on is that $\alpha =  {\rm Id}_{\Bbb R^2}^{-1}\circ\alpha\circ {\rm Id}_{\Bbb R}$. 
