Equivalent definition of "immersion"

Question

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 $p \in M$.

Example:

The curve $\alpha : \mathbb R \to \mathbb R^2$ given by $\alpha(t)=(t^3,t^2)$ is a differentiable mapping but is not an immersion.

And apparently it is because we need that $\alpha'(t) \not=0$, but $\alpha'(t)=0$ at $t=0$.

How is it that $\alpha'(t)\not=0$ for all $t \in \mathbb R$ is equivalent to the given definition of an immersion?

(I am new to the study of manifolds, so this is probably a basic question that I'm asking.)

Answer 1

For $\alpha: \Bbb{R}\to \Bbb{R}^2$, $d\alpha_p$ is injective iff its rank is $1$. $d\alpha_t$ is given by the matrix $(3t^2, 2t)^T$. At $t=0$, we see that $d\alpha_t = (0,0)^T$, which has rank $0$; thus, $d\alpha_0$ is not injective, and so $\alpha$ is not an immersion.

Now, in this case, it turns out that the map is injective iff it is nonzero (this is not true in general). That's simply because the dimension of the domain is $1$, and so the rank is either $0$ or $1$. In the former case, $d\alpha_p$ is zero; in the latter case, $\alpha$ is an immersion.

Answer 2

The differential of $\alpha:\mathbb{R} \longrightarrow \mathbb{R}^2$ at $t \in \mathbb{R}$ coincides with (i.e., may be "naturally identified" with) the usual derivative. In other words: $d\alpha_t = \alpha'(t)$. A linear map $\mathbb{R} \longrightarrow V$ from the real line into any vector space is injective precisely when it is nonzero. Thus, $\alpha$ is an immersion if for all $t$, we have $\alpha(t) \ne 0$.