Immersions into $\mathbb{R}^n$ The Problem: As clarified in the answer to this question , a smooth function $f:[a,b] \rightarrow \mathbb{R}^n$  is an immersion at the point $p \in [a,b]$ if the derivative $\partial f(p) \neq 0$. In the context of smooth manifolds $M$ and $N$, however, one says that that a smooth function $f:M \rightarrow N$ is an immersion at the point $p \in M$ if the differential at $p$,
$$f_{*,p}:T_pM \rightarrow T_{f(p)}N$$ 
is injective. I am trying to verify that this more general definition is equivalent to the previous one when working with curves in $\mathbb{R}^n$. According to the guidelines suggested in post on meta, I will be posting my proposed solution to this problem as an answer to the question. 
 A: My Approach: I posed on a very similar problem here and although some might consider this methodology overblown, I really see no other way to arrive at the solution so any constructive criticism about how to make the argument more elegant, or correct if it is in error, would be appreciated.
I claim that the differential $\partial f(p)$ of smooth function $f:[a,b]\rightarrow\mathbb{R}^n$ at the point $p \in [a,b]$ is injective if and only if $\partial f(p) \neq 0$. First, suppose that $\partial f(p)$ is injective. According to the rank/nullity theorem,
$$
\mathsf{dim}([a,b]) = \mathsf{dim}(\mathsf{Ker}(\partial f(p)) + \mathsf{dim}(\mathsf{Im}(\partial f(p)).
$$
Since $\partial f(p)$ is injective, the dimension of its kernel is $0$ and we know that the dimension of $[a,b]$ is 1. Therefore, the dimension of the image of $\partial f(p)$ must be $1$ and this precludes the possibility that $\partial f(p)$ is the zero-map whose image is a $0$-dimensional space. Conversely, if $\partial f(p)$ is not the zero-map the dimension $k$ of its image cannot be $0$. Thus, again by the rank/nullity theorem,
$$
1 = \mathsf{dim}(\mathsf{Ker}(\partial f(p)) + k.
$$
for some $k > 0$. However, the dimension of the kernel cannot exceed $1$ for the domain of $\partial f(p)$ is $1$-dimensional. The only possible choice for the dimension of the kernel is thus $0$ which proves that $\partial f(p)$ is injective.
