My background is in Physics - so it's possible I am missing something really obvious here, but I'm confused over this theorem:
The linear map $\phi$ : $T_{p}(\mathbb{R}^{n}) \rightarrow \mathcal{D}_{p}(\mathbb{R}^{n})$ is an isomorphism of vector spaces
[Where $ T_{p}(\mathbb{R}^{n}$ is the tangent space at a point $p\in \mathbb{R}^{N}$, $\mathcal{D}_{p}(\mathbb{R}^{n})$ is the set of all derivations at $p$]
The proof suggested is as follows:
Suppose $D_{v} = 0$ for $v \in T_{p}(\mathbb{R}^{n})$. Apply $D_{v}$ to the co-ordinate function $x^{j}$:[ $D_{v}$ is the directional derivative along $v$, below:]
$$ D_{v}(x_{j}) = \sum_{i} v^{i} \frac{\partial x^{j}}{\partial x^{i}} = \sum_{i}v^{i} \delta^{j}_{i} = v^{j} $$
But, since $D_{v} = 0$ $$ D_{v}(x_{j}) = 0 \implies v^{j} = 0, \forall j \implies v = 0 $$
So we've shown that the directional derivative is zero if we are along the direction of the zero vector (or - if we have a zero directional derivative, then the 0 vector must be a member of the tangent space)
That's fine - but I don't understand how this means that the map is injective - how does the fact that 0 is mapped uniquely mean that every single other member of the tangent space at $p$ must be mapped uniquely?
I guess the argument involves exploiting the vector space structure in some way I have missed... is there some kind of pigeonhole principle argument based off the fact that, since $\phi$ is a linear map, the dimension of the two spaces must be the same??? [i.e the fact that one element is mapped uniquely means that they all are - and if one weren't they all wouldn't be..]
$$ \begin{align} \phi :T_{p}(\mathbb{R}^{n}) &\rightarrow \mathcal{D}_{p}(\mathbb{R}^{n})\\ v&\mapsto D_{v} = \sum_{i} v^{i} \frac{\partial}{\partial x^{i}}\end{align} $$ is a linear map and $\phi$ is injective iff $\operatorname{ker}\phi=\{0\}$ and this is what they proved.