Relate to GP 1.3.9 - Differentiating $x_{i_1}, \dots, x_{i_k}$ result span($e_{i_1}, \dots, e_{i_k}$)? I start to think of this is question when I attempt exercise 1.3.9 on Guillemin and Pollack's Differential Topology
Consider the projection $\varphi: \mathbb{R}^N \rightarrow \mathbb{R}^k$:
$$(x_1, \dots, x_N) \mapsto (x_{i_1}, \dots, x_{i_k})$$
Differentiating:
$$d \varphi: T_x(X) \mapsto \text{ span}(e_{i_1}, \dots, e_{i_k})$$
So just to confirm - differentiating $x_{i_1}, \dots, x_{i_k}$ result span($e_{i_1}, \dots, e_{i_k}$)?
 A: A projection $\pi: \mathbb{R}^n \rightarrow \mathbb{R}^k$ defined by
$$ \pi (x) = \sum_{j=1}^k [x \cdot e_{i_j}]e_j $$
will push $\frac{\partial}{\partial x^{i_j}}$ to $\frac{\partial}{\partial y^{i_j}}$ under the differential $d\pi$. Let $y$ be the coordinate system $(y^{i_j})$ for $j=1, \dots k$. Consider:
$$ d\pi (\frac{\partial}{\partial x^{i_j}}) = \sum_{l=1}^k \frac{\partial y^{i_l}}{\partial x^{i_j}}\frac{\partial}{\partial y^{i_l}}$$
where $y = \pi (x)$ hence $y^l = x \cdot e_{i_l}= x^{i_l}$ and $\frac{\partial y^l}{\partial x^{i_j}}=\frac{\partial x^{i_l}}{\partial x^{i_j}} = \delta_{i_j,i_l} = \delta_{jl}$ thus,
$$ d\pi (\frac{\partial}{\partial x^{i_j}}) = \sum_{l=1}^k\delta_{jl}\frac{\partial}{\partial y^{i_l}} = \frac{\partial}{\partial y^{i_j}}$$
I think you want to identify $\frac{\partial}{\partial y^{i_j}}$ with $e_j$. Under that assumption I suppose your claim is true.
On the other hand, if you wish to view that span as the copy of $\mathbb{R}^k$ embedded in $\mathbb{R}^n$ in the natural manner by setting the complement of the $x^{i_j}$ coordinates to zero then the span is literally accurate. However, I'm not sure what you intend so I wrote this post. I suppose, the formula for the embedded case is just
$$ \pi (x) = (0,...0,x^{i_1},0,...,0,x^{i_k},0,...0) \in \mathbb{R}^n. $$
