If $v\in\mathbb{R}^N\setminus\mathbb{R}^{N-1}$, what is the projection $\pi_v$ with kernel $\mathbb{R}v$? I going over a lemma for the Whitney Embedding theorem which shows that an injective immersion of an $n$-manifold into $\mathbb{R}^N$ can actually be immersed in a lesser dimensional Euclidean space if $N>2n+1$. 
The lemma says for $v\in\mathbb{R}^N\setminus\mathbb{R}^{N-1}$, let $\pi_v\colon\mathbb{R}^N\to\mathbb{R}^{N-1}$ be the projection with kernel $\mathbb{R}v$. 
How is this map actually defined? Could one actually write down an explicit formula for what it does: $\pi_v(x_1,\dots,x^N)=\ ?$
Geometrically, $\mathbb{R}v$ is a line in $\mathbb{R}^N$, so does $\pi_v$ just project points onto $\mathbb{R}v$? If that's the case, why is the image in $\mathbb{R}^{N-1}$, that is, why is the last coordinate now $0$? 
Note* $\mathbb{R}^{N-1}$ is the subspace of $\mathbb{R}^N$ with last coordinate $0$.

Let $v \in \mathbb{R}^N\setminus\mathbb{R}^{N-1}$. As $\mathbb{R}^{N-1}$ is being identified with a subspace of $\mathbb{R}^N$, we know that $v \neq 0$ (because $0$ is in the subspace being identified with $\mathbb{R}^{N-1}$). We can therefore complete the linearly independent set $\{v\}$ to a basis $\{v, w_1, \dots, w_{N-1}\}$ for $\mathbb{R}^N$. Now the projection is given by $\pi_v(v) = 0$ and $\pi_v(w_i) = e_i$ for $i = 1, \dots, N-1$ where $\{e_i\}$ is the standard basis for $\mathbb{R}^{N-1}$. I'll leave it to you to check that $\operatorname{ker}\pi_v = \mathbb{R}v$.
