Question: [![ds,](https://i.stack.imgur.com/4cyw5.png)](https://i.stack.imgur.com/4cyw5.png)

I got stuck in the underlined sentence. I see that how Lemma 6.13 implies the existence of a vector $v\_{N+2n+1}$ such that $v\_{N+2n+1}$ is arbitrarily close to $e\_{N+2n+1}$ and $\pi\_{v\_{N+2n+1}}~|\_{\tilde M}~$ is an embedding. But what means that $\pi\_{v\_{N+2n+1}}~$ is arbitrarily close to $\pi\_{e\_{N+2n+1}}~$, and how can we show it?

Answer: 
Hint: The maps $\pi\_v$ are linear maps $\mathbb R^{N+1}\to\mathbb R^N$ and thus lie in a finite dimensional vector space, which has only one reasonable topology. (My preferred interpretation for all "arbitrarily close" statements would be in topological terms.) To verify the statement, you can simply compute the matrix representations of $\pi\_{e\_{N+2n+1}}$ and $\pi\_{v\_{N+2n+1}}$ and observe that they are close to each other (in your preferred norm). The main point is that $\pi\_v$ (more or less evidently) depends continuously on $v$ (just think about matrix entries).
Since the statement of the corollary is that $f$ can be uniformly approximated by embeddings, the point is to get the embedding uniformly close to $f$. The phrase "$\pi\_{v\_{N+2n+1}}$ is arbitrarily close to $\pi\_{e\_{N+2n+1}}$" is shorthand for "given any $\varepsilon>0$, $v\_{N+2n+1}$ can be chosen so that 
$$
\sup \_{x\in \widetilde M} \big|\pi\_{v\_{N+2n+1}}(x) - \pi\_{e\_{N+2n+1}}(x)\big|<\varepsilon."$$