As in the title. Here is Lee's proof:
And then "by applying the preceding lemma repeatedly, we can conclude that if an
n-manifold admits an injective immersion into some Euclidean space, then it ad-
mits one into $\mathbb R^{2n+1}$."
Wouldn't it be better just to define the relevant functions and go from there? Why use projective spaces and diagonals? I am self-studying this topic so there is probably something I am missing. Below I present my interpretation of this lemma and would like feedback. I am probably missing something important.
Let $i:M\to \mathbb R^N$ be inclusion. Define $f:M\times M\times \mathbb R\to \mathbb R^N$ by $f(x, y, t) = t[i(x) -i(y)]$ and $di:T(M)\to \mathbb R^N$ by $(x,v)\mapsto d_xi(v).$ Then, $0$ is in the image of both maps, and Sard's theorem implies that the set of vectors $v$ not in the image of $f$ or $di$ is dense in $\mathbb R^N:$ that is, since dim$(M\times M\times \mathbb R)=2n+1<N,\ d_{(x,y,t)}f$ is not surjective, for any $(x,y,t)$. Likewise, since dim $T(M)=2n<N$, the only regular values of $f$ and $di$ are those not in their range. Now applying Sard, we get that each image has measure zero in $\mathbb R^N$ so the complement is dense.
So, choose one such $v$ and let $H$ be the orthogonal complement of $v$ and $\pi_v$ the projection onto $H$. Then, $\pi_v\circ i$ is injective, for if not then $i(x)-i(y)=tv$ for some $0\neq t\in \mathbb R$, in which case $f(x,y,1/t)=v,$ which is a contradiction.
And $\pi_v\circ i$ is an immersion, for if not then there are $x\in M$ and and $0\neq w \in T_xM$ such that $d_x(\pi_v\circ i)(w) = \pi_v\circ d_xi(w)=0.$ This implies that $d_xi(w)=tv$ for some real number $t$. Now, $t\neq 0$ because $i$ is an immersion. But now we have a contradiction because $di\left (\frac{1}{t}w\right )=v.$
Edit: I think I (finally) understand Lee's proof: $\mathbb {RP}^{N-1}$ is the collection of hyperplanes through the origin in $\mathbb R$ and each one can be identified with a unit vector normal to it. Take such a hyperplane $P$ and let $\pi_P:M\to P$ be the orthogonal projection. (This corresponds to the vector $v$ in Prof Lee's proof). Now, if $p\neq q,$ then $\pi_P(p)=\pi_P(q)$ only if the $p-q$ is a vector perpendicular to $P$, so if we consider the smooth $f:M\times M\setminus \Delta\to \mathbb {RP}^{N-1}$ that sends $(p,q)$ to the hyperplane passing through the origin whose unit normal vector is $\frac{p-q}{\|p-q\|}$ then, by construction, the hyperplanes $P$ for which $\pi_P$ is injective lie in the complement of the image of $f$ and by Sard's Theorem, these are dense in $\mathbb {RP}^{N-1}$ and so the vectors $v$ that correspond to $P$ are dense in $\mathbb R^N$.
Now, if $(p,v)\in TM,$ then $d\pi_P(p,v)=0$ if and only if $v$ is a vector perpendicular to $P$ (using the same identification as the one in Lee's proof) and so if we consider the smooth $g:TM\setminus M_0 \to \mathbb {RP}^{N-1}$, that sends $(p,v)$ to the hyperplane passing through the origin perpendicular to $v$ it follows again by construction and Sard, that the collection of $P$ such that $d\pi_P$ is imjective is dense in $\mathbb {RP}^{N-1}$ and so the collection of the corresponding vectors $v$ is dense in $\mathbb R^N$.
Best Answer
Phrases like "... and using a dimensionality argument ..." raise eyebrows and doubts and skepticisms, so that's a gigantic problem with your proof.
Furthermore, given that you are working in $\mathbb R^N$ which has dimension $N$, rather than in $\mathbb R \mathbb P^{N-1}$ which has dimension $N-1$, I am not at all convinced that your unstated dimensionality argument will produce dimension bounds that are as tight as the ones given in Theorem 6.13. If you claim that they will, well... show me your dimensionality argument.
On top of all that, you have not even used the hypothesis $N > 2n+1$ in your proof, which should make those raised eyebrows go through the roof.