Why is the priemage of a regular value of $F(x,v)=df_x(v)$ finite? This comes up as part of a larger issue of showing a compact $n$-dimensional manifold can be immersed in $\mathbb{R}^{2n-1}$ except at finitely many points.
Suppose $X$ is a compact, $n$-dimensional manifold. Let $f\colon X\to\mathbb{R}^{2n}$ be an immersion. Say you have a map $F\colon T(X)\to\mathbb{R}^{2n}$ given by $F(x,v)=df_x(v)$, with a regular value $a$. I'm interested in why the preimage $F^{-1}(a)$ is necessarily finite. After this, I think I have an idea to show $X$ can be immersed into $\mathbb{R}^{2n-1}$ except on $F^{-1}(a)$.
So I think it suffices to show that $a$ has only finitely many preimages in $\{(x,v):|v|\leq 1\}$ in the tangent bundle $T(X)$. Towards the contrary, suppose there are infinitely many preimages $(x_i,v_i)$. By compactness, there is a convergent sequence such that $x_i\to x$ and $v_i/|v_i|\to w$. Apparently this implies $df_x(w)=0$, but how does this give a contradiction?
By linearity, I find the equation
$$
F(x_i,v_i/|v_i|)=df_{x_i}(v_i/|v_i|)=a/|v_i|
$$
Is the idea that we can without loss of generality assume that $|v_i|\to\infty$, so as $i\to\infty$, $df_x(w)=0$? But why does that show $F^{-1}(a)$ is finite? Thank you.
 A: First note that, since $f$ is an immersion, the map $F$ is (linear) injective and, hence, proper on each tangent space $T_xX$. Therefore, (by compactness if $X$) the map $F$ is proper, which means that preimage of a compact is compact. Now, you have a regular point $a\in {\mathbb R}^{2n}$ of $F$, meaning that for each $b\in F^{-1}(b)$ the derivative $DF_b$ is surjective. But $\dim TX= \dim {\mathbb R}^{2n}$, therefore, $DF_b$ is also injective. Thus, $F$ is a local diffeomorphism at each $b\in F^{-1}(a)$. Note that $F^{-1}(a)\subset TX$ is closed (actually, compact). If $F^{-1}(a)$ has an accumulation point $b$, we obtain a contradiction to the local injectivity of $F$ at each $b\in F^{-1}(a)$. Thus, $F^{-1}(a)$ is compact and discrete, hence, it is finite. qed
Edit. In order to prove properness of the map $F: TX\to {\mathbb R}^{2n}$:
Without loss of generality we may assume that $X$ is embedded in some ${\mathbb R}^N$ (maybe your book even defined differentiable manifolds this way). Suppose that $f: X\to {\mathbb R}^{2n}$ is an immersion, $x_i \in X$ is a sequence converging to some $x\in X$ and $v_i\in T_{x_i}X$ is a sequence of tangent vectors such that the norms
$$
df_{x_i}(v_i)
$$
are uniformly bounded. Then the norms of the vectors $v_i$ are also uniformly bounded. A hint for this exercise is to take a small neighborhood $U$ of $x$ such that $f|_U$ is an embedding and think about the derivative of the inverse map
$$
f^{-1}: f(U)\to X. 
$$
