I'm learning about embeddings and applications of Sard's Theorem in Guillemin & Pollack, and I'm trying to work out exercise 8.10:
Prove that every $k-$dimensional manifold $X$ may be immersed in $\mathbb{R}^{2k}.$
G&P call this The Whitney Immersion Theorem. I tried modeling my proof after the dimension reducing inductive argument provided in the proof for the weak embedding theorem, but after searching for an answer to something that had stumped me I found that I had essentially written out the proof of Theorem 5.9 found here. For completeness, this is:
First, we use the Whitney's embedding theorem for $\mathbb{R}^{2k+1}$ to find an embedding of $f:M^{k} \longrightarrow \mathbb{R}^{2k+1}$ (here $M^{k}$ is our $k-$manifold). Now, we define $g:TM^{k} > \longrightarrow \mathbb{R}^{m}$ where $m > 2k$ such that $g(x,v) = > df_{x}(v)$. Recall that $\dim TM^{k} = 2k$. Then, since $m > 2k$ we know that every point in $TM^{k}$ is a critical point of $g$. By applying Sard's theorem, we can pick up an $a \in \mathbb{R}^{m}$ such that $a \not \in g(TM^{k})$ and $a \neq 0$. Let $\pi$ be the projection of $\mathbb{R}^{m}$ onto the orthogonal complement of $a$, $H_{a}$. The composition $\pi \circ f: M^{k} \longrightarrow H_{a}$ is an immersion if $d(\pi \circ f)$ is injective.
To prove that $d(\pi \circ f)$ [the pdf says $f$ but I assume this is a typo] is injective, suppose for a contradiction that $v \neq 0$ and $v > \in T_{x}M^{k}$ such that $d(\pi \circ f)_{x}(v) = 0.$ Note that since $\pi$ is linear, $d(\pi \circ f)_{x} = \pi \circ df_{x}$ by the chain rule. So $\pi \circ df_{x}(v) = 0$. The projection of a vector $df_{x}(v)$ onto an orthogonal complement of $a$ is only zero if that vector is a scalar multiple of $a$. So $df_{x}(v) = ta$ for some $t > \in \mathbb{R}$. If $t = 0$, then $df_{x}(v) = 0$. This however cannot happen, because $0 \not \in g(TM^{k})$. So $t$ is not equal to zero. Therefore, $g(x,\frac{1}{t}) = a$ or $df_{x}(\frac{1}{t}) = a$ which is a contradiction since our choice of $a$ prohibits it from being part of the image of $g$. So $d(\pi \circ f)$ is injective. Note that $\pi \circ f$ creates an immersion into the $m-1$ dimensional subspace of $\mathbb{R}^{m}$. This process can be continued by induction on $m$ until $m = 2k$, so by repeating this process we will have found an immersion of $M^{k}$ into $\mathbb{R}^{2k}$.
My problem, specifically, is with the part:
If $t = 0$, then $df_{x}(v) = 0$. This however cannot happen, because $0 \not \in g(TM^{k})$.
Why can't this happen? To my understanding, we've only chosen for certain $a \not \in g(TM^{k})$, and even further so, shouldn't $0$ be in the image since the derivative is a linear map? I believe a patch to this is that we are taking $f$ to be an embedding, so its derivative is injective and thus its kernel is trivial, but wouldn't using this break the inductive argument at the end? As far as I understand, we are giving an immersion into the lower dimensional subspace, not an embedding necessarily.