Q:

Showing every knot has a regular projection using differential topology

Can we use differential topology to prove that every smooth knot has a regular projection? 
Here is some background:
Let $\gamma : S^1 \rightarrow \mathbb{R}^3$ be a smooth unit-speed imbedding.   For $v \in S^2$ we say that the projection, $\pi _v$, from $\mathbb{R}^3$ to the plane orthogonal to $v$ is a regular projection of $\gamma$ if the curve $\gamma _v := \pi _v\circ \gamma$ has the following properties:

$\gamma _v$ is an immersion
There exists a finite set  $I := \{a_1, a_2, ..., a_k, b_1, b_2,... b_k\} \subset S^1$ such that $\gamma _v |_{S^1-I}$ is injective, and $\gamma _v (a_i) = \gamma _v (b_i)$ are distinct points such that for each $i$, $\gamma _v'(a_i)$ and $\gamma _v' (b_i)$ are linearlly independent.  

Now fix a particular $\gamma$. To me, the task of proving the existence of a regular projection seems like a job for differential topology, and in particular, Sard's theorem.    For instance it follows immediately from Sard's theorem that the first condition is satisfied for almost all  $v \in S^2$ since the image of the map $S^1 \rightarrow S^2, s \mapsto \gamma '(s)$ has measure zero.  
I have been trying to prove something similar about the set of $v$ which satisfy the second condition but I haven't been successful.   To give an idea for the sorts of tools I was hoping to use I'll explain my attempts. 
One idea I had was to consider the map $$f: S^1 \times S^1 - \Delta \rightarrow \mathbb{R}, (s,t) \mapsto (\gamma '(s) \times \gamma ' (t)) \cdot (\gamma (s) - \gamma (t))$$   where $\Delta$ is the diagonal in the torus, $\times$ is the cross product and $\cdot$ is the dot product.  Then $f(s, t) = 0$ iff either {$\gamma (s) - \gamma (t)$ lies in the plane spanned by $\gamma ' (s)$ and $\gamma ' (t)$} or {$\gamma '(s)$ is parallel to $\gamma '(t)$}.  In either case, this means $\gamma (s)$ and $\gamma (t)$ must not project to the same point if the projection is to be regular.   This rules out the one dimensional subspace spanned by $\gamma (s) -\gamma (t)$.  Now if $0$ were a regular value for $f$ then $f^{-1}(0)$ would be a submanifold of dimension $1$.  We could then consider the smooth map $$f^{-1}(0) \rightarrow \mathbb{R}P^2, (s,t) \rightarrow [\gamma (s) -\gamma (t)]$$ and apply Sard's theorem to find that almost every $v \in S^2$ also satisfies condition 2.   However, as far as I can tell, there is no good reason $0$ should be a regular value for $f$.  Maybe we could pick an isotope of $\gamma$ with this property?  Moreover this still doesn't rule out the possibility that 3 or more points on the curve project to the same point. 

A:

Consider the chord map
$$\nu\colon S^1\times S^1 - \Delta \to S^2\,, \quad \nu(s,t) = \frac{\gamma(s)-\gamma(t)}{\|\gamma(s)-\gamma(t)\|}\,.$$
I'll let you compute that if $\nu(s,t) = v$, then [I think!] 
$$d\nu_{(s,t)}(w,z) = \frac1{\|\gamma(s)-\gamma(t)\|}\text{proj}_{T_v S^2}(w-z)\,.$$
So if  $\pm v$ are regular values of $\nu$, you are guaranteed that the projections of $\gamma'(s)$ and $\gamma'(t)$ onto the orthogonal complement of $v$ span. 
That there are finitely many points in the preimage of such $[v]\in\mathbb RP^2$ can be argued by making sure that $\pm v$ are never tangent to $\gamma$ and getting a compact $0$-dimensional submanifold of $C\times C - \Delta$.

