Trefoil knot cannot be injectively projected to a plane? I'm trying to do the following geometry qual problem from my university:
Give an example of a smooth embedding $f:S^1\rightarrow\mathbb{R}^3$, such that for each plane $P$ in $\mathbb{R}^3$, $\pi_P\circ f$ is not injective; here $\pi_P$ is the orthogonal projection of $\mathbb{R}^3$ onto $P$.
My guess is that the trefoil knot will work, but since I have zero knowledge of knot theory, can someone please let me know if there is a simple argument to see if (or why) this works? Alternatively, is there a simple example that I missed? Thank you!
Edit: Now I'm sort of confident that this is due to $R^3$-trefoil knot and $R^3-S^1$ have different fundamental groups. I'm still not sure about the details here. Can someone please help me finish the argument? Thank you!
 A: Your guess is correct. Assume $\pi_P\circ f$ is injective. Then, the image of $\pi_P\circ f$  is a simple closed curve, injective image of $S^1$, contained in a plane. The Jordan curve theorem tells us that this s.c.c. bounds a disk, i.e. represents the unknot. A knot is the unknot if there exists an unknotted diagram in some planar projection. Therefore, if $f$ is knotted, then there exists no injective projection.
A: You can prove this in a variety of ways using some machinery from knot theory, but here is a more elementary approach.  Let's think about the question more geometrically.  To say that $\pi_P\circ f$ is not injective means that there are two distinct points $p$ and $q$ in the image of $f$ such that the vector $p-q$ is orthogonal to $P$.  So, we just need to arrange that the function $g:S^1\times S^1\to\mathbb{R}^3$ given by $g(x,y)=f(x)-f(y)$ hits every possible direction in $\mathbb{R}^3$.  Or, letting $U=\{(x,y)\in S^1\times S^1:x\neq y\}$, we can restrict to $U$ and normalize to get a smooth map $h:U\to S^2$ given by $h(x,y)=\frac{f(x)-f(y)}{\|f(x)-f(y)\|}$, and we just need this map $h$ to be surjective.
Now the basic idea is, $U$ is a smooth 2-dimensional manifold, and so is $S^2$, so it shouldn't be too hard to make $h$ surjective: generically, $h$ will be open, and we just need enough little open patches in its image to cover all of $S^2$.  In particular, let's think about the local behavior of $h$ near a point $(a,b)\in U$, where we may assume without loss of generality that $f(a)=(1,0,0)$ and $f(b)=(0,0,0)$.  Note that if you change the second coordinate of $f(a)$ a bit and the third coordinate of $f(b)$ a bit, then the vectors $f(a)-f(b)$ will trace out an entire rectangle tangent to the unit sphere at $(1,0,0)$.  When you project this onto the sphere, it will cover an entire open neighborhood $V$ of $(1,0,0)$.
So we've just discovered that if we pick our $f$ so that $f(a)=(1,0,0)$ and $f(b)=(0,0,0)$ for some $a$ and $b$ and that $f$ is moving along certain straight paths near $a$ and $b$, then the image of $h$ near $(a,b)$ will contain a nonempty open set $V\subseteq S^2$.  Now by compactness, $S^2$ is covered by finitely many rotations of $V$.  So to get $h$ to be surjective, we just need $f$ to contain certain finitely many rotations of our pair of straight paths.  By translating them, we can easily arrange that these rotated paths are all disjoint from each other, and then we have to just interpolate smoothly between them to get our loop $f$.
