TITLE: Trefoil knot cannot be injectively projected to a plane?
QUESTION [8 upvotes]: 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!

REPLY [6 votes]: 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.