# A question about orthogonal projections of Smooth Embeddings of the circle.

Question: Let $f$ be a smooth embedding of $S^1\rightarrow \mathbb{R}^3$. Given an element $v\in S^2$ we have the orthogonal projection $\pi_v:\mathbb{R}^3\rightarrow P_v$ to the plane $P_v$ = the orthogonal complement of v. Show that for almost every $v\in S^2$, we have that $f_v=\pi_v\circ f$ is an immersion.

I'm fairly sure the strategy here will be to show that if $v$ is a regular value of $f$, then $\pi_v$ is an immersion and has injective derivative. Since the derivative of $f$ was injective, so too will the derivative of the the composition map be injective. Then by Sard's theorem, almost every v works.

The problem is when I write out the conditions on the tangent spaces, I don't see any reason why this would be true. if $v$ is regular, then that means that $df_x$ is surjective if $f(x)=v$. Since it is an embedding, that means the derivative is bijective. The projection is a submersion and also has surjective derivative, but there is no reason to suspect that the composition of a bijection and a surjection would be injective. I fear I've fallen into a paralysis of thought, and am a bit stuck in this rut of unproductive thinking.

Will uprate - thanks!

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The projection onto $P_v$ kills the component of the derivative that is parallel to $v$. Therefore, the entire derivative $df$ will be killed if and only if $$df=\lambda v \ \text{ for some } \ \lambda\in \mathbb R\tag1$$ We just have to choose $v$ so that (1) never happens.
To that end, rewrite (1) as $\dfrac{df}{|df|}=v$ and observe that the map $F(x)= \dfrac{df(x)}{|df(x)|}$ is a smooth map from the one-dimensional manifold $S^1$ to the two-dimensional manifold $S^2$. Its image has finite length, and therefore has zero area. Thus, we can choose $v\in S^2\setminus F(S^1)$, and this $v$ does what we want.