# Constructing smooth embedding of $M\subseteq \mathbb{R}^n$ into $\mathbb{R}^{n-1}$.

This material is from a class I am taking so some definition might be different from normal sense. So let me define some necessary concepts first and ask question.

Definition Let $$F:M\rightarrow N$$ be smooth map between two smooth manifolds.

Then $$F$$ is called a smooth embedding if it is an immersion(i.e, $$F_* : T_pM\rightarrow T_{F(p)}M$$ is injective $$\forall p\in M$$) and $$F(M)\subseteq N$$ is given with the subspace topology and $$F:M\rightarrow F(M)$$ is homeomorphism.

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$$S^{n-1}\subseteq \mathbb{R}^n$$ is a sphere.

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Setting : For any $$v\in S^{n-1}$$, there is a smooth map $$\pi_v:\mathbb{R}^n\rightarrow \mathbb{R}^{n-1}$$ such that $$(\pi_v)_*:T_x\mathbb{R}^n\rightarrow T_{\pi_{v}(x)}\mathbb{R}^{n-1} \hspace{3mm}\forall x\in \mathbb{R}^n.$$

with

$$ker(\pi_v)_*=span\{ v \}.$$

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My Question

There were two statement the professor did not prove.

(1) $$\pi_v|_M$$ is an immersion iff $$v\neq \frac{w}{|w|} \hspace{4mm}\forall w\in TM\setminus \{0\}.$$

(2) $$\pi_v|M$$ is $$1$$ to $$1$$ iff $$v\neq \frac{p-q}{|p-q|}\hspace{4mm}\forall p,q\in M\hspace{2mm}\text{ with } p\neq q.$$

I am not sure why the two state above are true. I understood that we are projecting our manifold in the direction of $$v$$. But I think there would be no problem even if $$TM$$ contains just one of tangent vector from $$span\{v\}$$. I will be thanking to any comments.

For $$v$$ a unit vector, the map $$\pi_v$$ is $$u \mapsto u - (u \cdot v) v$$ whose kernel is exactly $$H = span{v}$$. If $$T_xM$$, the tangent space to $$M$$ at $$x$$, intersects $$H$$ only in the zero-vector, then the restriction of $$\pi_v$$ to $$T_xM$$ has maximal rank, for it sends nothing but the zero-vector to zero.
That means that $$(\pi_v)_*$$, on each tangent space, $$T_xM$$, is an isomorphism, which meets your definition for "immersion."
The second claim seems clear. Suppose that $$\pi_v(P) = \pi_v(Q)$$ for points $$P, Q \in M \subset \Bbb R^n$$. Then
$$P - (P \cdot v) v = Q - (Q \cdot v)v$$ Doing a little algebra, we get $$(P - Q) = ((P-Q) \cdot v)v.$$ So $$P-Q$$ must be a multiple of $$v$$. And that means that $$p-q$$, when normalized, must actually be $$\pm v$$. If it's $$-v$$, then swapping the roles of $$P$$ and $$Q$$ gives you $$+v$$.
Similarly, if $$Q = P + v$$, then it's clear, by direct computations, that $$\pi_v(Q) = \pi_v(P)$$. So that covers both "if" and "only if".