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

Question

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.

Answer 1

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".