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$\newcommand{\wt}[1]{\widetilde{#1}}$ Hello, I just tried my hand at two exercises from John M Lee's book Riemannian Geometry and I would like to know whether my reasoning is sound or if I did something wrong. This is about exercise 2.3 (a) and (c).

Let $M^n\subset\widetilde{M}^m$ be an embedded submanifold.

(a) If $f\in C^\infty(M)$, show that $f$ can be extended to a smooth function on a neighborhood of $M$ in $\widetilde{M}$.

(c) Let $\widetilde{X}$ be a vector field on $\widetilde{M}$, i.e. $\widetilde{X}\in\mathfrak{X}(\widetilde{M})$. Show: $\widetilde{X}$ is tangent to $M$ $\iff$ If $f\in C^\infty(\widetilde{M})$ with $f_{|M}=0$, then $(\wt{X}f)_{|M}=0$

Okay, so here's what I did for (a).

Choose open sets $\wt{U}_\alpha\subset\wt{M}$ s.t. $M\subseteq\bigcup_\alpha \wt{U}_\alpha$ and such that they imply slice coordinates for the sets $U_\alpha=\wt{U}_\alpha\cap M$, and both form open covers of $M$. A note on the slice coordinates (abusing the notation a bit, but this is the way he does it in that particular book): If $\wt{U}_\alpha\leadsto x^1,...,x^m$, then $U_\alpha\leadsto x^1,...,x^n$.
Define $\wt{f}_\alpha\in C^\infty(\wt{U}_\alpha)$ via $\wt{f}_\alpha(x^1,...,x^m)=f(x^1,...,x^n)$ . This definition is obviously independet of the last $m-n$ coordinates.
Choose a partition of unity $p_\alpha$ subordinate to $\wt{U}_\alpha$, and set $F=\sum_\alpha p_\alpha\wt{f}_\alpha$. Then $F\in C^\infty\left(\bigcup_\alpha\wt{U}_\alpha\right)$ is an extension of $f$ in a neighborhood of $M$.

Now, for (c):

We again use the slice coordinates $x^i$ like above. Note first that if $f_{|M=0}$, then $D f_{|\wt{U}_\alpha}=\partial_i f\mathrm{d}x^i_{|\wt{U}_\alpha}$ with $n<i\leq m$, so let $f$ w.l.o.g. vanish on $M$.
$\wt{X}_{|M}$ is tangent to $M$.
$\iff$ In local coordinates of $\wt{U}_\alpha$, $\wt{X}_{|\wt{U}_\alpha}={X^i\partial_i}_{|\wt{U}_\alpha}$ with $X^i(p)=0$ for all $p\in U_\alpha$ and $n<i\leq m$.
$\iff$ $\wt{X}f_{|\wt{U}_\alpha}=0$, since-roughly spoken-there is no matching pair of indices in the $\partial_i$ and $\mathrm{d}x^j$. But this holds for all $\wt{U}_\alpha$, so for every point in $M$.

Personally, I can't think of any conflicts there, but that's why I'm asking here: I don't have too much experience. So - did I make a mistake somewhere, was I sloppy, etc?

Thanks!

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    In (c), I feel that you haven't made it obvious why if $\tilde{X}f|_M = 0$ for all $f$ vanishing on $M$ then $\tilde{X}$ is tangent to $M$. The problem in particular is the direction $\Leftarrow$ in the very last $\iff$ you claim, for which you haven't provided any clear argument. (You're also not very clear about quantifiers -- "for all $f$" etc. -- in this last statement.) – mollyerin Apr 02 '15 at 04:38
  • What I meant is that: If $f\big|M=0$ then $\tilde{X}f\big|_U=\mathrm{D}f\tilde{X}\big|_U=\partial_if\tilde{X}^i\big|_U=0$ iff $X^i{|U}=0$ for $n<i\leq m$. This is ensured by the comment made at the beginning of (c), about the $\mathrm{D}f$ in local coordinates. So in a sense, $\widetilde{X}$ can be seen as the canonical extension of some vector field $X$ on $M$ [that was part (b) of the exercise]. While showing (c), I assumed that $f$ was some smooth function that does vanish on $M$, other functions aren't of interest to us. Maybe that was a bit sloppy, I agree. – iWantToKnow Apr 02 '15 at 08:10
  • For a) Assume $\tilde{M}=\mathbb{R}$ and $M=(0,1)$. Let $f(x)=1/x$. Then clearly $f$ does not extend to smooth function on all of $\mathbb{R}$ – Shreedhar Bhat Jun 20 '20 at 06:47

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