Proving that I have an immersion in a (weak) version Whitney's embedding theorem for compact manifolds

Question

I am proving a weak version of Whitney's embedding theorem for compact manifolds :

Let $M$ be a compact smooth manifold of dimension $m$.

I can be proven that $M$ has a finite smooth atlas $A = \{(U_1, \varphi_1),...,(U_n, \varphi_n)\}$ Choose a smooth partition of 1, $\{θ_1,...,θ_n\}$ subordinate to $\{U_1,...,U_n\}$. For each $j$ consider the function $\tilde \varphi_j: M\rightarrow \Bbb R^m $ given by $\tilde \varphi(p) = 0$ for every $p \in M\setminus U_j$ and $\tilde \varphi_j(p) =\theta_j(p)\varphi_j(p) $ for all $p \in U_j$. It can be proven that $\tilde \varphi_j$ is smooth for every $j$

I have already proven these things and also that the smooth function $F:M \to \Bbb R^{n(m+1)}$

$F(p)=(\theta_1(p),...,\theta_n(p);\tilde \varphi_1(p),...,\tilde \varphi_n(p))$

is injective. I am having trouble proving that F is an immersion:

I can write $dF_p(X_p)=(X_pF^1,...,X_pF^{n(m+1)})$ ( as proven here

$F(p)=(F_1(p),...F_r(p))$ smooth map, Why is it true that $dF_p(X_p)=(X_p(F_1),...,X_p(F_r))$? )

So I just have to prove that the linear map $dF_p$ has trivial Kernel:

If $0=dF_p(X_p)=(X_pF^1,...,X_pF^{n(m+1)}) = (X_p\theta_1,...,X_p\theta_n,X_p\tilde\varphi _1,...,X_p\tilde\varphi_n)$

$\implies X_p\theta_j=0 \tag 1$

and $X_p\tilde\varphi _j\tag 2=0$

for every $j=1,...n$

$0=X_p\tilde\varphi _j=X_p(\theta_j\varphi _j)=\varphi_j X_p\theta_j+\theta_jX_p\varphi_j = 0 + \theta_jX_p\varphi_j=\theta_jX_p\varphi_j$

I am not sure whether I should descent to a "pointwise level" here:

$0=\theta_j(p)X_p\varphi_j(p)$ If $\theta_j(p)\neq 0$, then $X_p\varphi_j(p)=0$

But stil, how can I conclude from here that $X_p$ is the $0$-derivation ?

After that is proven, concluding that F is an embedding is straightforward because F is an immersion,smooth, bijective onto its image and it's a closed map (because a closed C set in M is compact , since M is compact, and F being continuous implies F(C) is compact in a T2 space, therefore closed in F(M))

Answer 1

You're almost there.

Notice that $\theta_j(p) > 0$ for at least one choice of $j$. For this choice of $j$, $p$ is contained in the domain of the chart $\varphi_j : U_j \to \mathbb R^m$.

Let $x^1, \dots x^m$ be the coordinate functions associated with the chart $\varphi_j : U_j \to \mathbb R^m$. Then $\{ \left( \partial_{x^1} \right)_p, \dots , \left( \partial_{x^m} \right)_p \}$ is a basis for the tangent space at $p$ - which is to say that any $X_p$ can be written in the form $$ X_p = a^1 \left( \partial_{x^1}\right)_p + \dots + a^m \left( \partial_{x^m} \right)_p $$ for some choice of constants $a^1, \dots, a^n$.

Since the function $\varphi_j$ has the coordinate representation $$ \widehat \varphi_j (x^1, \dots, x^m) = (x^1, \dots, x^m)$$ with respect to this chart, we have $$ \left( \partial_{x^1}\right)_p \varphi_j = (1, 0, \dots, 0),$$ $$ \left( \partial_{x^2}\right)_p \varphi_j = (0, 1, \dots, 0),$$ $$ \vdots $$ $$ \left( \partial_{x^m}\right)_p \varphi_j = (0, 0, \dots, 1).$$

So $$ X_p \varphi_j = (a^1, a^2, \dots, a^m).$$

If $X_p \varphi_j = 0$, then $a^1 = \dots = a^m = 0$, so $X_p = 0$, which is what you wanted to show.