The differential $\text d F_p$ is injective iff the pullback $F_p^*$ is surjective.

Question

I'm trying to prove the following claim:

Let $F\colon M \to N$ be a differentiable application beetween $C^\infty$ manifolds. Then the differential $\text dF_p\colon T_p M \to T_{F(p)}N$ is injective if and only if the pullback $F_p^*\colon C^\infty _N (F(p))\to C^\infty _M (p)$ is surjective.

Where $C^\infty _S (q)$ are the germs in $q$, $F_p^*(\mathbf g)=\mathbf g \circ F$ and $\text dF_p (X)=X\circ F_p ^*$.

I'll use Einstein's convention on summations. Let $(\varphi = (x^1,x^2,\dots,x^n),U)$ be a local chart in $p$.

Sufficiency: Proof by contrapositive: suppose that $\text d F_p(X^i \partial _i |_p)=O$ and $X^k\neq 0$. Then for all $\mathbf g \in C^\infty _N (F(p))$ $$0=\text d F_p(X^i \partial _i |_p)(\mathbf g)=X^i \partial _i |_p(F_p ^*(\mathbf g)).$$ This shows that $\mathbf x ^k$ is not in the image of $F_p ^*$, hence if $F_p^*$ is surjective then $\text d F _p$ is injective. []

Now for the converse, I was reasoning on these lines: if $F_p ^* (\mathbf h)\neq \mathbf g$ for all $\mathbf h \in C^\infty _N (F(p))$ and for some $\mathbf g \in C^\infty _M (p)$, then we can consider the ideal $$I=\{ \mathbf g \cdot \mathbf f_0 | \mathbf f _0 \in C^\infty _M (p)\}.$$ It is easily seen that $$\text {Im}F_p ^*\cap I= \{\mathbf 0 \}.$$ Now, if I could find a derivation such that $X(\mathbf g)=1$ and $X(C^\infty _M (p) - I)=\{0\}$ this would be done. My problem is to find such $X$.

Since it is not restrictive to assume $\mathbf g(p)=0$, more generally a derivation would act on $I$ in this way: $$X(\mathbf f)=X(\mathbf g \mathbf f _0)=X(\mathbf g)\mathbf f(p),$$ however I'm not sure if I can use this fact in any way.

Any help is appreciated; I'd prefer to complete my proof, but also a different one would be OK.

Answer 1

Since $dF_p$ is injective, $F$ is locally a immersion. With proper local coordinates, $F$ becomes $(x^1,\dots,x^m)\mapsto(x^1,\dots,x^m,0,\dots,0)$.

Now any $g\in C^\infty_M(p)$ is locally a function $g:(x^1,\dots,x^m)\mapsto \mathbb R$. Consider the map $f\in C^\infty_N(F(p))\to\mathbb R$ given by: $$ f=g\circ\pi, \pi:(x^1,\dots,x^n)\mapsto(x^1,\dots,x^m) $$ One immediately see that $g=f\circ F=F^*(f)$.

Answer 2

Here is a slightly different argument (from the one in the original question)

As you noticed, one direction is easy: suppose $F^*$ is surjective, and let $X$ be such that $dF(X)=0$. Then for any function $g$, there is a function $h$ such that $dh\circ dF = dg$, and hence $dg(X)=0$. Taking $g$ to be any coordinate function, as you did, you see that $X=0$, and hence $dF$ is injective.

For the converse direction, assume $dF$ is injective. Then one can choose local coordinates $x=(x_1,\dots,x_m)$ at $p$, and local coordinates of the form $(x,y)=(x_1,\dots,x_m,y_1,\dots,y_n)$ at $F(p)$ such that $F(x)=(x,0)$. This is a direct consequence of the implicit function theorem and sometimes called the "local immersion theorem". Then the result is easy: given any function $g=g(x)$ you can choose $h(x,y):=g(x)$ and you get $h\circ F = g$. Thus $F^*$ is surjective.

As a remark, notice that the first direction is "infinite dimension => finite dimension", and hence is easy, and the converse "finite dimension => infinite dimension" typically requires something like implicit function theorem (which contains the true difficulty of the question).

EDIT: sorry I didn't notice this was already answerd.