Does the map $F(x, y)=(f(x, y), y)$ induce an isomorphism $dF_{(a, b)}$?

Question

Suppose $f:\mathbb R^{n+m}\longrightarrow \mathbb R^n$ be a $C^p$ map such that $df_{(a, b)}:\mathbb R^{n+m}\longrightarrow \mathbb R^n$ is surjective for some $(a, b)\in\mathbb R^{n+m}$. Define $F:\mathbb R^{n+m}\longrightarrow \mathbb R^n\times \mathbb R^m$ setting $$F(x, y)=(f(x, y), y).$$ I'd like to show $dF_{(a, b)}:\mathbb R^{n+m}\longrightarrow \mathbb R^n\times \mathbb R^m$ is an isomorphism.

My Atempt: Since $df_{(a, b)}$ is surjective $B=\textrm{ker}(df_{(a, b)})$ has dimension $m$ so that we can write $$\mathbb R^{n+m}=A\oplus B$$ for some subspace $A\subseteq \mathbb R^{n+m}$ such that $\textrm{dim}(A)=n$. Than we might see $F$ as the map $$F:A\oplus B\longrightarrow \mathbb R^n\times B.$$ Then $dF_{(a, b)}:A\oplus B\longrightarrow \mathbb R^n\times B$ is given by $$dF_{(a, b)}(u, v)=(df_{(a, b)}(u, v), v).$$ It suffices showing $dF_{(a, b)}$ is surjective. In fact, given $(z, w)\in \mathbb R^n\times B$ then $z=df_{(a, b)}(z_1, z_2)$ for some $(z_1, z_2)\in A\oplus B$. Hence, taking $(z_1, w)\in A\oplus B$ we have

$$\begin{align*} dF_{(a, b)}(z_1, w)&=dF_{(a, b)}(z_1, z_2-z_2+w)\\ &=dF_{(a, b)}((z_1, z_2)+(0, -z_2+w))\\ &=dF_{(a, b)}(z_1, z_2)+dF_{(a, b)}(0, -z_2+w)\\ &=(df_{(a, b)}(z_1, z_2), z_2)+(df_{(a, b)}(0, -z_2+w), -z_2+w)\\ &=(z, z_2)+(0, -z_2+w)\\ &=(z, w), \end{align*}$$ what shows my statement.

Questions:

(i) Can I always consider the decomposition $\mathbb R^{n+m}=A\oplus B$.

(ii) Is it true that $df_{(a, b)}(0, -z+w_2)=0$? I know $-z_2+w\in B$.

(iii) If the above holds, is my proof right?

Answer 1

Define $F:\mathbb R^{n+m}\longrightarrow \mathbb R^n\times \mathbb R^m$ setting $$F(x, y)=(f(x, y), y).$$

When introducing $(x,y)$ here, you implicitly decomposed $\mathbb R^{n+m}$ as $\mathbb R^n\times \mathbb R^m$ but do not say how you did it. Not every split will work. Later in the proof you introduce a particular split (which works), but it's too late to do it there. One should have a precise claim followed by a proof, not a vague claim clarified in the process of proof.

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(i) Can I always consider the decomposition $\mathbb R^{n+m}=A\oplus B$.

Yes, it works. What you need for the proof to work is the bijectivity of the restriction of $df_{(a,b)}$ to $A$. Defining $B =\ker df_{(a,b)}$ and requiring $A$ to be a direct summand in $A\oplus B$ achieves precisely that.

(ii) Is it true that $df_{(a, b)}(0, -z+w_2)=0$? I know $-z_2+w\in B$.

Yes, since $B$ is the kernel of $df_{(a, b)}$.

(iii) If the above holds, is my proof right?

Yes, it is (except for the vague definition of $F$ in the claim, noted above).