I am reading Henri Cartan's "Elementary Theory of Analytic Functions of One or Several Complex Variables". On pp. 64-65, Cartan used a version of Implicit Function Theorem that I have never seen:
In the version found in most textbooks, an implicit fuction defined by the equation $f(x,y)=0$, where $f:\mathbb R^{m+n}\to\mathbb R^n$, is expressed locally as a function $y=g(x)$ where $g:\mathbb R^m\to\mathbb R^n$. However, Cartan's version looks totally different and I don't know what's going on. Would anyone please explain Cartan's version as well as his proof of the lemma in the images above? Thank you very much.
Suppose for the sake of concreteness that $\gamma_1'(t_0)\neq 0$. Now, consider the mapping $\delta:(a,b)\times \Bbb{R}\to\Bbb{R}^2$ defined as \begin{align} \delta(t,u)&:=\gamma(t)+\text{sign}(\gamma_1'(t_0))\cdot (0,u)= (\gamma_1(t), \gamma_2(t)+ \text{sign}(\gamma_1'(t_0))\cdot u) \end{align} Since $\gamma$ is a $C^1$ mapping by assumption, so is $\delta$. Now, the Jacobian matrix is \begin{align} \delta'(t,u)&= \begin{pmatrix} \gamma_1'(t) & 0\\ \gamma_2'(t) & \text{sign}(\gamma_1'(t_0)) \end{pmatrix} \end{align} So, in particular, at the point $(t_0,0)$, we have \begin{align} \delta'(t_0,0)&= \begin{pmatrix} \gamma_1'(t_0)& 0\\ \gamma_2'(t_0)& \text{sign}(\gamma_1'(t_0)) \end{pmatrix} \end{align} which is a matrix with determinant equal to $|\gamma_1'(t_0)|>0$. So, this means $\delta$ satisfies the assumptions of the inverse function theorem (and it is a standard exercise to show this is equivalent to the implicit function theorem). As a result:
It is clear by definition of $\delta$ that $\delta(t,0)=\gamma(t)$. So, point (i) of Cartan has been addressed. Now, if we shrink the open set $U'$ to a small enough open ball $U$, then (by continuity of the Jacobian determinant) by setting $V=\delta[U]$, we have that the further restricted mapping $\delta:U\to V$ is a $C^1$ diffeomorphism such that for all $(t,u)\in U$, we have $\det \delta'(t,u)>0$.
Note by the way that the mapping $\delta$ is not uniquely defined. For example, if $f:\Bbb{R}\to\Bbb{R}$ is any smooth map such that $f'(0)=\text{sign}(\gamma_1'(t_0))$, then the proof above can be applied word for word to the mapping $(t,u)\mapsto \gamma(t)+ (0,f(u))=(\gamma_1(t),\gamma_2(t)+f(u))$. This will of course yield different open sets $U,V$, but that's not relevant. All we care about here is that there exist some such diffeomorphism.
Edit:
Maybe I should have done this from the beginning, but the direct way to do this without any case distinction is to let $\xi\in\Bbb{R}^2$ be a vector such that $\{\gamma'(t_0),\xi\}$ is a basis of $\Bbb{R}^2$ (we can do this since $\gamma'(t_0)\neq 0$). Also, we may choose the vector $\xi$ such that $\det \begin{pmatrix}\gamma'(t_0)&\xi\end{pmatrix}>0$ (for example, $\xi$ could just be the vector $\gamma'(t_0)$ rotated counter clockwise by 90 degrees). Then, we simply consider the mapping $\delta:(a,b)\times \Bbb{R}\to\Bbb{R}^2$ defined as $\delta(t,u)=\gamma(t)+u\cdot\xi$, and apply the inverse function theorem to $\delta$ about the point $(t_0,0)$, and restrict the domain sufficiently to ensure the Jacobian determinant is strictly positive.