# Motivating the implicit function theorem

I'm trying to understand the proof of a special case of the implicit function theorem, of a mapping $f: \mathbb{R}^2 \longrightarrow \mathbb{R}$, from this link.

While I am able to follow the proof, I am unsure what motivates the mapping,

$H(x,y) = (x, f(x,y))$

to which one applies the inverse function theorem. I had trouble understanding the motivation behind the mapping defined in the inverse function theorem as well (omitted to maintain brevity), but I eventually got a feel for it; if the function were linear, the inverse of the linear map, determined by the first order Taylor polynomial, will sufficiently and exactly yield the inverse. If the mapping is non linear, then simply iterate and let the contraction mapping theorem take over.

I am unable to convince myself what motivates this choice of the map to prove the implicit function theorem.

First notice that the inverse function theorem requires to be a map from $\mathbb{R}^2 \to \mathbb{R}^2$. This means that your $f(x,y)$ is real-valued.
When regarding the implicit function theorem, you got two kinds of variables: first $x_1, \ldots, x_n$ and second $y_{n+1}, \ldots, y_{n+m}$ (in this case: $n=1$, $m =1$). In the implicit function theorem, you linearise the complicated system of equations with different dependencies and requiring that the second part of the jacobian is invertible yields that you can solve for the variables $y$ in dependence of $x$. In other words, you can find locally a function that expresses your $y$ in terms of $x$, denoted as $y=g(x)$. Probably a clearer explaination is written here by Zhen Lin.
This motivates the defintion of your function: first, you assume that the $x$ will remain constant (we don't want to solve for these variables, but express the $y$'s locally as a function of the $x$). The second component, namely $f(x,y)$, carries the dependence on both variables. With the inverse function, one derives conditions for a very specific choice in a neighbourhood of $x$ and $y$ for $f(x,y)$ that will give you the the desired $y=g(x)$.
• But what's exactly my question, in part: why must the second component carry dependence on $x$. Say, I define a mappy $G(x,y)$. If I desire a local function of the form $y = (x)$, then, denoting the function by a $h$, shouldn't I define the mapping $G(x, h(x))$. Why does $h$ depend on $y$ as well? Doesn't $h$ determine $y$? – Junaid Aftab Mar 24 '17 at 20:11
• @JunaidAftab, you don't know $h$. – Martín-Blas Pérez Pinilla Mar 24 '17 at 20:58
• @JunaidAftab, $f$ determines $y$ implicitly as a function of $x$. – Martín-Blas Pérez Pinilla Mar 24 '17 at 21:13