Thread: Implicit Function Theorem for f(x1,...,xm, y1,...,yn) = 0

1. Implicit Function Theorem for f(x1,...,xm, y1,...,yn) = 0

This is not a problem to be solved, but I could not understand something important from the text. Here it is.

Let $m,n$ be positive integers, $a \in E^m, b \in E^n,$ and let $f_1, f_2, ...,f_n$ be continuous real-valued functions on an open subset of $E^{m+n}$ that contains the point $(a,b)$ with $f_1(a,b) = ... = f_n(a,b) = 0$. Suppose that for each $i,j=1,...,n$

$\frac{\partial f_i(x_1,...,x_m,y_1,...,y_n)}{\partial y_j}$ exists and is continuous on the given open subset and that the $n \times n$ determinant

$det (\frac{\partial f_i}{\partial y_j}(a,b))$ is not zero.

Then there exists open subsets $U \subset E^m, V \subset E^n$ with $a \in U$ and $b \in V$ such that there is a unique function $\varphi :U \rightarrow V$ such that $f_i(x, \varphi (x)) = 0$ for each $i = 1,...,n$

....

There is some line written such that "the continuous function on $U \subset E^m$ whose value at $x$ is det $(\frac{\partial f_i}{\partial y_j}(x, \varphi (x)))$ is not zero at $a$

Question

1. Why does such a continuous function on $U$ whose value at $x$ equals det $(\frac{\partial f_i}{\partial y_j}(x, \varphi (x)))$?

I have no idea why a continuous function has value equals to det of something ?

2. It says "whose value at $x$ and "not zero at $a$". What is this? Why it mentioned two points , $x$ and $a$ ?

If one could explain without using linear algebra, I would be thankful.

Thank you very much.

2. It's just telling you to consider the function that maps $x\mapsto\mathrm{det}(\frac{\partial f_i}{\partial y_j}(x,\varphi(x)))$, which happens to be continuous. That's important because if the determinant is nonzero at a, then it is nonzero in a neighborhood of a, by continuity.