# Defining a function from R$^5$ tonR$^2$

There is a solved example in the textbook on implicit functions which goes as follows:

Let $f=(f_1,f_2)$ be a vector valued function defined from $\mathbb{R}^5$ to $\mathbb{R}^2$ where $f_1$ and $f_2$ are real valued functions defined on $\mathbb{R}^5$ by

$$f_1 (x_1,x_2,y_1,y_2,y_3)= 2e^{x_1} + x_2y_1 - 4y_2 +3$$ $$f_2 (x_1,x_2,y_1,y_2,y_3)= x_2\cos x_1 - 6x_1 + 2y_1 - y_3$$ Show that $f$ defines a unique function $g$ from $\mathbb{R}^3$ to $\mathbb{R}^2$ in a neighbourhood $T$ of the point $(3, 2, 7)$ such that $g(3, 2, 7)=(0,1)$ and $f (x_1,x_2,y_1,g(x_1,x_2,x_3)) = 0$

I have not understood in the first place as to how $f_1$ and $f_2$ are real valued functions defined on $\mathbb{R}^5$ to $\mathbb{R}^2$. Should'nt they be from $\mathbb{R}^5$ to $\mathbb{R}^4$. Request guide.

• Is my edit correct? Commented Dec 1, 2016 at 10:17
• I'm not sure about $f (x_1,x_2,y_1,g(x_1,x_2,x_3)) = 0$. it is correct or must be $f (x_1,x_2,y_1,g(y_1,y_2,y_3)) = 0$ ? Commented Dec 1, 2016 at 10:21
• Any function $\to \Bbb R^2$ can be decomposed into two functions $\to \Bbb R^1$, usually called components. It's basically looking at each of the coordinates in $\Bbb R^2$ separately. Commented Dec 1, 2016 at 10:22
• No, what is given in the textbook is f(x$_1$, x$_2$,y$_1$,g(x$_1$,x$_2$,x$_3$)) = 0
– SAK
Commented Dec 1, 2016 at 10:26

Both $f_1$ and $f_2$ are real valued functions from $\mathbb R^5$ to $\mathbb R$ and not to $\mathbb R^4$. E.g. $f_1$ takes the $5-$dimensional vector $(x_1,x_2,y_1,y_2,y_3)$ and sends it to the (one-dimensional) number $$\left(2e^{x_1}+x_2y_1-4y_2+3\right) \;\in \mathbb R$$ Plug in some easy values for $x_i,y_i$ to see this. The same with $f_2$. But $f=(f_1,f_2)$ and as such it "lives" in $\mathbb R^2$.