We say that $f$ maps $x$ to $u$, and that $u$ is the image of $x$ under $f$. If $S\subset D_f$, where $D_f$ is the domain of $f$, then the set \[ f(S) := \left\{ u : u = f(x), \,x\in S\right\} \] is the image of $S$ under $f$.

Remark: We will often denote the components of elements in $S$ by $x, y, z\ldots,$ and the components of elements in $f(S)$ by $u, v, w\ldots.$

Let $f:\R^2\to \R^2$ be the function defined by \[ f(x,y) := \left(x^2+y^2, x^2-y^2\right). \] Then

For $p=(x,y)\in \R^2,$ we have that

For $p=(x,y)\in \R^2,$ we have that

Also notice that

In this case, we can define a function $g$ on the range \[ R(f) := \left\{ u: u = f(x), \text{ for some } x\in D_f \right\} \] of $f$ by defining $g(u)$ to be the unique point in $D_f$ such that $f(g(u))=u.$

Moreover, $g$ is one-to-one. So we have

Moreover, $g$ is one-to-one. So we have

We say that $g$ is the inverse of $f$ and write

The relation between $f$ and $g$ is symmetric; that is, $f$ is also the inverse of $g,$ and we write

Let $f:\R^2\to \R^2$ be the function defined by \[ f(x, y) := \left(x-y, x+y\right). \]

Note that $f$ maps $(x,y)$ to $(u,v)$, where $ \; \left\{\begin{eqnarray*} u &=& x-y, \\ v &=& x+y. \end{eqnarray*} \right. $

Furthermore, observe that $R(f)= \R^2$ and $f$ is one-to-one, since for each $(u,v)\in \R^2 $ there is exactly one $(x,y)$ such that $f(x,y)= (u,v).$ This is so because the above system can be solved uniquely for $(x,y)$ in terms of $(u,v).$

Let $g:\R^2\to \R^2$ be the function defined by \[ g(x, y) := \left(x+y, 2x+2y\right). \]

Here $g$ maps $(x,y)$ to $(u,v)$ where $ \; \left\{\begin{eqnarray*} u &=& x+y, \\ v &=& 2x+2y. \end{eqnarray*} \right. $

In this case $f$ is not one-to-one, since every point on the line \[ x+ y = c, \quad (c \text{ constant}) \] is mapped onto the single point $(c,2c)$. Hence, $f$ does not have an inverse.

The crucial difference between the functions of Examples 13.2.1 and 13.2.1 is that the matrix representation of $f$ is non-singular while the matrix representation of $g$ is singular. In other words, considering

then $A$ is invertible (non-singular), and $B$ is not (singular).

Theorem 13.2.1. The function $f:\R^n\to \R^n$ defined by \[ f(x) := Ax, \quad A\in M_{n\times n} \] is invertible if and only if $A$ is nonsingular; in which case $R(f)= \R^n$ and \[ f^{-1}(u) = A^{-1}u. \]

Let $f:\R^2\to \R^2$ be the function defined by \[ f(x, y) := \left(e^x\cos y, e^x \sin y\right). \] This function is not one-to-one. Why?

However, it is one-to-one locally.

Let $f:\R^2\to \R^2$ be the function defined by \[ f(x, y) := \left(x^2-y^2, 2xy\right). \]

$f$ is not one-to-one since $ f(-x,-y) = f(x,y). $ However, $f$ is one-to-one on every set $S$ of the form \[ S= \left\{(x,y)~:~ ax+by >0 \right\}, \] where $a,b\in \R,$ not both zero. Geometrically $S$ is an open half-plane.

Let $f:\R^2\to \R^2$ be the function defined by \[ f(x, y) := \left(x^2-y^2, 2xy\right). \]

Then, $f$ is locally invertible at every $(x,y)\neq (0,0)$ since every such point lies in a half-plane of this form.

Therefore, $f$ is locally invertible on the entire plane with $(0,0)$ removed.

To motivate our study of this question, let us first consider the function $f$ as a linear transformation, that is, \[ f(x) = A x = \begin{bmatrix} a_{1,1} & a_{1,2} & \ldots & a_{1,n}\\ a_{2,1} & a_{2,2} & \ldots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n,1} & a_{n,2} & \ldots & a_{n,n} \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{bmatrix} \]

From Theorem 13.2.1, $f$ is invertible if and only if$ A$ is non-singular, in which case $R(f)= \R^n$ and

Also we have that \[ f'=A \quad \text{and}\quad \left(f^{-1}\right)'= A^{-1}. \]

Since $A$ and $A^{-1}$ are the differential matrices of $f$ and $f^{-1},$ respectively, we can say that a linear transformation is invertible if and only if its differential matrix $f'$ is nonsingular, in which case the differential matrix of $f^{-1}$ is given by

Because of this, it is tempting to conjecture that if $f:\R^n\to \R^n$ is continuously differentiable and $f'(p)$ is nonsingular, or, equivalently, $J_f(p)\neq 0,$ for $p\in \R^n,$ then $f$ is one-to-one on $\R^n.$ However, this is false!

Because of this, it is tempting to conjecture that if $f:\R^n\to \R^n$ is continuously differentiable and $f'(p)$ is nonsingular, or, equivalently, $J_f(p)\neq 0,$ for $p\in \R^n,$ then $f$ is one-to-one on $\R^n.$ However, this is false!

For example, if $ \,f(x,y):= \left(e^x \cos y , e^x \sin y \right), $ then

Because of this, it is tempting to conjecture that if $f:\R^n\to \R^n$ is continuously differentiable and $f'(p)$ is nonsingular, or, equivalently, $J_f(p)\neq 0,$ for $p\in \R^n,$ then $f$ is one-to-one on $\R^n.$ However, this is false!

For example, if $ \,f(x,y):= \left(e^x \cos y , e^x \sin y \right), $ then

$J_f(x,y)= e^{2x}\neq 0,$ but $f$ is not one-to-one on $\R^2.$

The best that can be said in general is that if $f$ is continuously differentiable and $J_f(p)\neq 0$ in an open set $V\subset D_f,$ then

1. $f$ is locally invertible on $V,$ and
2. the local inverses are continuously differentiable.

Intuitively, if a function is continuously differentiable, then it locally behaves like the derivative (which is a linear function).

The inverse function theorem establishes that if a function is continuously differentiable and the derivative is invertible, the function is (locally) invertible.

Let $f:\R^2\to \R^2$ be the function defined by \[ f(x,y) := \left(2 x + 3y, x + 4y\right). \]

Is $f$ invertible? If so, is it globally or locally invertible? To answer this questions first we need the Jacobian:

Since $J_f(x,y) \neq 0$ for all $(x,y)\in \R^2,$ then the inverse function theorem guarantees the existence of an open set $U$ where $f$ is invertible.

In fact, since $f$ is one-to-one and $R(f) = \R^2,$ then it is globally invertible. Also $f^{-1}$ is continuously differentiable and \[ \left(f^{-1}\right)'(u,v) = \left(f'(x,y)\right)^{-1}, \;f(x,y) = (u,v)\in \R^2. \]

Can we compute it inverse? 🤔 Yes, we just need to solve the system \begin{eqnarray*} u &=& 2x + 3 y, \\ v &=& x+ 4 y. \end{eqnarray*}

\begin{eqnarray*} u &=& 2x + 3 y, \\ v &=& x+ 4 y. \end{eqnarray*}

Thus we have that \[ f^{-1}(u,v) =\left(\frac{4u-3v}{5}, \frac{-u+2v}{5}\right). \] and

Is $\,f(x,y):= \left(e^x \cos y, e^x \sin y \right)\,$ invertible? If so, is it globally or locally invertible?

We know that $J_f(x,y) = e^{2x}\neq 0$ for any $(x,y)\in\R^2.$ The inverse function theorem tells us that there exists an open set $U\subset \R^2$ where $f$ is invertible. $f$ is not one-to-one, but it is locally invertible.

In particular, $f$ is one-to-one on the open set \[ U = \left\{(x,y): -\infty \lt x\lt \infty, 0\leq y\lt 2\pi \right\}. \]

Is $f(x,y)= \left(e^x \cos y, e^x \sin y \right)$ invertible? If so, is it globally or locally invertible?

In particular, $f$ is one-to-one on the open set \[ U = \left\{(x,y): -\infty \lt x\lt \infty, 0\leq y\lt 2\pi \right\}. \]

In this case we have that \[ \left(f|_U\right)^{-1}(u, v) = \left( \ln \left(u^2 + v^2\right)^{1/2} , \,\arg (u, v) \right), \] where $\; 0 \leq \arg (u,v) \lt 2\pi .\,$ (Not easy to find it! 😥)