Can a function have no inverse?

Question

Consider the implicit equation $ f^{-1} (x)=g(x)$. The function $g(x)$ is known and at least can be computed numerically. It may be piecewise continous or oscillating but it is always positive $ g(x) \ge 0 $. Here $ f(x) $ is not known.

Could it be that is there a function $ g(x) $ so it is never invertible and hence we cannot get $ f(x) $?

Answer 1

This might be useful for you:

DEFINITION: Let $f:A\to B$ and $g:B \to A$ be given. The function $f$ is called the inverse of $g$ and the function $g$ is called the inverse of $f$ if $g(f(a))=a$ for each $a \in A$ and $f(g(b))=b$ for each $b \in B$. In this event we sall also say that $f$ and $g$ are inverse functions and that each of the is invertible.

It is a consequence of this definition that if $f$ and $g$ are inverses, then both of them are one-one and onto:

1. $f$ is one-one if you have $x,y \in A$ then $f(x)=f(y)\Leftrightarrow x=y$.
2. $f$ is onto if $f(A)=B$.

As a general result, it is necessary and sufficient that $f$ is onto and one-one for $f$ to be invertible.

An example is the definition of $\arcsin x$. To define it, we must first change

$$f:\mathbb R \to \mathbb R\text{ ; } x\mapsto\sin x$$

to

$$f:\left[-\frac {\pi} 2, \frac {\pi} 2\right] \to [-1,1]\text{ ; } x\mapsto\sin x$$

Since in such definition, $\sin x$ is onto and one-one, it follows we can define

$$g: [-1,1]\to \left[-\frac {\pi} 2, \frac {\pi} 2\right] \text{ ; } x\mapsto\arcsin x$$

Directly answering your question. Let

1. $p$: $f$ is invertible.
2. $q$: $f$ is onto.
3. $r$: $f$ is one-one.

Then

$$(q\wedge r) \equiv p $$ or

$$(-q\vee -r) \equiv -p $$ I reccomend you read Chapter 1 of Introduction to Topology by Bert Mendelson.