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:
- $f$ is one-one if you have $x,y \in A$ then $f(x)=f(y)\Leftrightarrow x=y$.
- $f$ 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
- $p$: $f$ is invertible.
- $q$: $f$ is onto.
- $r$: $f$ 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.