Let $f:X\to Y$ be a function from sets $X$ to $Y$ . A quasi-inverse $g$ of $f$ is a function $g$ such that

1. $g:Z\to X$ where $\operatorname{ran}(f)\subseteq Z\subseteq Y$ , and
2. $f\circ g\circ f=f$ , where $\circ$ denotes functional composition operation.

Note that $\operatorname{ran}(f)$ is the range of $f$ .

Examples.

1. If $f$ is a real function given by $f(x)=x^2$ . Then $g(x)=\sqrt{x}$ defined on $[0,\infty)$ and $h(x)=-\sqrt{x}$ also defined on $[0,\infty)$ are both quasi-inverses of $f$ .
2. If $f(x)=1$ defined on $[0,1)$ . Then $g(x)=\frac{1}{2}$ defined on $\mathbb{R}$ is a quasi-inverse of $f$ . In fact, any $g(x)=a$ where $a\in [0,1)$ will do. Also, note that $h(x)=x$ on $[0,1)$ is also a quasi-inverse of $f$ .
3. If $f(x)=[x]$ , the step function on the reals. Then by the previous example, $g(x)=[x]+a$ , any $a\in[0,1)$ , is a quasi-inverse of $f$ .

Remarks.

• Every function has a quasi-inverse. This is just another form of the Axiom of Choice. In fact, if $f:X\to Y$ , then for every subset $Z$ of $Y$ such that $\operatorname{ran}(f)\subseteq Z$ , there is a quasi-inverse $g$ of $f$ whose domain is $Z$ .
• However, a quasi-inverse of a function is in general not unique, as illustrated by the above examples. When it is unique, the function must be a bijection:

  If $\operatorname{ran}(f)\ne Y$ , then there are at least two quasi-inverses, one with domain $\operatorname{ran}(f)$ and one with domain $Y$ . So $f$ is onto. To see that $f$ is one-to-one, let $g$ be the quasi-inverse of $f$ . Now suppose $f(x_1)=f(x_2)=z$ . Let $g(z)=x_3$ and assume $x_3\ne x_1$ . Define $h:Y\to X$ by $h(y)=g(y)$ if $y\ne z$ , and $h(z)=x_1$ . Then $h$ is easily verified as a quasi-inverse of $f$ that is different from $g$ . This is a contradition. So $x_3=x_1$ . Similarly, $x_3=x_2$ and therefore $x_1=x_2$ .

• Conversely, if $f$ is a bijection, then the inverse of $f$ is a quasi-inverse of $f$ . In fact, $f$ has only one quasi-inverse.
• The relation of being quasi-inverse is not symmetric. In other words, if $g$ is a quasi-inverse of $f$ , $f$ need not be a quasi-inverse of $g$ . In the second example above, $h$ is a quasi-inverse of $f$ , but not vice versa: $h(0)=0$ , but $hfh(0)=hf(0)=h(1)=1\ne h(0)$ .
• Let $g$ be a quasi-inverse of $f$ , then the restriction of $g$ to $\operatorname{ran}(f)$ is one-to-one. If $g$ and $f$ are quasi-inverses of one another, and $\operatorname{g}$ strictly includes $\operatorname{ran}(f)$ , then $g$ is not one-to-one.
• The set of real functions, with addition defined element-wise and multiplication defined as functional composition, is a ring. By remark 2, it is in fact a Von Neumann regular ring, as any quasi-inverse of a real function is also its pseudo-inverse as an element of the ring. Any space whose ring of continuous functions is Von Neumann regular is a P-space.

Bibliography

1

B. Schweizer, A. Sklar, Probabilistic Metric Spaces, Elsevier Science Publishing Company, (1983).