quasiinverse of a function
Let $f:X\to Y$ be a function from sets $X$ to $Y$. A quasiinverse^{} $g$ of $f$ is a function $g$ such that

1.
$g:Z\to X$ where $\mathrm{ran}(f)\subseteq Z\subseteq Y$, and

2.
$f\circ g\circ f=f$, where $\circ $ denotes functional^{} composition operation.
Note that $\mathrm{ran}(f)$ 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 $[0,\mathrm{\infty})$ and $h(x)=\sqrt{x}$ also defined on $[0,\mathrm{\infty})$ are both quasiinverses of $f$.

2.
If $f(x)=1$ defined on $[0,1)$. Then $g(x)=\frac{1}{2}$ defined on $\mathbb{R}$ is a quasiinverse 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 quasiinverse 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 quasiinverse of $f$.
Remarks.

•
Every function has a quasiinverse. 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 $\mathrm{ran}(f)\subseteq Z$, there is a quasiinverse $g$ of $f$ whose domain is $Z$.

•
However, a quasiinverse 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 $\mathrm{ran}(f)\ne Y$, then there are at least two quasiinverses, one with domain $\mathrm{ran}(f)$ and one with domain $Y$. So $f$ is onto. To see that $f$ is onetoone, let $g$ be the quasiinverse 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 quasiinverse 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 quasiinverse of $f$. In fact, $f$ has only one quasiinverse.
 •

•
Let $g$ be a quasiinverse of $f$, then the restriction^{} of $g$ to $\mathrm{ran}(f)$ is onetoone. If $g$ and $f$ are quasiinverses of one another, and $\mathrm{g}$ strictly includes $\mathrm{ran}(f)$, then $g$ is not onetoone.

•
The set of real functions, with addition^{} defined elementwise and multiplication defined as functional composition, is a ring. By remark 2, it is in fact a Von Neumann regular ring^{}, as any quasiinverse of a real function is also its pseudoinverse as an element of the ring. Any space whose ring of continuous functions is Von Neumann regular is a Pspace.
References
 1 B. Schweizer, A. Sklar, Probabilistic Metric Spaces, Elsevier Science Publishing Company, (1983).
Title  quasiinverse of a function 

Canonical name  QuasiinverseOfAFunction 
Date of creation  20130322 16:22:14 
Last modified on  20130322 16:22:14 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  11 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 03E20 
Synonym  quasiinverse 
Defines  quasiinverse function 