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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 $\operatorname{ran}(f)\subseteq Z\subseteq Y$, 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 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 $\operatorname{ran}(f)\subseteq Z$, 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 $\operatorname{ran}(f)\neq Y$, then there are at least two quasiinverses, one with domain $\operatorname{ran}(f)$ 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 $x_{3}\neq x_{1}$. Define $h:Y\to X$ by $h(y)=g(y)$ if $y\neq z$, 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 $\operatorname{ran}(f)$ is onetoone. If $g$ and $f$ are quasiinverses of one another, and $\operatorname{g}$ strictly includes $\operatorname{ran}(f)$, 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).
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