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I am looking for a two variable function (a surface), $g(x,b)$, with the following properties:

$$g: [0,pi] times (0,pi) to[0,pi] $$

$$g(0,b)=0$$
$$g(pi,b)=pi$$
$$g(b,b)=frac{pi}{2}$$
$$gleft( x,frac{pi}{2}right)=x$$

Where $xin[0,pi]$ and $bin(0,pi)$.

Can you define such a $g$? I have been looking for a while and am unable to find one. If you cannot define one, does such a $g$ even exist?

Something like
$$ g(x,b) = begin{cases} 0 & x=0\ arctanbigl(tan(x-tfracpi2)-tan(b-tfracpi2)bigr)+tfracpi2 & 0<x<pi \ pi & x=pi end{cases} $$
ought to work relatively smoothly, but this version with hyperbolic functions behaves nicer in some respects:
$$ g(x,b) = begin{cases} 0 & x=0\ tfracpi2 Bigl(tanhbigl(operatorname{arctanh}(2x/pi -1)-operatorname{arctanh}(2b/pi-1)bigr)+1Bigr) & 0<x<pi \ pi & x=pi end{cases} $$

"Nicer" is in the eye of the beholder, of course. These proposals both have the general form
$$ g(x,b) = begin{cases} 0 & x=0\ h^{-1}bigl(h(x)-h(b)bigr) & 0<x<pi \ pi & x=pi end{cases} $$
where $h$ is an increasing bijection $(0,pi)tomathbb R$ with $h(pi/2)=0$. This immediately guarantees the properties you specify, as well as continuity in your domain. Furthermore the function is increasing in $x$ and $b$ for fixed $b$ and $x$.

The $tan$ version has the property that at each fixed $b$ we have $g(varepsilon, b)approxvarepsilon$ for small enough $varepsilon$ (and similarly at the other three edges of the domain, mutatis mutandis). This probably is not what you want, but since I don't know what you want it for it could be exactly what you need.

On the other hand $operatorname{arctanh}$ gives the nice property that if you graph $g$ for constant $x$ or $b$, there are no inflection points. One would probably usually find that desirable.

For a purely algebraic option you might try $h(x)=frac{1}{pi-x}-frac{1}{x} $, which ought to behave more or less like the tangent option.

$$g(x,y)=2arctanfrac{tan(x/2)}{tan (y/2)}$$
I've started from $g(b,b)=fracpi2$. That means that somehow $g$ depends either on the difference $(x-y)$, or some ratio of similar functions for the coordinates $f(x)/f(y)$. The second form was hinted from the fact that $y$ is on an open interval, so $f(0),f(pi)in[0,pminfty]$. The fact that $g(0,b)=0$ means $f(0)=0$. But $g(pi,y)=pi$ means that I need to scale $f(pi)=infty$ back to $pi$ (that's where the ideas of $tan$ and $arctan$ functions came from). If $f(pi/2)=1$ you have $arctan tan x=x$.

Note: following the comment from @HenningMakholm, I can rewrite this in a more accurate way:
$$g(x,y)=lim_{uto x}2arctanfrac{tan(u/2)}{tan(y/2)}$$