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Assumptions: $f,g$ are real-valued. $g:[0,1]^2\to\mathbb R$. Functions $h_1:X\to\mathbb [0,1]$ and $h_2:Y\to\mathbb [0,1]$ are surjective continuous. $X,Y$ are connected separable. $f$ is continuous,

$f(x,y)=g(h_1(x),h_2(y))$.

Questions: Does the strict increasingness of $g$ in all variables imply the continuity of $g$?

Backgrounds: It is known the composite of continuous functions is continuous. What about the other way around?

What we know: even without increasingness, $g$ is continuous if one of the followings hold:

  1. X, Y are path-connected
  2. X,Y are compact Hausdorff
  3. $h_i$ are homomorphisms
  4. $h_i$ are quotient maps

To prove the claim, we only need to show that if $g$ is continuous restricted to one direction, then $g$ is continuous. Also, since $g$ is increasing, the restricted function of $g$ can only contain a jump discontinuity. (If we allow essential discontinuity, a simple counterexample exists, this is why the increasingness is powerful here). I've been trying to construct a counterexample with jump discontinuity for a while, with no luck.

High GPA
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    Presumably the domain of $g$ is meant to be $A = [0, 1] \times [0, 1]$ rather than $\mathbb R^2$. Otherwise you could introduce discontinuities outside of $A$ that'd never be detected by $h_1$ and $h_2$. – John Hughes Nov 28 '23 at 22:05
  • @JohnHughes Yes of course, I'll add that – High GPA Nov 28 '23 at 22:30
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    @TobyBartels I don't think they need to be ordered for this problem, since $g$ is the only function said to be increasing (in each variable separately), and $g$ maps $[0,1]^2\to \mathbb R$. – M W Nov 29 '23 at 00:32
  • @HighGPA Wait, now I'm confused. How does $h$ (presumably $h_1,h_2$) imply there needs to be an order? $h_i$ are just surjective continuous functions, there's no need for any order there. – M W Nov 29 '23 at 00:53
  • @MW You are right that there is no need to have an order. What I mean is that $X$ is orderable, in a trivial sense, that there is a real function $h_1$ defined on it. Say $x_1> x_2$ whenever $h_1(x_1)> h_1(x_2)$ and equivalent if equal. – High GPA Nov 29 '23 at 01:31
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    @MW : Oh, yes, you're right. – Toby Bartels Nov 29 '23 at 16:33

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