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Problem: Consider two continuous surjective functions $f,g:[a,b]→[a,b]$.If $$f(a)=f(b), g(a)=g(b),$$then find the minimum number of ordered pairs $(x,y)$ satisfying $h(x,y)= 0$ ,where $h(x,y)=f\circ g(x)-f\circ g(y)$; $x,y∈[a,b]$ and $x≠y$.

I thought the answer will be infinite since we're basically just counting the number of horizontal chords here and a quick sketch of $f\circ g$ shows that there are infinitely many such horizontal lines intersecting it at two points i.e., infinite solutions to $h(x,y)=0$.

But the answer is actually given as 6 with the following reasoning (open the link to get the image of the answer); (They use the intermediate value theorem very cleverly here) So where am I going wrong?

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I believe you are right and they are completely wrong (to be honest I did not understand their solution — I found it very confusing).

To give a quick proof of why you are right, consider the following: the function $f \circ g$ has domain the same as that of $g$ (i.e., $[a, b]$), and range the same as that of $f$ (i.e., $[a, b]$) — these claims follow from the fact that $g$ and $f$ are surjective. It is continuous because it is the composition of two continuous functions, and it is surjective because it is the composition of two surjective functions. Also, $f(g(a)) = f(g(b))$. In other words, $f\circ g : [a, b] \rightarrow [a, b]$ is a continuous surjective function with the property that $f(g(a)) = f(g(b))$ (the same properties of $f$ and $g$ described in the problem). If you are to draw $f \circ g$ you can easily use your argument (that of the horizontal lines).

If you want to be completely rigorous: assume $(f \circ g)(a)$ is not the max of $f \circ g$. Let $m \in [a, b]$ such that $(f\circ g)(m)$ is the maximum of $f \circ g$ (there may be many $m$'s, take the smallest). Apply IVT to interval $[a, m]$ and then to interval $[m, b]$. That is, if $u_1$ is a number between $(f\circ g)(a)$ and $(f\circ g)(m)$, there is a $c_1 \in (a, m)$ such that $(f\circ g)(c_1) = u_1$. Similarly, if $u_2$ is a number between $(f\circ g)(m)$ and $(f\circ g)(b)$, then there is a $c_2 \in (m, b)$ such that $(f\circ g)(c_2) = u_2$. Can you see how this proves that there are infinitely many solutions? Then just do the same thing with min after assuming $(f \circ g)(a)$ to be the max of $f \circ g$.

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  • $\begingroup$ This proves that there are two more solutions (c1, c2) & (c2, c1) for which h(x,y)= u1 - u1 = 0 or maybe u2 - u2 = 0 but uh how does it prove that there are infinitely many such c1s and c2s for which this occurs? $\endgroup$ Commented Dec 20, 2023 at 14:06
  • $\begingroup$ Also, could you help me understand where the book solution went wrong? I have a strong suspicion that in the calculation of delta, they assume invertibility of fog(x) and that's where the issue lies but then it also gets me thinking that even if it is not invertible, how are there infinitely many values of delta for which the final functional relation in the solution will hold? $\endgroup$ Commented Dec 20, 2023 at 14:16
  • $\begingroup$ Let be more clear: let $u \in [(f\circ g)(a), (f \circ g) (m)]$. Then by IVT there exists $c_1 \in (a, m)$ such that $(f\circ g)(c_1) = u$. But $(f\circ g)(a) = (f\circ g)(b)$, so $[(f\circ g)(a), (f \circ g) (m)] = [(f\circ g)(b), (f \circ g)(m)]$, so $u \in [(f\circ g)(b), (f \circ g)(m)]$ and by IVT there exists some $c_2 \in (m, b)$ such that $(f\circ g)(c_2) = u$. So $(c_1, c_2)$ is a solution. The thing to point out is that this is true for all values in the interval $[(f\circ g)(a), (f \circ g) (m)]$, so for a different $u'$ will you get different $(c_1', c_2')$. $\endgroup$ Commented Dec 20, 2023 at 18:19
  • $\begingroup$ Like I said, I found their solution to be extremely confusing, so if the issue is to find their mistake I'm just as clueless as you are :) $\endgroup$ Commented Dec 20, 2023 at 18:24
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    $\begingroup$ oh okay, so the IVT argument that you used, can be applied for arbitrarily many u between the interval (a,m) & (m,b), I understand now. $\endgroup$ Commented Dec 21, 2023 at 3:35

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