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I was reading about intersection points of $f(x)$ and $f^{-1}(x)$ in this site. (Proof: if the graphs of $y=f(x)$ and $y=f^{-1}(x)$ intersect, they do so on the line $y=x$) Then, I saw this statement was wrote by N. S.: "If the graphs of $f(x)$ and $f^{-1}(x)$ intersect at a single point, then that point lies on the line $y=x$.

It is also true that if the graphs of $f(x)$ and $f^{-1}(x)$ intersect at an odd number of points, then at least a point point lies on the line $y=x$. This follows immediately from the observation that the intersection points are symmetric with respect to that line..." I want to know whether it's true or not and if it's true how we can prove it algebraically?

My try: I tried many function and this statement was true but I can't prove it. (Or disprove.)

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The argument is mainly a counting argument involving just a little algebra on the functions themselves.

The functions $f(x)$ and $f^{-1}(x)$ intersect either at a finite number of points, or an infinite number of points. If the number of intersections is infinite, it is neither odd nor even. So we only need to consider a finite number of intersections.

If $(p,q)$ is one of the intersection points, that means $q = f(p) = f^{-1}(p).$ But from $q=f(p)$ we can deduce that $f^{-1}(q) = p,$ and from $q = f^{-1}(p)$ we can deduce that $f(q) = p,$ therefore $p = f(q) = f^{-1}(q),$ that is, and the two functions also intersect at $(q,p).$

So consider the set of intersection points that are above the line $y=x.$ Suppose there are $n$ of these points, where $n \geq 0.$ For each point $(p,q)$ above the line $y=x$ (that is, where $q>p$), there is a corresponding point $(q,p)$ below the line $y=x,$ and vice versa. Hence there are $n$ points below the line $y=x.$

Let the number of intersection points on the line $y=x$ be $m.$ Then the total number of intersection points is $n$ above the line, $n$ below the line, and $m$ on the line (where $m\geq 0$), for a total of $$ 2n + m. $$ Now, $m$ has the same parity as $2n+m.$ If the total number of intersections $2n+m$ is odd, it follows that $m$ is odd. But zero is not odd; all non-negative odd numbers are positive. So the total number of intersections on the line $y=x$ in that case is an odd positive number. In particular, it is at least $1.$

In this proof, we never assume there are any intersection points above the line $y=x,$ nor that there are any below the line or on the line. But we show that if there are an odd number of intersection points altogether, the number of intersection points on the line is positive.

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    Your proof is wonderful! But what about even intersection points ? What we can say about it ? – 2017-02-03
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    I think if number of intersection points is even , all of the points lie on the $y=x$ line. For example $f(x) = \sqrt x$ and then $f^{-1}(x) = x^2$ , $x \ge 0$ . Therefore intersection points are $(0,0)$ and $(1,1)$. – 2017-02-03
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    Try $f(1)=2, f(3)=0,$ and for all $x$ other than $0$ and $3,$ $f(x)=x-1.$ The intersection points are $(1,2)$ and $(2,1).$ If you restrict $f$ to continuous functions, however, I think it may only be possible to have a decreasing function with an odd number of intersections or an increasing function with all intersections on the line $y=x.$ – 2017-02-03
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    Okay consider continuous functions . Can you provide a function that number of intersection points with inverse is even and at least one of the points isn't locate on $y=x$ ? – 2017-02-03
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    If my thoughts about continuous functions are correct, the only choices are an odd number of intersections or all intersections on $y=x,$ so the answer would be no. I just don't have a complete proof in mind yet. (To be precise, I'm thinking about continuous functions on connected domains; if you allow a disconnected domain I think it may be possible to have an even number of intersections and none of them on $y=x.$) – 2017-02-03
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    Okay , thank you a lot for your answers . Great! – 2017-02-03