True or False : If $f(x)$ and $f^{-1}(x)$ intersect at an even number of points , all points lie on $y=x$ Previously I have discussed about odd number of intersect points (See : If the graphs of $f(x)$ and $f^{-1}(x)$ intersect at an odd number of points, is at least one point on the line $y=x$?)
Now , I want to know the even condition . For example $f(x) = \sqrt{x}$ and $f^{-1}(x) = x^2 , x\ge 0$ intersects each other in $(0,0)$ and $(1,1)$ points and these points located on $y=x$ line
Edit : Consider $f$ is continuous function.
 A: The continuous case:
If the domain is not an interval, we can still find a counterexample. Define
$$
f(x)=\cases{2x+2 & if $x\in(0,2)$,\cr
            x-3 & if $x\in(3,5)$,}
$$
then
$$
f^{-1}(x)=\cases{x/2-1 & if $x\in(2,6)$,\cr
                 x+3 & if $x\in(0,2)$,}
$$
and intersections are $(1,4)$, $(4,1)$.
However, if the domain is connected, it turns out to be true.
Proof of the theorem in the case of a connected domain (interval)
For contradiction, let us suppose that there is an even number of intersections and an intersection is out of diagonal.


*

*We assume that $f^{-1}$ exists which means that $f$ is injective. Moreover, $f$ is continuous and its domain is interval, so $f$ is increasing or decreasing.

*$f$ and $f^{-1}$ are symmetric by diagonal, therefore the set of intersections is symmetric by the diagonal.

*Let denote $(x_1, x_2)$ an intersection out of diagonal. By symmetry, there is an intersection $(x_2, x_1)$. We can therefore WLOG assume that $x_1<x_2$.

*$f(x_1) > x_1$ and $f(x_2) < x_2$, so by continuity, $f$ intersects the diagonal.

*There is an even number of intersections and by symmetry, there is an even number of intersections out of diagonal. So there is even number of intersections on the diagonal. So there are at least two of them: $f(x_3) = x_3$, $f(x_4) = x_4$

*$x_1<x_2$ and $f(x_1) > f(x_2)$, so $f$ is decreasing (on the whole, by 1).

*$x_3<x_4$ and $f(x_3) < f(x_4)$, so $f$ is increasing.


Contradiction :-)
A: Assume that $f$ is a continuous and invertible real function with a connected domain.   Then $f$ is either strictly decreasing or strictly increasing over its domain (or it would assume some value twice, and would not be one-to-one).  Consider these two cases:


*

*$f$ is decreasing.  Then, since $f(x)-x$ is also decreasing, $f$ can't cross $y=x$ more than once.  If it never crosses $y=x$, then it lies entirely above or below that line, and its inverse lies entirely on the other side; they never meet, and the theorem holds vacuously.  If, on the other hand, it crosses $y=x$ exactly once, then the total number of intersections between $f$ and $f^{-1}$ is odd, since off-diagonal intersections come in pairs.  The theorem holds in this case too.

*$f$ is increasing.  Then it cannot intersect $f^{-1}$ at any point off the line $y=x$.  Suppose it did, at a pair of points $(x,y)$ and $(y,x)$ with $x<y$: then we would have $f(x)=y > x=f(y)$, contradicting the fact that $f$ is increasing.  In this case, then, all intersections are on $y=x$, and the theorem holds.
Since the theorem holds whether $f$ is increasing or decreasing, it is true in general.
A: Your question, as it currently stands definitely does not hold, as you do not specify the domain, continuity or conditions on inverse. This answer does define an $f$ on the whole real line that has a proper inverse.
Consider the function:
$$
\begin{align}
f(x) = 
\begin{cases}
1 &\text{if $x = 0$}\\
0 &\text{if $x = 1$}\\
x^2 &\text{if $x > 0$ and $x\neq 1$}\\
-2x &\text{if $x < 0$}
\end{cases}
\end{align}
$$
which has inverse
$$
\begin{align}
f^{-1}(y) = 
\begin{cases}
1 &\text{if $y = 0$}\\
0 &\text{if $y = 1$}\\
\sqrt{x} &\text{if $x > 0$ and $x\neq 1$}\\
-\tfrac{x}{2} &\text{if $x < 0$}
\end{cases}
\end{align}
$$
and so $f(x) = f^{-1}(y)$ only at $x = 0$ and $x = 1$, but at these points $f(x) \neq x$.
A: For $f$ and $f^{-1}$ to intersect there must be a point(s) $(x,f(x)) = (x,f^{-1}(x))$. Assume $x \neq f(x)$. Because $f$ and $f^{-1}$ are symmetric around y=x then there must be a corresponding point(s) $(f(x),x) = (f^{-1}(x),x)$.
The line defined by the points $(x_0,f(x_0)$ and $(f(x_),x_0)$ is $$
(y-f(x_0) = \frac{f(x_0)-x_0}{x_0-f(x_0)}(x-x_0)
$$
and simplified becomes $$
y=-x+x_0+f(x_0)
$$
This line, with a slope of -1, is not parallel with $y=x$, therefore must cross $y=x$. Because the two points are symmetric around $y=x$, they must lie on different sides of $y=x$. By the intermediate value theorem both $f$ and $f^{-1}$ must cross $y=x$.   
Hence there is at least one point of $f$ on $y=x$
