# Continuous functions, its inverse (if exists) and intersections graphically

I have a question regarding graphical intersection between a continuous function and its inverse (if exists).

Suppose $f$ is a real continuous function and $f^{-1}$ exists.

Can anyone assist in proving the following problem:

If $f$ and $f^{-1}$ intersect graphically at some points, then the points must lie on the line $y=x$.(*)

For here, I have some trouble regarding the meaning of "some" points.

If it is uncountable many points of intersection, $y=\frac{1}{x}$ is a counterexample.

What if it is countable (including finite many or countably infinite)?

Intuitively, (*) is true when we draw the functions pictorially, but can anyone provide hints or steps to the proof?

If the statement is wrong at the first place, how should we amend?

• What do you mean by the intersection of two functions? Do you mean the intersection of their graphs? If so, please edit that into the body of the question. – Gerry Myerson Sep 4 '14 at 13:00
• Surely there must be some condition on f, otherwise let f be a function that swaps two given reals and fixes all the others – Belgi Sep 4 '14 at 13:01
• Thank you for the answer. Perhaps I change the question to continuous functions. – Novice Sep 4 '14 at 13:03
• @Belgi I'm missing how that is a counter example. As far as I can tell, all points $a$ such that $f(a)=f^{-1}(a)$ are in $\{(x,x)\colon x\in \mathbb R\}$. – Git Gud Sep 4 '14 at 13:09
• @GitGud the point is (a, f(a)), say f swaps 0 and 1, then f agrees with its inverse for all reals, but does not coincide with y=x(excuse me for the lack of tex, I'm using a mobile phone) – Belgi Sep 4 '14 at 13:18

The graphical proof can be algebra-fied: Let's look at a certain $x=x_1$ where $f(x_1)=x_1$ i.e. $f(x)$ intersects the $y=x$ at that $x_1$. Now since they intersect we have $x_1-f(x_1)=0 \Rightarrow x_1=f(x_1)$. Now we take the inverse $f^{-1}$ of both sides: $f^{-1}(x_1)=x_1 \Rightarrow f^{-1}(x_1)-x_1=0$ which means $f^{-1}(x)$ also intersects the $y=x$ line at that $x_1$ which means $f(x)$ intersects $f^{-1}(x)$ on the line $y=x$.
An alternative solution would be to say that in order for an intersection to take place we want $f^{-1}(x)=f(x)$ and the $x$ which satisfies this is where the intersection happens. Taking the $f$ of both sides gives us $x=f(f(x))$ which is a functional equation. Now $f(x)=x$ is a solution to that functional equation which indicates that the intersections happen at $y=x$. The only issue is that I'm not sure how to prove that there are no other solutions to that functional equation.
• True, I tried to show that if $f(x)$ intersects $y=x$ then at that point we also have $f(x)$ intersecting $f^{-1}(x)$. – Sheheryar Zaidi Sep 4 '14 at 14:06
• Your second question is the crux. Pictorially it is evident, but $\frac{1}{x}$ is the spoiler. – Novice Sep 4 '14 at 14:09
• Such a function is an involution (I found on Wikipedia). Apparently, $y=-x$ is an involution. The function coincides with its inverse, yet the intersections is the line itself, not $y=x$. – Novice Sep 4 '14 at 14:14
• Which is why we probably cannot show that all intersections take place on the line $y=x$. Interesting problem I must say. – Sheheryar Zaidi Sep 4 '14 at 14:17