TITLE: Proof: if the graphs of $y=f(x)$ and $y=f^{-1}(x)$ intersect, they do so on the line $y=x$
QUESTION [6 upvotes]: This came out of a textbook problem, and as Lubin pointed out below, it's not actually true as originally stated. I'm guessing it should be restated as: If the graphs of $y=f(x)$ and $y=f^{-1}(x)$ intersect at some value(s) of $x$, but not at all values of $x$ in their respective domains ($f$ is not it's own inverse), then any points of intersection fall on the line $y=x$.
Or maybe it could just say: If the graphs of $y=f(x)$ and $y=f^{-1}(x)$ intersect at a finite number of points, then any points of intersection fall on the line $y=x$.
I'm sure its staring me right in the face. It's totally intuitive and I can see it on the graph, but when I try to prove it analytically I end up chasing my tail. Here's what I've got so far:
Suppose $f$ is an invertible function. Then it's necessarily one-to-one, so for any $a,b$ in the domain of $f$, it is true that $f(a)=f(b)$ if and only if $a=b$. The same is true of $f^{-1}$.
Now suppose a point $(a,b)$ lies on both $y=f(x)$ and $y=f^{-1}(x)$. Then $b=f(a)$, which implies $a=f^{-1}(b)$. Also, $b=f^{-1}(a)$, which implies $a=f(b)$.
So we know,
$$f(a)=f^{-1}(a)=b$$
and,
$$f(b)=f^{-1}(b)=a$$
and I'd think it would be relatively simple to combine these and prove that $a=b$, but I keep going in circles and proving stuff like $a=a$. I've tried making use of $a=f^{-1}(f(a))$, but still no dice. Any help would be much appreciated.

REPLY [2 votes]: The right statement should probably be:
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 a n 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...