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.

  • 4
    $\begingroup$ How about $f(x)=-1/x$? $\endgroup$
    – Lubin
    Mar 20 '13 at 18:13
  • $\begingroup$ Wow, good point. So it's a bogus premise from the start. @Lubin $\endgroup$
    – ivan
    Mar 20 '13 at 18:35
  • $\begingroup$ Well, it’s not completely bogus. The special property of $-1/x$ is that it’s an involution: it’s its own inverse. Maybe something interesting can be salvaged. $\endgroup$
    – Lubin
    Mar 20 '13 at 18:42
  • $\begingroup$ I don't understand: "don't intersect at every point". What do you mean? $\endgroup$
    – Julien
    Mar 20 '13 at 22:27
  • $\begingroup$ @julien - I mean they intersect at some value(s) of $x$, but not for every value of $x$ ($f$ is not its own inverse). I'll try to clarify that in the question. Thanks. $\endgroup$
    – ivan
    Mar 20 '13 at 22:54

Our intuition is biased towards continuous functions, especially on the real line, but you posed the question (and attempted to answer it) in an abstract, purely set-theoretical way. The story seems to be that your intuition is correct for continuous one-to-one functions defined on $\mathbb R$, and onto $\mathbb R$, but my example of $-1/x$ was continuous all right, but on the disconnected set $\mathbb R\setminus\{0\}$.

Let’s look first at some simple counterexamples. First the involution $1\mapsto2\mapsto1$ on the set $\{1,2\}$. The graph is $\bigl\lbrace(1,2),(2,1)\bigr\rbrace$, no points on the diagonal. Next, a noninvolution on a finite set, \begin{align} f&: 1\mapsto2\mapsto3\mapsto1\,,\,4\mapsto5\mapsto4\\ g=f^{-1}&:1\mapsto3\mapsto2\mapsto1\,,\,4\mapsto5\mapsto4 \end{align} on the set $\{1,2,3,4,5\}$. In words, $f$ permutes $1$, $2$, and $3$ cyclically, and interchanges $4$ and $5$. Again, no fixed points, but the graphs of the function and its inverse have $(4,5)$ in common.

Finally, an infinite example, defined on $\mathbb R$: \begin{align} \forall x\notin\mathbb Z\,,\, &f(x)=x+1\,,\,\text{but}\,\,\forall n\in\mathbb Z\,,\, f(n)=1-n\\ \forall x\notin\mathbb Z\,,\, &g(x)=x-1\,,\,\text{but}\,\,\forall n\in\mathbb Z\,,\, g(n)=1-n\,, \end{align} discontinuous at all integers. Again, I cooked it up so that $f$ was an involution on a subset.

Let’s try to show that a continuous surjective one-to-one function $f\colon\mathbb R\to\mathbb R$, with inverse $g$, has the property that if the graphs of $f$ and $g$ intersect, then $f$ has a fixed point $x_0$, i.e. $f(x_0)=x_0$ for some $x_0\in\mathbb R$. And we want to use only the most basic calculus-level mathematics. What we’re assuming about $f$ is that there is a pair of points $(\alpha,\beta)$ and $(\beta,\alpha)$ on the graph of $f$. In other words $f(f(\alpha))=\alpha$ for some $\alpha\in\mathbb R$.

In case $\alpha=\beta$, we’re done, so assume that $\alpha<\beta$. Then the function $f(x)-x$ is positive at $\alpha$ and negative at $\beta$, so must be zero somewhere between, by the Intermediate Value Theorem, and the proof is done.


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...


I don't think you can say much (unless maybe you assume that $f$ is increasing). A decreasing function can easily have points of period 2, for example, $f(x) = 1 - x^2$ on $[0,1]$, $a = 0 , b = 1$, without being an involution.

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    – leo
    Mar 20 '13 at 20:04

I see what you mean by chasing your tail. I think the key to this is to go back to a definition of what the inverse function is. The way the inverse of a function is defined on Wikipeida is that $f(x)=y \Leftrightarrow f^{-1}(y)=x$. So, we have two implications here, $f(x)=y$ implies that $f^{-1}(y)=x$ and $f^{-1}(y)=x$ implies that $f(x)=y$.

Using this definition makes the "run around" go away.

Given, $y_1$ = $f(x_1)$ = $f^{-1}(x_1)$

We have $y_1 = f(x_1)$ given. By the definition we then know that $f^{-1}(y_1) = x_1$

Substituting that back into the original equation, $x_1 = f^-1(y_1) = f^-1(x_1)$. But, don't forget, that equation is still equal to $y_1$ because $y_1 = f^-1(x_1)$. So, $y_1 = x_1$.

The point $(y_1,x_1)$ certainly lies on the line $y=x$.

One last thing, I'm assuming here that the inverse function exists for $x_1$ and $y_1$. So, if there is some sort of domain restriction on $f^{-1}$ to make it an actual function, then $x_1$ can't be out of that domain.

  • $\begingroup$ Is there a mistake here, or am I missing a step?: "Substituting that back into the original equation, x1 = f^-1(y1) = f^-1(x1)." I see that $x_1=f^{-1}(y_1)$, but how do you then equate them to $f^{-1}(x_1)$? $\endgroup$
    – ivan
    Mar 20 '13 at 22:51
  • $\begingroup$ @ivan, looks like there is a mistake here. I'll go hang my head in shame now, and maybe I'll eat some worms. $\endgroup$
    – beeflobill
    Mar 20 '13 at 23:23
  • $\begingroup$ dang! I was hoping you'd cracked it. No shame in trying, it's as much as I've been able to do :) $\endgroup$
    – ivan
    Mar 20 '13 at 23:28

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