3

My conjecture is that a continuous function is invertible only if it is strictly monotonic.

I assume that this is true, because if a continuous function $f$ is not strictly monotonic, then there must be a point $b$ where for some point $a<b$, and for some point $c>b$, we have $f(a)<f(b)$ and $f(c)<f(b)$. Then because of the intermediate value theorem, there must be some point $d$, where $a<d<b$, and some point $e$, where $b<e<c$, such that $f(d)=f(e)$. (or the inequalities are flipped) In other words, if it is not strictly monontonic, then $f$ must go down after having gone up (or the other way around), and therefore there must be outputs that share the same input.

Is this correct?

user56834
  • 12,925

1 Answers1

1

For precisely the reasons you state, an arbitrary function $f:\mathbf{R}\to \mathbf{R}$ which is strictly monotonic will be injective. Indeed, if $f(x)=f(y)$, then we must have that $x=y$, else $x>y$ and $f(x)>f(y)$ or $x<y$ and $f(x)<f(y)$. Notice how this doesn't need continuity. This does give us left invertibility, i.e. there exists a function $g$ so that $g\circ f=\text{Id}$. We might ask, however, when we can get that our function is invertible in the stronger sense - i.e., when our function is a bijection.

If we promote our function to being continuous, by the Intermediate Value Theorem, we have surjectivity in some cases but not always. Suppose that $f$ is strictly increasing, since the case of strictly decreasing is basically the same argument. Suppose that for all $M\in \mathbf{R}_{>0},$ we can find an $x$ sufficiently large so that $f(x)>M$. Suppose, in addition, that we can find a $z$ sufficiently small (i.e. negative) so that $f(z)<-M$. Then, we know that $[-M,M]\subset f(\mathbf{R})$ for all $M\in \mathbf{R}_{>0}$, by the IVT. This implies that $f$ is surjective, and injective, and hence a bijection.

Notice how $f:\mathbf{R}\to \mathbf{R}$ can be continuous and strictly increasing without being surjective. Take, for example, $f:\mathbf{R}\to \mathbf{R}$ given by $f(x)=\arctan(x)$.

Alekos Robotis
  • 26,184