The inverse of a continuous one-to-one function that is defined on a connected set is not always continuous

Question

I am trying to find a function $f:B \subset \Bbb R^n \rightarrow \Bbb R^m$ for $B$ a connected set that is continuous, one-to-one where $f^{-1} = f(B) \rightarrow B$ is discontinuous. The hint I have been given in my textbook is to choose $m>1$. I know that the image will be a connected set. The only idea I had is to send an angle to the unit circle $\theta \rightarrow (\cos(\theta),\sin(\theta)), \theta \in [0, 2\pi)$ and then take the inverse to be $f^{-1}:(x,y)\rightarrow (\arctan(y/x))$ but I feel that this function is continuous and not appropriate. Any hints appreciated.

I have seen a similar question here: Inverse function that takes connected set to non-connected set

I don't know if this is explicit enough for you, but what if you picture wrapping a line segment up to form a circle? The inverse function tears the circle apart, so it's not continuous. — CJD, Jun 02 '18 at 03:53
@JuliánAguirre I thought arctan was continuous on its domain? — IntegrateThis, Jun 02 '18 at 06:41
Oops yeah. Intuitively the discontinuity is at the point where the circle loops around. — Osama Ghani, Jun 02 '18 at 07:37
@OsamaGhani But the function isn't even defined at that point... Ok will work on this more tomorrow. — IntegrateThis, Jun 02 '18 at 08:01
You can find some examples here: Are continuous self-bijections of connected spaces homeomorphisms? Maybe also some of the posts linked there might be interesting for you. — Martin Sleziak, Jun 04 '18 at 13:20
And this also rather close to your question, since there is no requirement for the function to be bijective: Functions which are Continuous, but not Bicontinuous. And again, you might have a look also at some of the linked posts. BTW exactly the example with circle which you mentioned in your question is given there as one of the answers. — Martin Sleziak, Jun 04 '18 at 13:22
@IntegrateThis $f^{-1}$ is defined at $(1, 0)$: $f(0) = (1, 0)$, so $f^{-1}(1, 0) = 0$. — ploosu2, Jun 04 '18 at 13:50

score 1 · Accepted Answer · answered Jun 04 '18 at 13:46

The example you have is good! The inverse $f^{-1}$ isn't continous: It takes the point $A=(1, 0)$ to $0\in[0, 2\pi)$. But think of a point on the circle very close to the point $A$ lying on the half-plane $\{y<0\}$. It will be mapped to very close to $2\pi$. But $0$ and $2\pi$ are far apart; hence $f^{-1}$ isn't continous.

(You don't need to come with a formula for the inverse, just think how it maps.)

score 0 · Answer 2 · answered Jun 04 '18 at 12:17

I was working on this, managed to prove the opposite (If wrong, please point out the mistake)

I claim that $f^{-1}$ is continuous. Let's call it $g$ for convenience of notation, i.e. $g\circ f (x) = x$ $\forall x \in B$

Enough to show that $g^{-1}(U)$ is open in $f(B)$ if $U$ is open in $B$

Now, $g^{-1}(U) = \{ x : g(x) \in U , x \in f(B)\}$ and

$g^{-1}(U)$ is open $\iff$ $f(U)$ is open, as both are the same

Let $U' = f(U)$. Since $f$ is continuous, $U'$ is open $\iff$ $f^{-1}(U')$ is open

$f^{-1}(U') = \{ y : f(y) \in U' , y \in B\} = \{ y : f(y) \in f(U) , y \in B\}$. Now, as $f$ is one-one, $f(y) \in f(U) \iff y \in U$.

So, $f^{-1}(U') = U$.

Thus, $U' = f(U) = g^{-1}(U)$ is open $\iff$ $U$ is open.

P.S : Apologies if this proof is wrong.

You claim that: "$U'$ is open $\iff$ $f^{-1}(U')$ is open". This is true for quotient maps. But in general, it is not true for continuous maps. — Martin Sleziak, Jun 04 '18 at 13:32

The inverse of a continuous one-to-one function that is defined on a connected set is not always continuous

2 Answers2