In the course of trying to invert a particularly nasty hormeomorphism candidate, I decided to look for theorems that can tell me when an inverse is continuous without actually having to invert. I have discovered what appears to me to be some sort of a conflict or redundancy between the definition of a homeomorphism and a theorem. The definition of a homeomorphism I'm using is that f is a homeomorphism if

  1. $f$ is a bijection
  2. $f$ is continuous
  3. $f$ has a continuous inverse.

The theorem I have found is:

If f is continuous and injective on an interval (a,b) then $f^{-1}$ is also continuous.

I am working with $f:A \subset \mathbb{R} \to \mathbb{R}$ with the usual topology on $A$ and $\mathbb{R}$.

My question is if part 1 of the definition is true why isn't part 3 redundant?

My guess is that the theorem applies to subsets of $\mathbb R$ and that the definition is more general and allows for functions of the type $g:X \to Y$ where $X$ and $Y$ have different topologies. But that is just a guess because I'm also aware of the unit circle as not being homeomorphic to an interval. Another guess I have is that because the theorem came from a section on monotone functions, that the circle example fails the theorem due to a lack monotonicity of the homeomorphism candidate function. These are just my guesses.

For what it's worth, I am trying to show that $(0,1)$ is homeomorphic to the real line using a function other than the usual examples, but that is not my question.

  • $\begingroup$ Your guess is right $\endgroup$ Jul 11, 2016 at 4:59
  • $\begingroup$ Consider $f:[0,2\pi)\rightarrow \mathbb{C}, \ f(t)=\exp(it)$ as the typical example. The map $f:[0,2\pi)]\rightarrow f([0,2\pi))$ is bijective and continuous, but the inverse is not. So for the "found" theorem it is crucial that $(a,b)$ is open.... $\endgroup$
    – Alex
    Jul 11, 2016 at 5:04

2 Answers 2


We want the definition of a homeomorphism to be applicable in any case where there are topologies involved$^*$. To give an example of a function that is bijective and continuous without continuous inverse, consider the map $$ \rho:[0,2\pi)\to S^1\subset\mathbb{C}$$ $$ \rho:\theta\mapsto e^{i\theta}.$$ To be clear, $S^1$ denotes the unit circle in the complex plane. This map is a bijection, and it is continuous, but its inverse is not continuous.

This is because points that wrap around the circle to meet each other at $(1,0)$ are torn apart by the inverse map $-$ so to speak.

$*$ Here assume the topologies are the usual ones on the half-open interval and $S^1$.

  • $\begingroup$ Thanks. Could you explain a little more why the inverse is not continuous in your example? $\endgroup$
    – jeo15
    Jul 11, 2016 at 5:39
  • 1
    $\begingroup$ To avoid being overly formal, think about it this way: a function is continuous if points that are near each other in the domain are sent to points near each other in the image. If we wrap the interval around the circle in this manner, we can do this continuously. If we try to invert this map, it will involve peeling the interval off of the circle. We have to start peeling at a certain point; $(1,0)$ to be precise. The points on the circle on either side of $(1,0)$ are very close together, but yet their images under $\rho^{-1}$ end up on opposite sides of the interval. (Try drawing it). $\endgroup$ Jul 11, 2016 at 5:43

Not a direct answer. I am not sure, but this might be interesting for you anyhow, so I took the liberty. It is about a continuous injective map that appears to be a homeomorphism under extra conditions that are not very demanding.

Theorem: Let $f:K\to H$ be a continuous function where $K$ is compact and $H$ is Hausdorff. If $f$ is injective on $E\subseteq K$ and the sets $f\left[E\right]$ and $f\left[\overline{E}-E\right]$ are disjoint then the restriction $f:E\rightarrow f\left[E\right]$ is a homeomorphism.

Proof: A set closed in (compact) $K$ is compact, so $f$ will send it to a compact set in (Hausdorff space) $H$, in which every compact set is closed. This makes $f$ a closed map. It is enough to prove that $f$ sends a set $A$ closed in $E$ to a set closed in $f\left[E\right]$. We have $A=E\cap F$ for some closed $F$. If $F$ 'works' then so does $\overline{E}\cap F$, so we can choose here for an $F$ with $F\subseteq\overline{E}$. If $z\in f\left[E\right]\cap f\left[F\right]$ then $x\in E$ and $y\in F$ exist with $z=f\left(x\right)=f\left(y\right)$. If $x\neq y$ then the injectivity of $f\upharpoonleft E$ implies that $y\notin E$ and consequently $y\in\overline{E}-E$. This however leads to the conclusion that $z$ belongs to the empty intersection of $f\left[E\right]$ and $f\left[\overline{E}-E\right]$. This contradiction shows that $x=y$ and consequently $z\in f\left[A\right]$. Proved is now that $f\left[A\right]=f\left[E\right]\cap f\left[F\right]$ with $f\left[F\right]$ closed and consequently $f\left[A\right]=f\left[E\right]\cap f\left[F\right]$ closed in $f\left[E\right]$.

Unfortunately the disjointness of $f\left[E\right]$ and $f\left[\overline{E}-E\right]$ cannot be missed as condition.


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