Conflict between homeomorphism definition and continuity of inverse theorem 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


*

*$f$ is a bijection

*$f$ is continuous 

*$f$ 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.
 A: 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 $S^1$.
A: 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.
