Show that the function $f: [0, 2 \pi) \to S^1$ defined by $f(t)=e^{it}=(\cos t, \sin t)$ is not homeomorphic.
Answer:
This function is bijective and continuous. But $f^{-1}$ is not continuous at $(1,0)$.
But how to show that $f^{-1}$ is not continuous at $(1,0)$.
From definition of continuous function, we call $f:X \to Y$ a continuous function if for $x \in X$ there are neighborhoods $U$ in $X$ and $V$ in $Y$ with $x \in U$ and $f(x) \in V$, then $f(U) \subset V$.
Following this definition,
We have $f^{-1}(1,0)=0 $.
Let $U$ be any open set containing $(1,0)$ and $V$ be any open set containing $0$, we have to show that $f^{-1}(U) \not\subset V$.
But I can not show it.
Can someone show the work with some details that $f^{-1}$ is not continuous at $(1,0)$ ?
There is a explanation in Wikipedia as below:
The function $f^{-1}$ is not continuous at the point $( 1 , 0 )$ because although $ f^{-1}$ maps $( 1 , 0 )$ to $0$, any neighbourhood of this point also includes points that the function maps close to $2 \pi $, but the points it maps to numbers in between lie outside the neighborhood.
But I could not understand it clearly. How they choosing neighborhood?
$A = [0,2\pi)$ and $C = S^1$ are not homeomorphic because:
1. $A$ is not compact, $C$ is compact.
2. $A$ has many cut points, $C$ has none.
3. $A$ is contractable, $C$ is not.
4. $A$ is simply connected, $C$ is not.
As you were showing if $f:A \to C$ is homeomorphism,
then $f^{-1}$ is not continuous. It cannot be continuous
since continuous images of compact spaces are compact and
since $C$ is compact, $A$ would have to be compact which it isn't.