Proposition: Let $f:\left [ 0,1 \right ) \rightarrow S^{1}$ be given by $f\left ( x \right )=x$. Then, f is not a homeomorphism.
I do understand the proof. What I really mean is, I do not understand the theorems and motivation to which the author is making at each juncture of his proof.
Proof:
Consider $\left [ 0,1 \right )$ as a subset of both $\left [ 0,1 \right )$ and $S^{1}$. In the first case, $\left [ 0,1 \right )=\left ( \frac{-1}{2},\frac{1}{2} \right )\cap \left [ 0,1 \right )$ so $\left [ 0,\frac{1}{2} \right )$ is open in the subspace topology on $\left [ 0,1 \right )$. On the other hand, let $\pi:\mathbb{R}\rightarrow S^{1}$ be the projection map defined $\pi\left ( x \right )=\left [ x \right ]$, then $\pi^{-1}\left [ 0,1 \right )=\cup _{n\in\mathbb{Z}}\left [ n,n+\frac{1}{2} \right )$ which is clearly not open in the standard topology on $\mathbb{R}$. Hence, $\left [ 0,\frac{1}{2} \right )$ is not open in the quotient topology on $S^{1}$
A step-by-step clear explanation would be greatly appreciated.
Thanks in advance.