To show that a mapping $$\phi:[0,1)\to S^1 $$ is not a homeomorphism.
The conditions for being a homeomorphism are firstly the mapping should be bijective and second the maps $\phi$ and $\phi^{-1}$ should be continuous.
The first condition of bijectivity is of one to one and onto. The circle could be thought of $$\phi(t)=(\cos2\pi t,\sin2\pi t)$$ How to show that $\phi ^{-1}$ $$\phi^{-1}(u,v)=\bigg(\frac{\cos^{-1} u}{2\pi},\frac{\sin^{-1} v}{2\pi}\bigg)$$ is not continuous?
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$\begingroup$By definition, a function $\phi:X\to Y$, between two topological spaces is a homeomorphism if it is bijective, continuous, with continuous inverse $\phi^{-1}:Y\to X$. Recall also, that for any bijective function we have: $$ \phi:\textrm{continuous } \Leftrightarrow \phi^{-1}:\textrm{open}, \ \ \textrm{ and } \ \ \phi:\textrm{open } \Leftrightarrow \phi^{-1}:\textrm{continuous} $$ so we can equivalently state that: a function $\phi:X\to Y$, between two topological spaces is a homeomorphism if it is bijective, continuous and open.
The function $\phi: [0,1) \rightarrow \mathcal{S}^{1}$, given by $\phi(t) = (\cos2 \pi t, \sin2 \pi t)$ is continuous and bijective but it is not open (equivalently: its inverse function $\phi^{-1}: \mathcal{S}^{1} \rightarrow [0,1)$ is not continuous):
Proof:To see that, let us recall that here, the space $Y=\mathcal{S}^{1}$ is equipped with its relative topology as a subset of $\mathbb{R}^2$ and the space $X=[0,1)$ is equipped with its relative topology as a subset of $\mathbb{R}$. Notice that in this relative topology, $[0,t)$ for $0<t<1$, is an open subset of $[0,1)$. Now, it is easy to see that the image $\phi\big([0,t)\big)\subseteq\mathcal{S}^1$ is not an open subset of $\mathcal{S}^1$ (in its relative topology). Therefore, $\phi$ is not open or equivalently $\phi^{-1}$ is not continuous.
So, $\phi$ is not a homeomorphism.
$\endgroup$ $\begingroup$The space $S^1$ is compact but $[0,1)$ is not compact, hence these spaces are not homeomorphic.
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