= Solution
Use three compactness facts: a closed subset of a compact space is compact; a continuous image of a compact space is compact; and a compact subset of a <Hausdorff space> is closed. For a closed $C\subseteq X$, these facts show that $f(C)$ is closed in $Y$. Therefore the bijection $f$ is a closed map, and $(f^{-1})^{-1}(C)=f(C)$ proves <continuity> of its inverse. \b[Thus $f$ is a <homeomorphism>.] This proves the <compact-to-Hausdorff continuous bijection theorem> without requiring $X$ itself to be Hausdorff.
Let $q:[0,1]\to[0,1]/\sim$ be the quotient map and $h(t)=(\cos2\pi t,\sin2\pi t)$. The map $h$ is continuous and identifies precisely the equal parameters and the endpoint pair. Hence it induces a bijection $\bar h$ from the <quotient space> to the unit circle, with $h=\bar h\circ q$; the definition of the <quotient topology> makes $\bar h$ continuous. The quotient is compact as a continuous image of $[0,1]$, while the circle is Hausdorff as a subspace of $\mathbb R^2$. Applying the preceding theorem gives
$$
\boxed{[0,1]/\sim\ \cong S^1.}
$$
Back to article page
