= Two stereographic charts on the circle
{title2=$S^1$}
Take the <circle> $S^1=\{(x,y)\in\mathbb R^2:x^2+y^2=1\}$ with its <subspace topology>. Write $W=(-1,0)$ for its west point and $E=(1,0)$ for its east point. Two <coordinate charts> obtained by <stereographic projection> are
$$
U=S^1\setminus\{W\},\qquad \varphi(x,y)=t=\frac{y}{1+x},\qquad V=S^1\setminus\{E\},\qquad \psi(x,y)=u=\frac{y}{1-x}.
$$
Both take values in the entire <real line>. Geometrically, join $W$ to $(x,y)$ and intersect the resulting line with $x=0$: the height is $t$. Projecting from $E$ instead gives height $u$. Each projection omits its own starting point, but the other <coordinate chart> includes that point. Hence $U\cup V=S^1$.<Topology I: smooth manifolds, University of Toronto MAT1300HF, 2012/Week 1 course notes>[Topology I: smooth manifolds, Week 1, Example 1.6, equations (4)–(5), page 2; specialize the sphere charts to dimension one]{ref}
\Image[/circle-stereographic-charts-unfolding.png]
{title=Two stereographic charts cover the circle and unfold it onto two real lines}
To verify the <homeomorphisms>, solve the displayed equations for $(x,y)$. The <inverse functions> are
$$
\varphi^{-1}(t)=\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right),\qquad \psi^{-1}(u)=\left(\frac{u^2-1}{1+u^2},\frac{2u}{1+u^2}\right).
$$
Each formula lies on the <circle>, avoids its excluded point, and is <continuous> for every real argument. Substitution into the projection formulas returns $t$ or $u$; substitution in the other direction returns $(x,y)$. Thus these are <continuous> <bijections> with <continuous> <inverse functions>, as required for <coordinate charts>. Since the <Euclidean plane> is <Hausdorff> and <second countable>, the <subset> $S^1$ with its <subspace topology> is too. This checks all three conditions for a <topological manifold>, rather than just drawing a smooth-looking curve.
On the overlap $U\cap V$, both $x=\pm1$ are excluded, so $y\neq0$. Using $y^2=1-x^2$, the <topological transition map> is
$$
u=\frac{y}{1-x}=\frac{1+x}{y}=\frac1t,\qquad t\in\mathbb R\setminus\{0\}.
$$
For example, $(x,y)=(3/5,4/5)$ has $t=1/2$ and $u=2$. The two <connected components> of the overlap, the upper and lower arcs, correspond to the positive and negative parts of the punctured <real line>. The values $t=0$ and $u=0$ represent $E$ and $W$, respectively, and belong to only one chart each. The reciprocal coordinate change and its <inverse function> are <smooth functions> away from zero, so this <topological atlas> also supplies a <smooth atlas>.
Why can neither chart be enlarged to a single <coordinate chart> on the whole <circle>? The <circle> is a <closed set> and a <bounded set> in the <Euclidean plane>, hence <compact> by the <Heine-Borel theorem>. A global <coordinate chart> would give a <homeomorphism> to a nonempty <open set> $O\subseteq\mathbb R$. Its image would be <compact> and therefore <closed> in the <real line>. By <connectedness> of the <real line>, this forces $O=\mathbb R$, which is not <compact>. This is the one-dimensional instance of <no compact manifold has a single Euclidean chart>. Locally the <circle> is indistinguishable from the <real line>; globally <compactness> obstructs a single chart.
Back to article page
