The Interval with Its Endpoints Identified
When the two ends of an interval are joined, points near the left endpoint and points near the right endpoint become neighbors of the same point. The result should be a circle, but a picture does not establish that the topologies agree. We will define the quotient topology on the identified interval and prove that the natural map from this quotient to the circle is a homeomorphism.
Let \(I=[0,1]\) have its usual subspace topology from \(\mathbb{R}\). Define an equivalence relation \(\sim\) on \(I\) by declaring \(s\sim t\) when either \(s=t\), or \(\{s,t\}=\{0,1\}\). Thus the class of either endpoint is \([0]=[1]=\{0,1\}\), and every other class is a singleton. Write \(Q=I/{\sim}\), and let \(q:I\to Q\) send each point to its equivalence class.
Let \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\), with the subspace topology inherited from the Euclidean topology on \(\mathbb{R}^2\). Consider the map \(g:I\to S^1\) given by \(g(t)=(\cos(2\pi t),\sin(2\pi t))\). It is continuous because its coordinate functions are continuous. Also \(g(0)=g(1)=(1,0)\), so \(g\) takes equal values on equivalent points.
Proof. Define \(\phi:Q\to S^1\) by \(\phi([t])=g(t)\). This is well-defined because equivalent points have equal \(g\)-values. It is surjective: every point of \(S^1\) has the form \((\cos\theta,\sin\theta)\) for some \(\theta\in[0,2\pi]\), and then it equals \(g(\theta/(2\pi))\).
To check injectivity, suppose \(\phi([s])=\phi([t])\). Then \(g(s)=g(t)\). Using the cosine difference identity, \(\cos(2\pi(s-t))=\cos(2\pi s)\cos(2\pi t)+\sin(2\pi s)\sin(2\pi t)=1\). Thus \(s-t\) is an integer. Since \(s,t\in[0,1]\), their difference lies in \([-1,1]\), so \(s-t\) is \(-1\), \(0\), or \(1\). If it is \(0\), then \(s=t\); if it is \(1\) or \(-1\), the two points are \(0\) and \(1\). In every case \([s]=[t]\). Therefore \(\phi\) is a bijection.
For continuity, let \(U\) be open in \(S^1\). The definition of \(\phi\) gives \(q^{-1}(\phi^{-1}(U))=g^{-1}(U)\), which is open in \(I\) because \(g\) is continuous. By the definition of the quotient topology, \(\phi^{-1}(U)\) is open in \(Q\). Hence \(\phi\) is continuous.
It remains to prove that this continuous bijection is a homeomorphism; continuity and bijectivity alone do not suffice. By the Heine–Borel theorem, \(I\) is compact. Since \(q\) is continuous and surjective, \(Q\) is compact: for any open cover of \(Q\), its preimages under \(q\) form an open cover of \(I\); a finite subcover of \(I\) gives a finite subcover of \(Q\).
The space \(S^1\) is Hausdorff. Indeed, distinct points \(p,r\in S^1\) have Euclidean distance \(d=\|p-r\|>0\); the open balls in \(\mathbb{R}^2\) of radius \(d/3\) centered at \(p\) and \(r\) are disjoint, and their intersections with \(S^1\) are disjoint open neighborhoods in \(S^1\).
We use the fact that a compact subset \(K\) of a Hausdorff space \(Y\) is closed. To verify it, fix \(y\in Y\setminus K\). For each \(k\in K\), choose disjoint open neighborhoods \(U_k\) of \(k\) and \(V_k\) of \(y\), using the Hausdorff property. The sets \(U_k\) cover \(K\), so compactness gives a finite subcover \(U_{k_1},\ldots,U_{k_n}\). The open neighborhood \(V_{k_1}\cap\cdots\cap V_{k_n}\) of \(y\) misses every \(U_{k_i}\), and therefore misses \(K\). Every point outside \(K\) thus has an open neighborhood disjoint from \(K\), so \(Y\setminus K\) is open and \(K\) is closed.
Now let \(C\) be closed in \(Q\). Its complement is open, so \(C\) is compact: an open cover of \(C\), together with \(Q\setminus C\), covers the compact space \(Q\), and a finite subcover restricts to a finite cover of \(C\). The continuous image \(\phi(C)\) is compact, and the fact just proved makes \(\phi(C)\) closed in \(S^1\). Thus \(\phi\) is a closed bijection. For any closed \(C\subseteq Q\), the preimage of \(C\) under \(\phi^{-1}\) is \(\phi(C)\), which is closed; hence \(\phi^{-1}\) is continuous. Therefore \(\phi\) is a homeomorphism. \(\square\)
What the Quotient Topology Does at the Join
The theorem proves more than a match between the points of \(Q\) and the points of the circle. It proves that a subset of \(Q\) is open precisely when its full preimage under \(q\) is open in the interval. The condition is important near the identified point: an open neighborhood there must include a portion of the interval near each endpoint.
Worked Example: An Open Arc Through the Identified Point
Consider \(U=\{(x,y)\in S^1:x>1/2\}\), an open subset of \(S^1\) containing \((1,0)\). To find its preimage, we solve \(\cos(2\pi t)>1/2\) for \(t\in[0,1]\). On this interval the inequality holds when \(t\in[0,1/6)\) or \(t\in(5/6,1]\): the boundary angles are \(\pi/3\) and \(5\pi/3\), corresponding to \(t=1/6\) and \(t=5/6\), and the inequality is strict, so those two boundary points are excluded.
Thus \(g^{-1}(U)=[0,1/6)\cup(5/6,1]\). This set is open in the subspace \(I\): it is the intersection of \(I\) with \((-\!1/6,1/6)\cup(5/6,7/6)\). It is saturated because it contains both \(0\) and \(1\). Consequently its image \(W=\phi^{-1}(U)\) is open in \(Q\), by the quotient-topology criterion.
Check: \(q^{-1}(W)=[0,1/6)\cup(5/6,1]\), which is open in \(I\), and \(\phi(W)=U\). The endpoints represent one point, so the two pieces of the preimage together form a single neighborhood of that point.
Worked Example: Identifying the Ends of a Different Interval
Take \([2,5]\) with its usual subspace topology, and identify \(2\) with \(5\), leaving all other points in singleton classes. The affine map \(h:[2,5]\to I\) given by \(h(t)=(t-2)/3\) is a homeomorphism: it is continuous, has continuous inverse \(h^{-1}(u)=2+3u\), and maps \(2\) to \(0\) and \(5\) to \(1\). It therefore matches the endpoint equivalence classes on the two intervals.
The circle map on this interval is \(G(t)=(\cos(2\pi(t-2)/3),\sin(2\pi(t-2)/3))\). At \(t=2\), its value is \((\cos 0,\sin 0)=(1,0)\); at \(t=5\), it is \((\cos 2\pi,\sin 2\pi)=(1,0)\). At the midpoint \(t=7/2\), the normalized parameter is \(((7/2)-2)/3=(3/2)/3=1/2\), so \(G(7/2)=(\cos\pi,\sin\pi)=(-1,0)\).
Since \(h\) carries the endpoint classes and the quotient topology to those of \(I/{\sim}\), the homeomorphism already proved gives a homeomorphism from this new quotient to \(S^1\). The midpoint check also confirms that the middle of the interval maps to the point opposite the identified endpoint.
Worked Example: A Neighborhood Away from the Join
Consider the open subset \(V=\{(x,y)\in S^1:x<-1/2\}\), centered at \((-1,0)\). Its preimage under \(g\) is determined by \(\cos(2\pi t)<-1/2\). The cosine is below \(-1/2\) for angles strictly between \(2\pi/3\) and \(4\pi/3\); dividing these bounds by \(2\pi\) gives \(1/3<t<2/3\). Therefore \(g^{-1}(V)=(1/3,2/3)\).
This is open in \(I\) and contains neither endpoint, so it is saturated: all of its points are singleton classes. Hence \(\phi^{-1}(V)\) is open in the quotient. Unlike the neighborhood in the first example, this one does not cross the join, and its interval preimage is one ordinary open interval.
Check: \(g(1/3)=(\cos(2\pi/3),\sin(2\pi/3))=(-1/2,\sqrt{3}/2)\), and \(g(2/3)=(\cos(4\pi/3),\sin(4\pi/3))=(-1/2,-\sqrt{3}/2)\). Both boundary points have first coordinate exactly \(-1/2\), so the strict inequality defining \(V\) excludes both, as does the computed preimage.
A Common Pitfall: Forgetting the Quotient Topology
A frequent error is to imagine that the identified point has only one interval-side neighborhood. For the arc \(U\) in the first example, using \((0,1/6)\cup(5/6,1)\) as its preimage is a wrong calculation: it omits \(0\) and \(1\), even though both map to \((1,0)\in U\). The correction is \(g^{-1}(U)=[0,1/6)\cup(5/6,1]\). The endpoint pieces are open relative to \(I\), and together they express the two-sided neighborhood at the identified point.
Another tempting shortcut is to assume every continuous bijection is a homeomorphism. Here the compactness of \(Q\) and the Hausdorff property of \(S^1\) supplied the missing argument. The need for a hypothesis can be seen on a two-point set \(\{a,b\}\): give the domain the discrete topology and the codomain the indiscrete topology \(\{\varnothing,\{a,b\}\}\). The identity map from the discrete space to the indiscrete space is a continuous bijection, because the only open sets in the codomain have open preimages. Its inverse is not continuous: the preimage of the domain-open set \(\{a\}\) is \(\{a\}\), which is not open in the indiscrete codomain. Thus the bijection is not a homeomorphism.
In the endpoint-identification construction, the quotient topology is exactly what makes the identification compatible with openness. The induced bijection sends the quotient class of the endpoints to \((1,0)\), and compactness and Hausdorffness then verify that no open-set information is lost in either direction.
Check Your Understanding
Use the quotient criterion and the map \(g(t)=(\cos(2\pi t),\sin(2\pi t))\) to answer these questions.
- Which points of \(I\) have the same image under \(g\), and why do they define exactly the classes of \(\sim\)?
- For \(U=\{(x,y)\in S^1:y>0\}\), compute \(g^{-1}(U)\). Is it saturated and open in \(I\)?
- Explain where compactness of \(Q\) and Hausdorffness of \(S^1\) enter the proof that \(\phi\) is a homeomorphism.
- Why does the set \((0,1/6)\cup(5/6,1)\) fail to be the preimage of the arc \(x>1/2\)?
- For an interval \([a,b]\) with \(a<b\), what normalized parameter replaces \(t\) in the circle map, and what values does it take at the two endpoints?