What Makes a Cover Useful?
Seifert–van Kampen computations begin by splitting a space into pieces. The pieces must overlap in a way that lets information about loops pass between them. A cover is not suitable merely because its sets are easy to describe: the sets must be open, and the pieces and their overlap must satisfy specific path-connectedness conditions.
Throughout, a path in a space \(X\) from \(x\) to \(y\) means a continuous map \(\gamma:[0,1]\to X\) with \(\gamma(0)=x\) and \(\gamma(1)=y\). Here \([0,1]\) has its usual subspace topology from \(\mathbb{R}\). The space \(X\) is path-connected if every pair of its points can be joined by a path. A family of open sets covers \(X\) if their union is \(X\).
These conditions concern the topology of the sets themselves. Openness is relative to \(X\): a set open in \(X\) need not be open in some larger space in which \(X\) sits. Also, the fact that \(U\) and \(V\) each connect their own points does not guarantee that their intersection connects its points. That last condition must be checked separately.
The Seifert–van Kampen theorem is the standard result that uses such a cover to relate the fundamental group of \(X\) to the fundamental groups of the two pieces and their overlap. Its full algebraic statement and proof are treated in the later material on the Seifert–van Kampen theorem. Here the task is to construct and verify the cover, not to carry out the group calculation.
A Useful Path-Connectedness Test
Proof. Choose \(c\in A\cap B\). Let \(x,y\in A\cup B\). If both points lie in \(A\), a path in \(A\) joins them; if both lie in \(B\), a path in \(B\) joins them. In the remaining case, after interchanging \(x\) and \(y\) if necessary, suppose \(x\in A\) and \(y\in B\). Path-connectedness gives a path \(\alpha:[0,1]\to A\) from \(x\) to \(c\), and a path \(\beta:[0,1]\to B\) from \(c\) to \(y\). Define \(\gamma:[0,1]\to A\cup B\) by \(\gamma(t)=\alpha(2t)\) for \(0\le t\le 1/2\) and \(\gamma(t)=\beta(2t-1)\) for \(1/2\le t\le1\). Both formulas give \(c\) at \(t=1/2\), so the paths join continuously there. Thus \(\gamma\) joins \(x\) to \(y\). Every pair of points in \(A\cup B\) can be joined by a path, proving the claim. \(\square\)
This proposition is a practical construction tool. If each piece is assembled from path-connected parts, it is enough to check that the parts meet. For the theorem’s overlap condition, however, one must apply the same kind of check to \(U\cap V\) itself.
Worked Example: Overlapping Relative Open Intervals
Let \(X=[0,2]\) have the subspace topology inherited from the standard topology on \(\mathbb{R}\). Consider \(U=[0,5/4)\) and \(V=(3/4,2]\).
Openness. We have \(U=(-1,5/4)\cap X\), so \(U\) is open in \(X\). Also \(V=(3/4,3)\cap X\), so \(V\) is open in \(X\). The square brackets at \(0\) and \(2\) do not prevent relative openness: those endpoints belong to \(X\), and each set is the intersection of \(X\) with an open set of \(\mathbb{R}\).
Cover and overlap. The endpoint \(0\) belongs to \(U\), the endpoint \(2\) belongs to \(V\), and the intervals overlap. More precisely, \(U\cup V=[0,2]=X\), and \(U\cap V=(3/4,5/4)\), since points in the intersection must satisfy both \(x\ge0\), \(x<5/4\), \(x>3/4\), and \(x\le2\).
Path-connectedness. For any \(a,b\in U\), the path \(\gamma(t)=(1-t)a+tb\) stays in \(U\): it lies between \(a\) and \(b\), both of which are at least \(0\) and strictly less than \(5/4\). The same linear path stays in \(V\) for any \(a,b\in V\), since both are strictly greater than \(3/4\) and at most \(2\). The same argument applies to \(U\cap V=(3/4,5/4)\). Thus all three sets are path-connected. For example, \(1\in U\cap V\) is an available basepoint. This is a suitable cover.
Check: \(3/4<1<5/4\), so \(1\) really is in the overlap. The cover includes both endpoints of \(X\), and its intersection is nonempty.
A Circle Cover That Fails the Overlap Test
Worked Example: Removing Two Points from the Circle
Give \(S^1=\{(\cos\theta,\sin\theta):\theta\in\mathbb{R}\}\) the subspace topology inherited from \(\mathbb{R}^2\). Let \(p=(1,0)\), \(q=(-1,0)\), \(U=S^1\setminus\{p\}\), and \(V=S^1\setminus\{q\}\).
Singletons are closed in \(\mathbb{R}^2\), so \(\{p\}\) and \(\{q\}\) are closed in \(S^1\); hence \(U\) and \(V\) are open in \(S^1\). Since \(p\ne q\), no point is removed from both sets, and \(U\cup V=S^1\).
Each piece is path-connected: removing one point from the circle leaves an open arc, which can be parametrized continuously by angles in an open interval of length \(2\pi\). But \(U\cap V=S^1\setminus\{p,q\}\) consists of the upper and lower open semicircles. These are disjoint, nonempty relatively open subsets of the intersection, and together they exhaust it. No path in the intersection can join a point on one semicircle to a point on the other: under the angle parametrization, any such path would have to pass through one of the two removed points. Thus the intersection is not path-connected.
This pair covers the circle with two open, path-connected sets, but it is not a suitable two-set cover for the theorem. The failure is exactly the overlap condition; having a cover and having path-connected pieces are not enough.
Check: \(p\notin U\) but \(p\in V\), and \(q\in U\) but \(q\notin V\), confirming that neither removed point is missing from the union. On the other hand, both \(p\) and \(q\) are missing from \(U\cap V\), separating its two semicircles.
A Suitable Cover of a Wedge
For a space built by identifying points, openness is often easiest to verify from the definition of the quotient topology. If \(q:Z\to X\) is a quotient map, a subset \(W\subseteq X\) is open precisely when \(q^{-1}(W)\) is open in \(Z\). We use this test in the next construction.
Worked Example: Two Circles Joined at One Point
Let \(C_1\) and \(C_2\) be copies of \(S^1\), with the subspace topology just specified. Form \(X=C_1\vee C_2\) by identifying their points \((1,0)\), and give \(X\) the quotient topology from the disjoint union \(C_1\bigsqcup C_2\). Write \(b\) for the identified point. In each circle let \(J_i\) be the open arc around \((1,0)\) parametrized by \(-1/8<t<1/8\), using \((\cos(2\pi t),\sin(2\pi t))\). Define \(U\) to contain all of \(C_1\) and the arc \(J_2\); define \(V\) to contain \(J_1\) and all of \(C_2\), with the circles understood as their images in the quotient.
Openness. Let \(q:C_1\bigsqcup C_2\to X\) be the quotient map. The preimage \(q^{-1}(U)\) is \(C_1\bigsqcup J_2\), open in the disjoint union because \(C_1\) is open as a component and \(J_2\) is open in \(C_2\). Therefore \(U\) is open in \(X\). Similarly \(q^{-1}(V)=J_1\bigsqcup C_2\) is open, so \(V\) is open.
The cover and overlap. The sets \(U\) and \(V\) together contain both full circles, so \(U\cup V=X\). Their intersection consists of \(J_1\) and \(J_2\) joined at \(b\): its preimage is \(J_1\bigsqcup J_2\). Each arc is path-connected, and they share the identified point \(b\), so the proposition shows that the intersection is path-connected. The same proposition shows that \(U\) is path-connected: the circle \(C_1\) and the arc \(J_2\) meet at \(b\). It also shows that \(V\) is path-connected, since \(J_1\) and \(C_2\) meet at \(b\).
Conclusion. This is a suitable cover, and \(b\in U\cap V\) is a basepoint for a Seifert–van Kampen computation. The two pieces retain the two circles, while their overlap connects through a small arc around the joining point.
Check: each chosen arc includes \(t=0\), so it contains \(b\); its endpoints \(t=-1/8\) and \(t=1/8\) are excluded. The overlap is therefore nonempty, and its two arc portions meet at \(b\), rather than being separated as in the circle example.
A Cover-Checking Routine
For a proposed pair \(U,V\), check each requirement directly rather than inferring one from another. In particular, use the topology of \(X\) to establish openness, and inspect the intersection as its own space. The following order keeps the checks separate:
Show that every point of \(X\) lies in \(U\) or \(V\). A point omitted from both sets means there is no cover.
Use the subspace definition when \(X\) sits inside another space, or the quotient definition when \(X\) is a quotient. Openness in a larger ambient space is not the test by itself.
Give paths, or decompose each set into path-connected pieces and check that the pieces meet.
Compute \(U\cap V\), establish that it is nonempty, and show that paths between its points remain inside it.
Select \(x_0\in U\cap V\); a point outside the overlap cannot serve as the common basepoint for the standard two-set setup.
Common pitfall. In the circle example, it is wrong to argue that the overlap is path-connected because \(U\) and \(V\) are each path-connected. Their intersection is \(S^1\setminus\{p,q\}\), not either piece, and it has two components. The correction is to calculate the intersection first and then test paths within that intersection. If it fails the test, choose a different cover or use a theorem whose hypotheses fit the situation; do not apply the two-set version as though the missing condition were automatic.
A useful cover is therefore a piece of mathematical design. It balances two aims: the pieces should be simple enough to analyze, and the overlap should be connected enough for the theorem’s hypotheses. A clean verification of each condition is what makes the subsequent group computation legitimate.
Check Your Understanding
For each proposal, identify which cover conditions hold and justify any failure.
- In \(X=[0,3]\) with its subspace topology from \(\mathbb{R}\), test \(U=[0,2)\) and \(V=(1,3]\). Is this a suitable cover?
- For the circle cover \(U=S^1\setminus\{p\}\), \(V=S^1\setminus\{q\}\) with distinct \(p,q\), which hypothesis fails?
- In the wedge construction, why is it not enough to show that \(q^{-1}(U)\) is open in just one circle component?
- State and prove the condition in the path-connected-union proposition that allows paths to be joined through a common point.
- Why must a basepoint for the standard two-set setup lie in \(U\cap V\)?