Why the Intersection Matters
The Seifert–van Kampen theorem turns a space covered by two open sets into a computation with the fundamental groups of those sets and their intersection. A key hypothesis in its standard two-set form is that the intersection is path-connected. The circle provides a small, exact example of what can go wrong when this condition is dropped: two open arcs cover it, both arcs are contractible, and their intersection has two components. The circle nevertheless has a nontrivial fundamental group.
We use the standard circle \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\), with its subspace topology from the Euclidean plane. A path in a space \(X\) from \(x\) to \(y\) is a continuous map \(\gamma:[0,1]\to X\) with \(\gamma(0)=x\) and \(\gamma(1)=y\). A space is path-connected if every two of its points can be joined by a path. A loop based at \(x_0\) is a path with both endpoints \(x_0\); the fundamental group \(\pi_1(X,x_0)\) consists of based loops up to based homotopy, with concatenation as its operation.
The result is not a formula that can be applied after simply omitting a failed hypothesis. We will verify exactly which hypothesis fails in a cover of \(S^1\), and then exhibit the loop that a naive computation would miss. We use the standard calculation \(\pi_1(S^1,i)\cong\mathbb{Z}\), established in the earlier material on the fundamental group of the circle.
A Cover by Two Open Arcs
Let \(1=(1,0)\), \(-1=(-1,0)\), \(i=(0,1)\), and \(-i=(0,-1)\) be points of \(S^1\). Define \(U=S^1\setminus\{1\}\) and \(V=S^1\setminus\{-1\}\). These are open in \(S^1\): a singleton is closed in the metric space \(S^1\), so its complement is open. They cover \(S^1\), since no point can be both \(1\) and \(-1\).
Worked Example: Verify the Two-Arc Cover
Use the parametrization \(c(t)=(\cos t,\sin t)\). The map \(c\) on \(0<t<2\pi\) parametrizes \(U\), and on \(-\pi<t<\pi\) it parametrizes \(V\). Each is an open arc: it is the circle with one point removed. Each is path-connected, since the corresponding interval is path-connected and the parametrization is a homeomorphism onto the arc.
The intersection removes both \(1\) and \(-1\). Its points have nonzero second coordinate, so \(U\cap V=C_+\cup C_-\), where \(C_+=\{(x,y)\in S^1:y>0\}\) and \(C_-=\{(x,y)\in S^1:y<0\}\). These are disjoint, nonempty, relatively open subsets whose union is the intersection. Thus the intersection is disconnected. Each piece is itself an open arc: \(C_+\) is parametrized by \(0<t<\pi\), and \(C_-\) by \(\pi<t<2\pi\); in particular, each piece is path-connected.
To check that these are exactly the connected components, a connected subset of \(U\cap V\) cannot meet both \(C_+\) and \(C_-\). If it did, its intersections with those two relatively open sets would separate it into two disjoint nonempty relatively open subsets, contradicting connectedness. Since each \(C_+\) and \(C_-\) is connected, they are the two components.
Check: \(i\in C_+\) and \(-i\in C_-\), so both components are nonempty. Also \(U\cup V=S^1\), while \(U\cap V\) has exactly two components. Thus the cover is genuinely by two open, path-connected arcs, but it fails the path-connected-intersection hypothesis.
Each of \(U\) and \(V\) is contractible. For example, in the coordinate \(t\in(0,2\pi)\) on \(U\), fix \(t_0=\pi\) and define \(H(t,s)=(1-s)t+st_0\) for \(0\leq s\leq1\). The value remains in \((0,2\pi)\), and this continuously contracts the coordinate interval to \(t_0\). Transporting this homotopy through the parametrization contracts \(U\); the same construction in the coordinate interval \((-\pi,\pi)\) contracts \(V\). In particular, their fundamental groups are trivial. One can also see this directly: a loop in either coordinate interval contracts to its basepoint by linear interpolation, with its endpoints fixed throughout.
The Naive Computation and Its Failure
If the standard theorem applied, then the group calculation would use the fundamental groups of \(U\), \(V\), and \(U\cap V\), with maps from the latter group into the first two. The problem is not that these groups are difficult to find. The problem is that the standard theorem does not apply: \(U\cap V\) is disconnected.
Worked Example: The Invalid Free-Product Calculation
Both \(U\) and \(V\) are contractible, so \(\pi_1(U,i)\) and \(\pi_1(V,i)\) are trivial groups. The point \(i\) lies in \(C_+\), so \(\pi_1(U\cap V,i)\) refers to loops based in the component containing \(i\); it is also trivial, since that component is an open arc.
A tempting but invalid calculation is to insert these trivial groups into the usual formula and conclude that \(\pi_1(S^1,i)\) is trivial. The resulting algebraic expression would indeed be trivial: a free product of two trivial groups, with a trivial group amalgamated, is trivial. But this is not a valid application of the theorem, because the theorem requires all of \(U\cap V\) to be path-connected, whereas \(U\cap V=C_+\cup C_-\).
The contradiction is concrete. The established calculation for the circle gives \(\pi_1(S^1,i)\cong\mathbb{Z}\), not the trivial group. The mistake is treating a theorem with a missing hypothesis as if its conclusion still followed. The based group \(\pi_1(U\cap V,i)\) sees loops based in \(C_+\); it does not by itself record how the separate component \(C_-\) supplies another way to pass between the two covering arcs.
Check: the trivial group has one element, while \(\mathbb{Z}\) has distinct elements \(0\) and \(1\). Therefore the naive answer cannot be isomorphic to the actual fundamental group.
How the Second Component Supplies a Loop
The two components of the intersection contain two useful points: \(i\in C_+\) and \(-i\in C_-\). There is a path from \(i\) to \(-i\) lying in \(U\), and a different path between the same points lying in \(V\). Following the first path and returning along the second forms a loop in the union. The loop exists because the two intersection components give distinct locations at which paths can pass between \(U\) and \(V\).
Worked Example: Recover the Missing Loop
Define \(\alpha:[0,1]\to U\) by \(\alpha(t)=(\cos(\frac{\pi}{2}+\pi t),\sin(\frac{\pi}{2}+\pi t))\). Its angle runs from \(\pi/2\) to \(3\pi/2\), so \(\alpha(0)=i\) and \(\alpha(1)=-i\). The path avoids \(1\), hence lies in \(U\).
Define \(\beta:[0,1]\to V\) by \(\beta(t)=(\cos(\frac{3\pi}{2}+\pi t),\sin(\frac{3\pi}{2}+\pi t))\). Its endpoints are \(\beta(0)=-i\) and \(\beta(1)=i\). It avoids \(-1\), hence lies in \(V\). Concatenate \(\alpha\) with \(\beta\). For \(0\leq t\leq1/2\), the concatenated loop has angle \(\pi/2+2\pi t\); for \(1/2\leq t\leq1\), it has angle \(3\pi/2+2\pi(t-1/2)\). Both expressions equal \(\pi/2+2\pi t\). Thus the whole loop is \(t\mapsto(\cos(\pi/2+2\pi t),\sin(\pi/2+2\pi t))\): it starts at \(i\), makes one full turn, and returns to \(i\).
This is a generator of \(\pi_1(S^1,i)\cong\mathbb{Z}\), using the standard identification in which one counterclockwise turn represents \(1\). The check is visible in the angle: the final angle is \(\pi/2+2\pi\), exactly one full turn beyond the initial angle \(\pi/2\). The two paths each lie in a contractible set, but their concatenation need not contract in the union.
There is also a useful bookkeeping picture for this particular cover. Make one vertex for each covering set, \(U\) and \(V\), and one edge for each component of their intersection. Here there are two vertices and two edges, one for \(C_+\) and one for \(C_-\). The two distinct edges provide two distinct ways to switch between the covering sets; going one way across one component and returning across the other produces a cycle. The explicit loop above realizes that cycle in \(S^1\).
This picture explains the role of multiple intersection components in this example; it is not a replacement for the hypotheses or conclusion of the two-set theorem. More generally, when an intersection is disconnected, a group based at a single point of it does not record all the ways its components connect the covering sets. A groupoid form of van Kampen keeps track of multiple basepoints and paths between them. For the present purpose, the concrete loop is enough to see what the naive group calculation discarded.
A Common Pitfall
The central pitfall is to reason, “Both covering sets are contractible, so their union must have trivial fundamental group.” Contractibility of the individual sets says that loops contained entirely in either one can be contracted there. It says nothing by itself about loops that travel through one set from one intersection component to another and return through the other set.
In the circle cover, the path \(\alpha\) travels inside \(U\) from the upper component to the lower component; \(\beta\) returns inside \(V\). Neither path alone is a loop, so the triviality of \(\pi_1(U)\) and \(\pi_1(V)\) does not contract their concatenation. The disconnected intersection is precisely why the ordinary two-set free-product-with-amalgamation formula cannot be used to conclude otherwise.
Check Your Understanding
For each question, distinguish what the cover directly shows from what the standard theorem permits you to conclude.
- For \(U=S^1\setminus\{1\}\) and \(V=S^1\setminus\{-1\}\), identify the two components of \(U\cap V\) and give one point in each.
- Which hypothesis of the standard two-set Seifert–van Kampen theorem fails for this cover?
- Why does the based group \(\pi_1(U\cap V,i)\) only refer to one component of the intersection?
- Describe paths from \(i\) to \(-i\) in \(U\) and from \(-i\) to \(i\) in \(V\). What loop results when they are concatenated?
- Explain why knowing that \(U\) and \(V\) are contractible does not imply that their union is simply connected.