From an Overlap to a Free Product
The Seifert–van Kampen theorem computes the fundamental group of a space assembled from open pieces. In general, loops in the overlap identify elements from the two pieces, producing a free product with amalgamation. When the overlap is simply connected, it has no nontrivial loops to identify, and the amalgamation reduces to an ordinary free product. The basepoint must lie in the overlap so that the two inclusion maps compare loops based at the same point.
For a space \(X\) and a point \(x_0\in X\), \(\pi_1(X,x_0)\) is the group of homotopy classes of loops based at \(x_0\), with multiplication given by concatenation. A space is path-connected if every two points can be joined by a continuous path. It is simply connected if it is path-connected and every loop is homotopic, keeping its basepoint fixed, to the constant loop.
Here \(G*H\) denotes the free product of groups \(G\) and \(H\). It comes with homomorphisms from \(G\) and \(H\) and has this defining property: for any group \(K\) and homomorphisms \(f:G\to K\) and \(g:H\to K\), there is a unique homomorphism \(G*H\to K\) whose restrictions to the two factors are \(f\) and \(g\). The formula says that, under the stated hypotheses, the loops contributed by \(U\) and \(V\) combine without extra identifications.
Why Simple Connectedness Removes the Amalgamation
The Seifert–van Kampen theorem, established in the earlier material on the fundamental group and the Seifert–van Kampen theorem, gives the appropriate general calculation. In the path-connected open-cover case with \(x_0\in W\), it identifies \(\pi_1(X,x_0)\) with the free product of the two groups, subject to the relations identifying the images of every element of \(\pi_1(W,x_0)\) under the two inclusion maps.
The hypotheses have distinct jobs. Openness makes the open-cover form of van Kampen applicable. Path connectedness lets loops and basepoints in each piece be compared using paths; choosing \(x_0\in W\) makes the two inclusion-induced homomorphisms based at the same point. Simple connectedness makes the group of the overlap trivial, so no nontrivial overlap loops impose relations. In examples, these conditions must be checked before applying the formula.
Worked Punctured-Plane Covers
Worked Example: Removing Two Points
Let \(X=\mathbb{R}^2\setminus\{(-2,0),(2,0)\}\), with the subspace topology inherited from the Euclidean plane. Set \[ U=X\cap\{(x,y):x<1\},\qquad V=X\cap\{(x,y):x>-1\}. \] We compute \(\pi_1(X,x_0)\) using \(x_0=(0,1)\).
Check the cover and intersection. Both half-planes are open in \(\mathbb{R}^2\), so \(U\) and \(V\) are open in \(X\). Every point of \(\mathbb{R}^2\) has first coordinate either less than \(1\) or greater than \(-1\); hence \(U\cup V=X\). Their intersection is \[ W=U\cap V=\{(x,y):-1<x<1\}. \] Neither removed point lies in this strip. It is path-connected, and it is contractible: the homotopy \((x,y,t)\mapsto ((1-t)x,(1-t)y)\), for \(0\le t\le1\), stays in the strip and contracts it to \((0,0)\). Thus \(W\) is simply connected, and \(x_0\in W\).
Compute the groups of the pieces. The set \(U\) is the half-plane \(x<1\) with the single point \((-2,0)\) removed. The coordinate change \((x,y)\mapsto(-\ln(1-x),y)\) is a homeomorphism from the half-plane \(x<1\) to \(\mathbb{R}^2\); it sends the removed point to \((-\ln 3,0)\). So \(U\) is homeomorphic to a plane with one point removed. Likewise, \((x,y)\mapsto(\ln(x+1),y)\) carries the half-plane \(x>-1\) homeomorphically to \(\mathbb{R}^2\) and sends \((2,0)\) to \((\ln 3,0)\). Thus \(V\) is also homeomorphic to a plane with one point removed. Both are path-connected. The standard calculation that a plane with one point removed has fundamental group \(\mathbb{Z}\) follows from its deformation retraction onto a circle and the fundamental group of the circle. Therefore \(\pi_1(U,x_0)\cong\mathbb{Z}\) and \(\pi_1(V,x_0)\cong\mathbb{Z}\).
Apply the formula. All the hypotheses have been checked, so \[ \pi_1(X,x_0)\cong\mathbb{Z}*\mathbb{Z}. \] The two factors record loops around the two removed points, one contributed by each half-plane.
Worked Example: Removing Three Points
Let \(X=\mathbb{R}^2\setminus\{(-4,0),(-3,0),(4,0)\}\). Take \[ U=X\cap\{(x,y):x<2\},\qquad V=X\cap\{(x,y):x>-2\}. \] Their union is \(X\), since the two half-planes together cover \(\mathbb{R}^2\). Their intersection is the strip \(-2<x<2\), which contains none of the three removed points. As in the previous example, this strip is path-connected and contracts to \((0,0)\) by straight-line contraction. Choose \(x_0=(0,1)\) in the strip.
The set \(U\) is a half-plane with two points removed. The homeomorphism \((x,y)\mapsto(-\ln(2-x),y)\) sends the half-plane \(x<2\) to the plane and carries its two removed points to two distinct points. By the two-point calculation above and a homeomorphism of the plane carrying any pair of distinct points to any other such pair, \(\pi_1(U,x_0)\cong\mathbb{Z}*\mathbb{Z}\). The set \(V\) is a half-plane with one point removed, so the same coordinate-change argument as before gives \(\pi_1(V,x_0)\cong\mathbb{Z}\). Both pieces are path-connected: for a finite set of removed points, any two remaining points can be joined by two line segments through an intermediate point chosen off the finitely many lines that would make either segment hit a removed point.
The free-product formula now gives \[ \pi_1(X,x_0)\cong(\mathbb{Z}*\mathbb{Z})*\mathbb{Z} =\mathbb{Z}*\mathbb{Z}*\mathbb{Z}. \] The last expression just groups the three free factors together; it introduces no additional relation.
Worked Example: A Simply Connected Piece Adds No Factor
Let \(X=\mathbb{R}^2\setminus\{(0,0)\}\). Define \(U=\{(x,y):x<-1\}\) and \(V=X\cap\{(x,y):x>-2\}\). These are open subsets of \(X\), and \(U\cup V=X\): every point with \(x\le -1\) lies in \(U\), while every point with \(x>-1\) lies in \(V\). The overlap is the strip \(-2<x<-1\), which is open, path-connected, and contracts to \((-3/2,0)\) by straight-line contraction. Choose \(x_0=(-3/2,1)\) in that strip.
The half-plane \(U\) is convex, hence path-connected and contractible, so its fundamental group is trivial. The set \(V\) is a half-plane with the origin removed. The coordinate map \((x,y)\mapsto\ln(x+2)\) in the first coordinate, leaving \(y\) unchanged, identifies that half-plane with \(\mathbb{R}^2\); it takes the removed point to a point of the plane. Thus \(\pi_1(V,x_0)\cong\mathbb{Z}\). The formula gives \[ \pi_1(X,x_0)\cong\{1\}*\mathbb{Z}\cong\mathbb{Z}. \] Indeed, a homomorphism out of \(\{1\}*\mathbb{Z}\) is determined solely by its homomorphism from \(\mathbb{Z}\), so the trivial factor contributes no generators or relations. This agrees with the standard computation for a punctured plane.
A Hypothesis to Keep: Connected Overlap
Simple connectedness of the overlap includes path connectedness by definition. It is not enough for each open piece to be simply connected: the overlap must also be path-connected for this formula. For a concrete warning, give \(S^1\) its subspace topology from \(\mathbb{R}^2\), and set \(U=S^1\setminus\{(-1,0)\}\), \(V=S^1\setminus\{(1,0)\}\). These are open in \(S^1\), cover \(S^1\), and each is homeomorphic to an open interval, so each is contractible. But \(U\cap V\) is the circle with two points removed, which consists of two disjoint open arcs and is not path-connected. The circle has fundamental group \(\mathbb{Z}\), not the trivial group that one would incorrectly obtain by taking the free product of the two trivial groups. The cover does not satisfy the connected-overlap hypothesis, so that conclusion is not licensed.
Common pitfall. A simply connected overlap does not mean that the fundamental group of the union is the direct product of the groups of the pieces. The theorem gives a free product. For example, the two-point punctured plane above has group \(\mathbb{Z}*\mathbb{Z}\), with no relation equating the two generators. A direct product would impose the additional relation that the generators commute. That relation has not been supplied by the simply connected overlap; the overlap supplies no nontrivial loop relations at all.
Check Your Understanding
For each question, identify which hypotheses are relevant before computing.
- In the two-point punctured-plane example, why does the overlap contain no removed point, and why is the chosen basepoint in it?
- Suppose \(X=U\cup V\) satisfies the theorem and \(\pi_1(U,x_0)\) is trivial. What group does the formula give?
- Why can the two-point punctured-plane computation not be replaced by a direct-product calculation using the stated theorem?
- For the circle cover in the warning, which hypothesis fails, and what is the fundamental group of the overlap?
- In the three-point punctured-plane example, which piece contributes two copies of \(\mathbb{Z}\), and which contributes one?