From an Open Cover to a Fundamental Group
The fundamental group records loops based at a chosen point, up to homotopy while keeping the basepoint fixed. Its definition is local to loops, but a space may be easier to understand as the union of two open pieces. The Seifert–van Kampen theorem says exactly how the loop information on those pieces and their overlap determines the group of the union.
For a space \(X\) and basepoint \(x_0\in X\), a loop is a continuous map \(\alpha:I\to X\), where \(I=[0,1]\), with \(\alpha(0)=\alpha(1)=x_0\). Two loops represent the same element of \(\pi_1(X,x_0)\) if they are homotopic through loops based at \(x_0\). The group operation is concatenation of loops, followed by passage to homotopy classes. In the theorem below, each space has the subspace topology inherited from \(X\).
Here the maps are induced by inclusions: for example, \(i_*\) sends a based loop in \(W\) to the same loop regarded as a loop in \(U\). The common basepoint matters because the fundamental group is based. Since \(x_0\in W\), it belongs to all four spaces, and these inclusion maps do not require a change-of-basepoint path.
The pushout notation describes a group built from the two groups while identifying the two images of every element from the overlap. More precisely, it has homomorphisms from \(\pi_1(U,x_0)\) and \(\pi_1(V,x_0)\) that agree after precomposition with \(i_*\) and \(j_*\). It is universal with this property: any other group receiving homomorphisms from those two groups that agree on the overlap receives a unique homomorphism from the pushout compatible with both. This is the free product with amalgamation discussed in the earlier algebraic material.
The Presentation Form
The same conclusion can be written using generators and relations. Suppose presentations are available: \(\pi_1(U,x_0)=\langle S_U\mid R_U\rangle\), \(\pi_1(V,x_0)=\langle S_V\mid R_V\rangle\), and \(\pi_1(W,x_0)=\langle S_W\mid R_W\rangle\). For each generator \(c\in S_W\), choose words \(u_c\) in the generators \(S_U\) and \(v_c\) in \(S_V\) representing \(i_*(c)\) and \(j_*(c)\), respectively. Then the theorem gives
The relations \(R_W\) do not need to be added separately: because \(i_*\) and \(j_*\) are homomorphisms, each relation in \(R_W\) already becomes the identity under both substitutions. The added relations say that the two images of each overlap generator are equal in the combined group.
Worked Applications
Worked Example: An Interval Covered by Two Intervals
Let \(X=(-1,2)\subset\mathbb{R}\), with the subspace topology from the standard topology on \(\mathbb{R}\). Take \(U=(-1,3/5)\) and \(V=(0,2)\), also with their subspace topologies. These are open in \(X\), and \(U\cup V=X\): points at most \(0\) lie in \(U\), points greater than \(0\) lie in \(V\), and every point of \(X\) is covered. Their intersection is \(W=(0,3/5)\), which is nonempty.
Each of \(U,V,W\) is an interval, so any two points in any one of them can be joined by the straight-line path between them. Thus all three are path-connected. Each interval is contractible, so its fundamental group is trivial: a contraction takes every based loop to the constant loop. Consequently, both maps from \(\pi_1(W,x_0)\) are maps from the trivial group to the trivial group, for any \(x_0\in W\). The pushout is trivial, and the theorem gives \(\pi_1(X,x_0)=\{1\}\).
Check: \(X\) itself is an interval and contracts to \(x_0\) by the straight-line homotopy \(H(x,t)=(1-t)x+tx_0\). This independently confirms that every based loop in \(X\) is null-homotopic.
Worked Example: A Disk-Like Union
Let \(X=U\cup V\subset\mathbb{R}^2\), where \(U=B((0,0),2)\) and \(V=B((1,0),2)\), with the Euclidean topology and the subspace topology on each set. Both \(U\) and \(V\) are open in \(X\), since each is the intersection of \(X\) with an open Euclidean ball. Their union is \(X\) by definition. Their intersection \(W=U\cap V\) is nonempty, since \((0,0)\) lies in both balls.
Each of \(U,V,W\) is convex: if two points lie in a ball, every point on the line segment joining them lies in that ball, and an intersection of convex sets is convex. A nonempty convex subset of \(\mathbb{R}^2\) is path-connected by straight-line paths. It is also contractible to any chosen point in it using the straight-line homotopy. Thus all three fundamental groups are trivial. Taking \(x_0=(0,0)\in W\), the theorem gives \(\pi_1(X,x_0)=\{1\}\).
Check: \(X\) is itself convex, because it is the union of these particular overlapping balls? That assertion does not follow from convexity of the pieces, so it is not needed. The theorem only uses the convexity and contractibility of \(U,V,W\), all of which have been verified directly.
Worked Example: Reading the Presentation Relations
Suppose a cover satisfying the theorem has groups with presentations \(\pi_1(U,x_0)=\langle a\mid a^4=1\rangle\), \(\pi_1(V,x_0)=\langle b\mid b^6=1\rangle\), and \(\pi_1(W,x_0)=\langle c\mid c^2=1\rangle\). Suppose the inclusion maps send \(c\) to \(a^2\) and \(b^3\), respectively. These assignments respect the overlap relation: \((a^2)^2=a^4=1\), and \((b^3)^2=b^6=1\), so both assignments define homomorphisms from the stated group of \(W\).
To form the presentation, retain the relations \(a^4=1\) and \(b^6=1\), and equate the two images of \(c\), giving \(a^2=b^3\). Therefore
Check: the overlap relation \(c^2=1\) maps to \(a^4=1\) on the \(U\) side and \(b^6=1\) on the \(V\) side. Both are already among the displayed relations, as required.
Why the Hypotheses Matter
The theorem is useful because it converts a topological cover into algebraic data: compute the groups of the pieces and the overlap, determine the inclusion homomorphisms, and then combine the presentations. But the hypotheses are not optional. In particular, if the overlap is disconnected, the stated version does not apply. The following common mistake shows why simply ignoring that condition can give a wrong answer.
Common pitfall. Cover \(S^1\), with its subspace topology from \(\mathbb{R}^2\), by the two open subsets \(U=S^1\setminus\{(1,0)\}\) and \(V=S^1\setminus\{(-1,0)\}\). Each is an open arc and is contractible, so each has trivial fundamental group. However, \(W=U\cap V\) consists of two disjoint open arcs, and is not path-connected. A tempting but invalid calculation is to apply the theorem using only the trivial groups of \(U\) and \(V\), concluding that \(\pi_1(S^1)\) is trivial. This contradicts the established computation \(\pi_1(S^1)\cong\mathbb{Z}\). The correction is that this cover fails the path-connected-intersection hypothesis, so the theorem stated here gives no conclusion for it. The disconnected-intersection case requires separate care.
| Data from the cover | Role in the theorem |
|---|---|
| \(X=U\cup V\), with \(U,V\) open | Allows loops in the union to be analyzed through the two pieces |
| \(U,V\) path-connected | Ensures the groups based at \(x_0\) capture the pieces |
| \(W=U\cap V\) nonempty and path-connected | Provides one based overlap group and the two inclusion maps |
| The maps \(i_*,j_*\) | Specify which elements must be identified in the pushout |
Check Your Understanding
For each question, state which open-cover hypotheses or induced maps are being used.
- Why must the chosen basepoint \(x_0\) lie in \(U\cap V\) for the theorem as stated?
- Suppose \(\pi_1(U,x_0)=\langle a\mid a^3=1\rangle\), \(\pi_1(V,x_0)=\langle b\mid b^4=1\rangle\), and a generator \(c\) of \(\pi_1(W,x_0)\) maps to \(a\) and \(b^2\). Write the resulting presentation, assuming these maps respect all relations of the overlap group.
- In the interval example, identify the contraction that shows each piece has trivial fundamental group.
- Which hypothesis fails for the stated two-arc cover of \(S^1\), and why does that prevent use of this version of the theorem?
- In the pushout conclusion, what equality must the two homomorphisms out of \(\pi_1(U,x_0)\) and \(\pi_1(V,x_0)\) satisfy on elements coming from \(\pi_1(W,x_0)\)?