The Bouquet and Its Loop Generators
A bouquet of circles is a graph formed by identifying one chosen point on each circle to a single common point. Loops can travel around any circle, then return to the common point and travel around another. The Seifert–van Kampen theorem makes this description precise: each circle contributes a generator, and the generators have no relations among them.
Let \(S^1\) have its standard topology, with a chosen point \(v\). For \(n\geq 1\), define the \(n\)-circle bouquet \(X_n\) to be the quotient of \(n\) copies of \(S^1\) obtained by identifying all their chosen points. Give \(X_n\) the quotient topology, and write \(v\) for the identified point. For \(n=0\), let \(X_0=\{v\}\). A loop based at \(v\) is a continuous map \(\gamma:I\to X_n\) with \(\gamma(0)=\gamma(1)=v\). Its fundamental-group element is its homotopy class, where the homotopy keeps both endpoints at \(v\); multiplication is concatenation of loops.
On the \(i\)th circle choose an orientation and let \(a_i\) denote the loop that travels once around that circle in the chosen direction. Traveling once in the opposite direction represents \(a_i^{-1}\). A finite sequence of such traversals gives a word in the symbols \(a_i\) and \(a_i^{-1}\). Adjacent symbols \(a_i a_i^{-1}\) or \(a_i^{-1}a_i\) can be cancelled. A word with no such adjacent inverse pair is called reduced.
The point of the computation is not merely to find loops that generate the group. We must also show that no additional relations hold: in particular, a nonempty reduced word must not represent the identity.
An Open Cover for Van Kampen
We use the Seifert–van Kampen theorem in the following standard form, established in the earlier material on that theorem. If a space \(X\) is the union of two open, path-connected sets \(U,V\), their intersection is path-connected, and the basepoint lies in the intersection, then the fundamental group of \(X\) is the free product of the groups of \(U\) and \(V\), with the inclusions identifying the images of loops in the intersection. In particular, if \(U\cap V\) is contractible, then its fundamental group is trivial and \(\pi_1(X,v)\cong\pi_1(U,v)*\pi_1(V,v)\). We also use the standard earlier computation \(\pi_1(S^1,v)\cong\mathbb{Z}\), with a once-around loop as generator.
To apply the theorem to \(X_n\), represent each circle by a parameter \(t\in[0,1]\) with \(0\) and \(1\) identified at \(v\). Choose \(0<\delta<1/2\). The short open arc around \(v\) on a circle consists of the parameters \(t\) with \(0\leq t<\delta\) or \(1-\delta<t\leq1\), with the endpoints \(0\) and \(1\) understood as the same point.
For \(n\geq2\), let \(U\) contain the first \(n-1\) circles in full and only this short open arc on the \(n\)th circle. Let \(V\) contain the \(n\)th circle in full and only the short open arc on each of the first \(n-1\) circles. These sets are open in the quotient topology: their preimages in each circle are either the whole circle or an open arc around its identified point. They cover \(X_n\), since \(U\) contains each of the first \(n-1\) circles and \(V\) contains the last circle. Their intersection is the common open star consisting of the short arc on every circle. It is path-connected and contracts to \(v\) by moving each point along its arc toward \(v\).
The set \(U\) deformation retracts onto the bouquet of the first \(n-1\) circles: on the short arcs of the last circle, move each point toward \(v\), while fixing the first \(n-1\) circles. The set \(V\) deformation retracts onto the \(n\)th circle by the corresponding contraction on the short arcs of the first \(n-1\) circles. These contractions are continuous in the quotient topology because they are continuous on each circle and agree at the identified point. Both sets are therefore path-connected, and their fundamental groups are those of the indicated subbouquets.
Proof. We use induction on \(n\). For \(n=0\), every loop in the one-point space \(X_0\) is the constant loop, so its fundamental group is trivial, which is the free group on no generators. For \(n=1\), \(X_1=S^1\), and the standard circle computation gives \(\pi_1(X_1,v)\cong\mathbb{Z}=F(a_1)\).
Now let \(n\geq2\), and suppose the result holds for \(n-1\). Use the open sets \(U,V\) just constructed. Their intersection is contractible, so the Seifert–van Kampen theorem gives \(\pi_1(X_n,v)\cong\pi_1(U,v)*\pi_1(V,v)\). The deformation retractions identify \(\pi_1(U,v)\) with \(\pi_1(X_{n-1},v)\cong F(a_1,\ldots,a_{n-1})\), and identify \(\pi_1(V,v)\) with \(\pi_1(S^1,v)\cong\mathbb{Z}\), generated by the \(n\)th circle loop \(a_n\). Thus \(\pi_1(X_n,v)\cong F(a_1,\ldots,a_{n-1})*\mathbb{Z}\).
To identify this free product, recall its universal property: homomorphisms from the factors into any group \(H\) extend uniquely to a homomorphism from their free product into \(H\). A homomorphism from \(\mathbb{Z}\) to \(H\) is uniquely specified by the image of \(1\), and a homomorphism from \(F(a_1,\ldots,a_{n-1})\) is uniquely specified by the images of its generators. Therefore a homomorphism from \(F(a_1,\ldots,a_{n-1})*\mathbb{Z}\) to \(H\) is uniquely specified by an arbitrary choice of \(n\) elements of \(H\). This is exactly the defining universal property of \(F(a_1,\ldots,a_n)\). The isomorphism sends each generator to the corresponding circle loop, proving the claim. \(\square\)
Reading and Reducing Loop Words
Worked Example: One Circle
For \(X_1\), identify the group and the class of a loop that travels twice around the circle in the chosen direction and then once in the opposite direction.
The standard computation \(\pi_1(S^1,v)\cong\mathbb{Z}\) assigns \(1\) to one positive traversal and \(-1\) to one negative traversal. The loop in question has integer \(1+1+(-1)=1\), so its class is \(a_1\). In word notation, \(a_1a_1a_1^{-1}\) cancels to \(a_1\).
Check: the resulting word has one positive traversal, so its integer is \(1\), agreeing with the sum of the three traversal integers.
Worked Example: Two Circles and a Reduced Loop
For \(X_2\), determine the fundamental group and reduce the loop that travels around the first circle, then the second circle, then the first circle in reverse, then the second circle in reverse.
The theorem gives \(\pi_1(X_2,v)\cong F(a_1,a_2)\). The loop is represented by \(a_1a_2a_1^{-1}a_2^{-1}\). Each adjacent pair has different circle labels, so there is no adjacent inverse pair to cancel. The word is already reduced and is not the identity, since it is nonempty.
Check: applying the cancellation rule to each of the three adjacent pairs gives no cancellation: \(a_1\) is not \(a_2^{-1}\), \(a_2\) is not \(a_1\), and \(a_1^{-1}\) is not \(a_2^{-1}\). Thus the reduced word remains \(a_1a_2a_1^{-1}a_2^{-1}\).
Worked Example: Cancelling Traversals in a Four-Circle Bouquet
In \(X_4\), reduce the loop with word \(a_1a_2a_2^{-1}a_3a_1^{-1}a_1a_4a_4^{-1}\).
First cancel the adjacent pair \(a_2a_2^{-1}\). The word becomes \(a_1a_3a_1^{-1}a_1a_4a_4^{-1}\). Next cancel the adjacent pair \(a_1^{-1}a_1\), leaving \(a_1a_3a_4a_4^{-1}\). Finally cancel \(a_4a_4^{-1}\), leaving \(a_1a_3\). This last word is reduced, because its two letters have different circle labels.
Check: the cancellations removed one positive and one negative traversal of circle 2, one negative and one positive traversal of circle 1, and one positive and one negative traversal of circle 4. The remaining traversals are precisely one positive traversal of circle 1 and one positive traversal of circle 3, giving \(a_1a_3\).
Why the Free Group, Not a Direct Product?
A common mistake is to treat the circles as independent coordinates and conclude that the group is \(\mathbb{Z}^n\), where multiplication is commutative. The loops share a basepoint, so they can be traversed in any order, but that does not make different orders homotopic. For two circles, the words \(a_1a_2\) and \(a_2a_1\) are both reduced and are different words in \(F(a_1,a_2)\). By the reduced-word description, they represent different group elements. Thus the group is not the direct product \(\mathbb{Z}^2\), whose elements commute.
The cover explains why there are no extra relations: at each induction step the two pieces meet in a contractible set, so there are no nontrivial loops in the intersection whose images would be identified. The resulting group is a free product, and each new circle contributes a new independent generator. This is useful whenever a space can be divided into pieces with connected, simply connected overlap: the overlap determines whether generators from different pieces acquire relations.
Check Your Understanding
Use the open-cover computation and reduced-word description to answer the following.
- What is \(\pi_1(X_0,v)\), and why does the case \(n=0\) have no generators?
- In \(X_3\), write the word for a loop that travels once negatively around circle 2, then positively around circle 1.
- Reduce \(a_1a_2a_2^{-1}a_3a_3^{-1}a_1^{-1}\) and state whether the resulting word represents the identity.
- Why is the intersection in the van Kampen cover contractible, and which hypothesis of the theorem does this verify?
- Explain why \(a_1a_2\) and \(a_2a_1\) are distinct elements of \(\pi_1(X_2,v)\).