Why a Wedge Produces a Free Product
When two connected surfaces are joined at one chosen point, a loop can travel through either surface and return to the joining point. The two surfaces share that point, but their other points and loops are not identified. The Seifert–van Kampen theorem captures this by making the fundamental group of the wedge a free product: loops from each surface retain their own relations, with no additional relations imposed between the two sets of loops.
All surfaces below have their usual topology and are based at specified points. For based spaces \((X,x_0)\) and \((Y,y_0)\), their wedge \(X\vee Y\) is the quotient of the disjoint union \(X\bigsqcup Y\) in which \(x_0\) and \(y_0\) are identified to one point. The wedge has the quotient topology: a set is open exactly when its preimage in each summand is open. The identified point is the basepoint of \(X\vee Y\).
We use the standard presentations for closed connected surfaces. The sphere \(S^2\) has trivial fundamental group. The closed orientable surface \(\Sigma_g\) of genus \(g\geq1\) has presentation with generators \(a_1,b_1,\ldots,a_g,b_g\) and the single relator \([a_1,b_1]\cdots[a_g,b_g]\), where \([a,b]=aba^{-1}b^{-1}\). The closed nonorientable surface \(N_k\) of genus \(k\geq1\) has generators \(c_1,\ldots,c_k\) and relator \(c_1^2\cdots c_k^2\). These surface presentations are standard results established by the polygon and van Kampen computations for surfaces.
The Wedge Theorem
Proof. Choose open disk neighborhoods \(D_x\subseteq X\) of \(x_0\) and \(D_y\subseteq Y\) of \(y_0\). In the wedge let \(U=X\cup D_y\) and \(V=D_x\cup Y\), where each union uses the identified basepoint. These sets are open: their preimages in the two summands are respectively \((X,D_y)\) and \((D_x,Y)\), all of which are open. They cover \(X\vee Y\), and \(U\cap V=D_x\vee D_y\).
The radial contraction of \(D_y\) to \(y_0\), leaving \(X\) fixed, descends through the quotient and gives a deformation retraction of \(U\) onto \(X\). The radial contraction of \(D_x\) gives a deformation retraction of \(V\) onto \(Y\). Contracting both disks to their centers contracts \(D_x\vee D_y\) to the wedge point, so \(U\cap V\) is path-connected and simply connected. The Seifert–van Kampen theorem for this open cover therefore identifies \(\pi_1(X\vee Y)\) with the free product of the groups of \(U\) and \(V\), since the intersection has trivial fundamental group. The deformation retractions identify those groups with \(\pi_1(X,x_0)\) and \(\pi_1(Y,y_0)\). This proves the claimed isomorphism.
For the presentation statement, take disjoint generator sets \(S_X,S_Y\) and relator sets \(R_X,R_Y\) presenting the two groups. The union presentation has generators \(S_X\cup S_Y\) and relators \(R_X\cup R_Y\). The algebraic reason it presents a free product is proved below. \(\square\)
The argument also applies to any finite wedge of path-connected closed surfaces: apply the two-summand result repeatedly. If a summand is \(S^2\), its group is trivial and contributes no generators or relators. The theorem concerns based wedges; it does not say that arbitrary surfaces joined along a circle have free-product groups.
Why the Union of Presentations Works
Proof. Let \(P=\langle S\cup T\mid R\cup Q\rangle\). The generator assignment from \(S\) to its elements in \(G\) and from \(T\) to its elements in \(H\) respects every relator in \(R\cup Q\). It therefore gives a homomorphism \(P\to G*H\). In the other direction, the inclusions of the generators into \(P\) respect \(R\) and \(Q\), so they induce homomorphisms \(G\to P\) and \(H\to P\). By the defining universal property of the free product, these induce a homomorphism \(G*H\to P\).
The composite \(P\to G*H\to P\) fixes every generator in \(S\cup T\), so it is the identity on \(P\). The composite \(G*H\to P\to G*H\) restricts to the identity on both \(G\) and \(H\); the universal property says it is the identity on \(G*H\). Thus the homomorphisms are inverse isomorphisms, proving the presentation claim. \(\square\)
This proposition is what makes the calculation systematic: retain each summand’s relators, and do not add a relator involving generators from different summands. The free product allows arbitrary alternating words from the two groups unless their own relators reduce them.
Worked Example: A Torus Wedged with a Projective Plane
Let \(T^2\) be the torus based at a point and let \(\mathbb{RP}^2=N_1\) be the projective plane based at a point. Compute \(\pi_1(T^2\vee\mathbb{RP}^2)\).
The torus presentation is \(\langle a,b\mid aba^{-1}b^{-1}\rangle\). The projective-plane presentation is \(\langle c\mid c^2\rangle\). Their generator sets are disjoint, so the wedge theorem and the union-presentation proposition give
Thus the relations are \(aba^{-1}b^{-1}=1\), equivalently \(ab=ba\), and \(c^2=1\). There is no relation requiring \(c\) to commute with \(a\) or \(b\). In free-product notation this group is \(\mathbb{Z}^2*C_2\), where \(C_2\) is the cyclic group of order two.
Check: the first two generators with their relator present precisely the torus group, and the third generator with \(c^2=1\) presents the projective-plane group. The union presentation has both original relators and no others, so the proposition verifies the free-product calculation.
Worked Example: Two Projective Planes
For \(\mathbb{RP}^2\vee\mathbb{RP}^2\), call the generators of the two summands \(c\) and \(d\). Each projective-plane factor has one generator whose square is the identity. The union presentation is therefore
For example, the word \(cdcd\) is a reduced alternating word: neither \(c^2\) nor \(d^2\) occurs in it, and no relation in the presentation cancels the adjacent letters \(cd\). In particular, the presentation does not include \(cd=dc\). The free-product description records that the only reductions arise from the relations within each individual factor.
Check: setting \(d\) equal to the identity leaves \(\langle c\mid c^2\rangle=C_2\), and setting \(c\) equal to the identity leaves \(\langle d\mid d^2\rangle=C_2\). These are exactly the two factor groups embedded by the free-product construction.
Worked Example: A Genus-Two Surface Wedged with a Klein Bottle
Let \(\Sigma_2\) be the closed orientable genus-two surface and let \(N_2\) be the closed nonorientable genus-two surface, also called the Klein bottle. The standard presentations are \(\pi_1(\Sigma_2)=\langle a_1,b_1,a_2,b_2\mid [a_1,b_1][a_2,b_2]\rangle\) and \(\pi_1(N_2)=\langle u,v\mid u^2v^2\rangle\). Use separate letters for the two summands, so their generator sets do not overlap. The wedge group is
Expanding the first relator gives \(a_1b_1a_1^{-1}b_1^{-1}a_2b_2a_2^{-1}b_2^{-1}=1\). The second remains \(uuvu=1\), since \(u^2v^2=uuvu\). Neither expansion introduces a letter from the other surface’s generator set.
Check: setting \(u=v=1\) in the displayed presentation leaves exactly the genus-two orientable presentation; setting \(a_1=b_1=a_2=b_2=1\) leaves exactly the Klein-bottle presentation. The wedge calculation preserves both sets of relations.
A Common Pitfall: Combining Relators into One
A tempting but incorrect shortcut is to multiply relators from the two summands and keep only that product as a relation. For the torus and projective-plane example, replacing the two relators by the single relation \([a,b]c^2=1\) does not impose each original relation separately. Here is a concrete check. In the group of symmetries of a square, write \(r\) for a rotation by a quarter-turn and \(s\) for a reflection. They satisfy \(r^4=s^2=1\) and \(srs=r^{-1}\). Assign \(a=s\), \(b=r\), and \(c=r\). Then
So this assignment satisfies the proposed single product relation, but \([a,b]=r^2\ne1\) and \(c^2=r^2\ne1\). It does not satisfy the torus and projective-plane relators separately. The actual wedge presentation must retain both relators. The free-product presentation is built from the union of the relator sets, not from an arbitrary product of relators.
The calculation matters because it preserves information about how loops in the separate surfaces interact. A free product is not generally a direct product: a direct product would additionally require every element from one factor to commute with every element from the other. In a wedge, the shared basepoint alone does not impose those commutation relations.
| Wedge summands | Presentation contributed | Wedge-group operation |
|---|---|---|
| \(\Sigma_g\) and \(N_k\) | Keep the surface relator from each summand | Free product |
| \(T^2\) and \(\mathbb{RP}^2\) | \([a,b]\) and \(c^2\) | \(\mathbb{Z}^2*C_2\) |
| \(S^2\) and a based surface \(Y\) | The sphere contributes no generators or relators | \(1*\pi_1(Y)\cong\pi_1(Y)\) |
Check Your Understanding
For each question, keep the generators of distinct summands separate and retain each summand’s relators.
- Write a presentation for the fundamental group of \(\Sigma_2\vee\mathbb{RP}^2\).
- What is the fundamental group of \(S^2\vee N_2\), in terms of a presentation for \(N_2\)?
- Write the free-product presentation for \(N_1\vee N_1\), and identify its two relators.
- Explain why the wedge theorem applies to the open sets \(U=X\cup D_y\) and \(V=D_x\cup Y\).
- In the torus–projective-plane presentation, why is \([a,b]c^2=1\) alone not a valid replacement for the two separate relators?