The Surface with One Boundary Component
Removing an open disk from a closed orientable surface changes the fundamental group computation in an important way: the resulting surface retracts onto a graph, so its group is free. The boundary circle is not an extra relation in that free group. Instead, when the boundary is followed with a suitable orientation, its loop traces a particular product of commutators.
Let \(g\) be a nonnegative integer, and let \(S_{g,1}\) denote a compact, connected, orientable surface of genus \(g\) with one boundary component, with its usual topology. It can be obtained by removing the interior of a disk from a closed orientable genus-\(g\) surface. A basepoint is a chosen point \(x_0\in S_{g,1}\). The fundamental group \(\pi_1(S_{g,1},x_0)\) consists of based loops modulo homotopies that keep the basepoint fixed; its operation is concatenation of loops.
We use the standard polygon model for the closed orientable genus-\(g\) surface: its \(4g\) sides, in order around the polygon, are labelled \(a_1,b_1,a_1^{-1},b_1^{-1},\ldots,a_g,b_g,a_g^{-1},b_g^{-1}\), and sides with the same generator label are identified in the direction indicated by the labels. The side identifications produce one vertex and \(2g\) edges. The word read around the polygon is \(W_g=[a_1,b_1]\cdots[a_g,b_g]\).
Why the Fundamental Group Is Free
Proof. First take \(g\geq1\). In the polygon model, remove a small open disk from the interior of the polygon before making the side identifications. The polygon with this disk removed is an annulus: its outer boundary is the polygon perimeter and its inner boundary becomes the boundary of \(S_{g,1}\). The annulus deformation retracts onto its outer boundary. For example, identify it with a round annulus and move every point radially to the outer circle; throughout the deformation the outer circle is fixed pointwise. Because the deformation fixes the polygon perimeter, it respects all the identifications made along that perimeter, and therefore gives a deformation retraction of \(S_{g,1}\) onto the quotient of the polygon perimeter.
That quotient has one vertex, since all polygon corners are identified, and one loop for each pair of identified sides. It is therefore a wedge of \(2g\) circles. The standard computation of the fundamental group of a wedge of finitely many circles gives the free group on the loops represented by those \(2g\) edges. Thus \(\pi_1(S_{g,1},x_0)\cong F(a_1,b_1,\ldots,a_g,b_g)\); changing the basepoint identification uses a path and gives an isomorphic group.
To read the boundary, follow the inner boundary of the annulus through the deformation retraction. With the orientation chosen to match the direction around the polygon perimeter, the resulting loop reads the side labels in order, hence has word \(a_1b_1a_1^{-1}b_1^{-1}\cdots a_gb_ga_g^{-1}b_g^{-1}=W_g\). The induced orientation on the inner boundary of the oriented annulus is opposite to the outer perimeter orientation, so with that induced orientation the word is \(W_g^{-1}\). Choose the basepoint so its retraction lands at the polygon vertex; for other basepoints, a path to this point transports the based loop and the generators together. Finally, if \(g=0\), \(S_{0,1}\) is a disk. A disk contracts to a point, so its fundamental group is trivial, in agreement with the free group on zero generators; its boundary loop is null-homotopic. \(\square\)
The basepoint and orientation qualifications matter. A loop based at a different boundary point can be transported to the standard basepoint along the boundary; the corresponding based group element depends on this transport. The free-group description is canonical only up to choices of generators, while the geometric boundary circle itself is well defined up to reversing its direction.
Reading and Checking the Boundary Word
Worked Example: Genus Zero
For \(g=0\), the surface \(S_{0,1}\) is a disk. There are no polygon edges to identify and no generators, so the group is the free group on zero generators, namely \(\{1\}\). The product of commutators has no factors and is the identity element.
Check: Model the disk as the unit disk in \(\mathbb{R}^2\). The homotopy \(H(x,t)=(1-t)x\), for \(0\leq t\leq1\), stays in the disk and contracts it to its center. In particular, the boundary loop contracts, as required by the trivial fundamental group.
Worked Example: A Once-Punctured Torus
For \(g=1\), the group is \(F(a_1,b_1)\), the free group on two generators. The oriented boundary loop reads \([a_1,b_1]=a_1b_1a_1^{-1}b_1^{-1}\). Writing \(a=a_1\) and \(b=b_1\), its word is \(aba^{-1}b^{-1}\).
This word is reduced: its adjacent pairs are \(a,b\), then \(b,a^{-1}\), then \(a^{-1},b^{-1}\), and none consists of a letter followed by its inverse. Its length is four, so it is not the identity in the free group. The reverse traversal has word \(b a b^{-1}a^{-1}\), obtained by reversing the order and inverting each letter.
Check: The boundary is a single loop, and the four sides of the square model read \(a,b,a^{-1},b^{-1}\). The same four letters occur in the commutator, in the same order. No cancellation occurs, confirming that this boundary element is nontrivial in the punctured surface group.
Worked Example: Genus Two
For \(g=2\), there are four free generators \(a_1,b_1,a_2,b_2\). The group is \(F(a_1,b_1,a_2,b_2)\), and the oriented boundary word is
Expanding the two factors gives the eight displayed letters. The first four are \(a_1,b_1,a_1^{-1},b_1^{-1}\); the next four are \(a_2,b_2,a_2^{-1},b_2^{-1}\). Within each block, consecutive letters are not inverses. At the join, \(b_1^{-1}\) is followed by \(a_2\), also not its inverse. The word is therefore reduced and has length eight.
Check: Reversing the traversal gives \(b_2a_2b_2^{-1}a_2^{-1}b_1a_1b_1^{-1}a_1^{-1}\). This is the inverse of the displayed word, since the inverse of a product reverses the factor order and inverts each factor. The boundary element is nontrivial because its reduced word has positive length.
Why Filling the Boundary Changes the Group
A frequent error is to impose the boundary word as a relation while computing the fundamental group of \(S_{g,1}\). That would incorrectly identify the boundary loop with the identity. The polygon model explains the distinction: the punctured polygon retracts to its perimeter graph, but gluing a disk into the hole adds a two-dimensional cell. That additional disk makes the boundary loop null-homotopic.
Proof. Before the disk is attached, the surface retracts to the wedge of \(2g\) circles, whose fundamental group is free on the edge generators. The boundary loop represents \(W_g\). Attaching a disk along a loop makes that attaching loop null-homotopic; the disk-attachment form of the Seifert-van Kampen theorem says that the resulting group is the original group modulo the normal subgroup generated by the attaching-loop element. Thus the new group has exactly the stated presentation. The relation belongs to the filled surface, not to \(S_{g,1}\) itself. \(\square\)
For genus one, the distinction is especially visible. The punctured torus has group \(F(a,b)\), and its boundary word \(aba^{-1}b^{-1}\) is a nonempty reduced word, so it is not the identity. After filling the boundary, the relation \(aba^{-1}b^{-1}=1\) is imposed. Multiplying this equation on the left by \(a^{-1}\) and on the right by \(b\) gives \(ba=ab\). The resulting group is therefore generated by two commuting elements, unlike the free group of the punctured torus.
The essential computation is thus twofold: the graph spine supplies \(2g\) unconstrained generators, and the route around the single boundary component records how those generators are arranged. That route is a product of commutators, but it becomes a relation only when a disk is attached.
Check Your Understanding
Use the polygon model and keep track of whether a disk has been attached.
- What is \(\pi_1(S_{3,1},x_0)\), and how many free generators does it have?
- Write the oriented boundary word for \(S_{1,1}\), and write the word for the reverse traversal.
- Explain why the boundary word for \(S_{2,1}\) is reduced and nontrivial in its free group.
- What relation is imposed on the generators when the boundary of \(S_{2,1}\) is filled with a disk?
- Why does the genus-zero case have a trivial fundamental group even though its boundary is a circle?