Attaching a Disk Imposes a Relation
Attaching a disk along a loop fills that loop with a disk. For the fundamental group, this has a precise algebraic effect: the loop becomes trivial. In general, however, a group relation does not merely remove one element from a list of generators. It also forces every conjugate of that element to become trivial. The Seifert–van Kampen theorem explains why the correct quotient is by a normal subgroup.
Let \(X\) be a path-connected topological space, let \(x_0\in X\), and let \(f:S^1\to X\) be a continuous loop with \(f(1)=x_0\), where \(S^1\) has its usual topology. Write \(D^2=\{z\in\mathbb{C}:|z|\leq 1\}\) with its subspace topology. The space formed by attaching a disk along \(f\) is the quotient \(Y=(X\bigsqcup D^2)/{\sim}\), where each \(z\in\partial D^2=S^1\) is identified with \(f(z)\in X\), and there are no other identifications except those forced by this rule. The quotient has the quotient topology.
The Open Cover Behind the Computation
Choose radii \(1/2\) and \(3/4\). Let \(q:X\bigsqcup D^2\to Y\) be the quotient map, and set \(U=q(X\bigsqcup\{z\in D^2:|z|>1/2\})\) and \(V=q(\{z\in D^2:|z|<3/4\})\). These are open in \(Y\): their preimages under \(q\) are open in the disjoint union, and each preimage contains every point equivalent to any of its points. They cover \(Y\). The set \(U\) contains \(X\) and an outer collar of the disk; \(V\) contains the disk’s inner part. Their intersection is the annulus \(1/2<|z|<3/4\), so it is path-connected and has the homotopy type of \(S^1\).
The outer collar in \(U\) can be pushed radially to the boundary, fixing \(X\). Thus \(U\) deformation retracts onto \(X\). The set \(V\) is an open disk and is contractible. A circle going once around \(U\cap V\), followed by this retraction to \(X\), gives the attaching loop \(f\), up to the choice of orientation. In \(V\), that circle is null-homotopic.
Proof. Use the open cover \(Y=U\cup V\) just constructed. The intersection is path-connected, \(V\) is simply connected, and \(U\) deformation retracts onto \(X\). The Seifert–van Kampen theorem, in its standard open-cover form, identifies \(\pi_1(Y,x_0)\) with the quotient of \(\pi_1(U,x_0)\) by the normal closure of the image of \(\pi_1(U\cap V)\). Here and below, basepoints in the intersection are connected to \(x_0\) by paths in \(U\), as required in that theorem. The group \(\pi_1(U\cap V)\) is infinite cyclic, generated by a circle going once around the annulus. Its image in \(\pi_1(U,x_0)\cong\pi_1(X,x_0)\) is \([f]\) (or \([f]^{-1}\) if the opposite orientation is chosen). The normal closures of an element and its inverse agree. Therefore the quotient is by \(\langle\!\langle[f]\rangle\!\rangle\), as claimed. \(\square\)
The use of a normal closure is essential: a quotient of a group by a subgroup is a group only when that subgroup is normal. The theorem says that the disk kills the attaching loop and all its conjugates. If a loop is described by a word \(w\) in generators, the resulting presentation adds the relator \(w=1\). Replacing \(w\) by a conjugate \(hwh^{-1}\) does not change the normal closure, so it gives the same group.
Worked Computations
Worked Example: A Degree-Four Attaching Loop
Start with \(X=S^1\), based at \(1\), and attach a disk using \(f:S^1\to S^1\) given by \(f(z)=z^4\). The fundamental group of the circle is infinite cyclic, \(\pi_1(S^1,1)=\langle a\rangle\cong\mathbb{Z}\), where \(a\) goes once counterclockwise around the circle. The loop \(f\) goes around four times, so \([f]=a^4\). The theorem gives \(\pi_1(Y,1)=\langle a\rangle/\langle\!\langle a^4\rangle\!\rangle\).
Since \(\langle a\rangle\) is abelian, its normal closure of \(a^4\) is the subgroup \(\langle a^4\rangle\). Writing the infinite cyclic group additively as \(\mathbb{Z}\), this subgroup is \(4\mathbb{Z}=\{4k:k\in\mathbb{Z}\}\). The quotient consists of the four residue classes \(0,1,2,3\) modulo \(4\), with addition modulo \(4\). Hence \(\pi_1(Y,1)\cong\mathbb{Z}/4\mathbb{Z}\).
Check: the attaching loop represents \(a^4\), which is zero in the quotient because \(4=0\) modulo \(4\). The class of \(a\) is not zero, and its successive multiples give the four distinct residue classes, so the quotient has exactly the stated cyclic group.
Worked Example: A Relator Conjugate to a Generator
Let \(X=S^1\vee S^1\), the wedge of two circles with common basepoint \(x_0\). Give the circles generators \(a\) and \(b\); then \(\pi_1(X,x_0)=F(a,b)\), the free group on those generators. Attach a disk along the loop represented by \(w=aba^{-1}\). The theorem gives \(\pi_1(Y,x_0)=F(a,b)/\langle\!\langle aba^{-1}\rangle\!\rangle\).
In the quotient, the relation \(aba^{-1}=1\) implies \(b=1\): multiply the equation on the left by \(a^{-1}\) and on the right by \(a\), obtaining \(b=1\). Equivalently, the normal closure of \(aba^{-1}\) is the normal closure of \(b\), because \(aba^{-1}\) is a conjugate of \(b\), and \(b=a^{-1}(aba^{-1})a\) is a conjugate of \(aba^{-1}\). Thus the quotient kills \(b\) and leaves the unrestricted generator \(a\), giving \(\pi_1(Y,x_0)\cong F(a)\cong\mathbb{Z}\).
Check: in the resulting group, \(aba^{-1}\) evaluates to \(a\cdot 1\cdot a^{-1}=1\), as the attaching loop requires. The generator \(a\) has no imposed relation, so the remaining group is infinite cyclic.
Worked Example: A Word Relator on Two Circles
Again let \(X=S^1\vee S^1\) with generators \(a,b\) and basepoint \(x_0\). Attach a disk along the word \(w=a^2b^{-3}\), meaning the loop that traverses the \(a\)-circle twice in its chosen direction and then traverses the \(b\)-circle three times in the reverse direction. The calculation is \(\pi_1(Y,x_0)=F(a,b)/\langle\!\langle a^2b^{-3}\rangle\!\rangle\). The relator equation is \(a^2b^{-3}=1\); multiplying on the right by \(b^3\) gives \(a^2=b^3\). Conversely, if \(a^2=b^3\), multiplying on the right by \(b^{-3}\) gives \(a^2b^{-3}=1\). So this attaching loop imposes precisely that relation and its conjugates. This presentation describes the result without asserting that the group simplifies to a familiar group.
Check: substituting the equivalent equation \(a^2=b^3\) into the relator gives \(a^2b^{-3}=b^3b^{-3}=1\). The reverse multiplication recovers the displayed relator, so the relation has neither lost nor added an inverse.
A Common Pitfall: Quotienting by Too Little
A tempting but incorrect calculation is to quotient a nonabelian group by the ordinary subgroup generated by the relator, without checking normality. For example, attach a disk to \(S^1\vee S^1\) along \(a^2\). The correct answer is \(F(a,b)/\langle\!\langle a^2\rangle\!\rangle\), not a quotient by merely \(\langle a^2\rangle\). The subgroup \(\langle a^2\rangle\) is not normal in \(F(a,b)\): the conjugate \(ba^2b^{-1}\) is a reduced word containing \(b\), and so is not a power of \(a^2\). Thus the ordinary subgroup is not a valid kernel for a group quotient. The disk forces \(a^2=1\) and also forces \(ba^2b^{-1}=1\), along with every other conjugate. That is exactly what the normal closure records.
More generally, one should not infer that the resulting group is trivial just because the attaching loop becomes trivial. The theorem kills the element represented by that loop, not automatically every generator of \(\pi_1(X,x_0)\). The quotient retains whatever elements survive the imposed normal relation.
When the Relation Is Already Trivial
If \([f]\) is the identity in \(\pi_1(X,x_0)\), then its normal closure is the trivial subgroup, and the theorem gives \(\pi_1(Y,x_0)\cong\pi_1(X,x_0)\). This includes the case of a constant attaching loop. The disk attachment changes the space, but it does not change its fundamental group: the boundary loop already represented the identity. In general, the theorem separates the geometric operation—adding a disk—from its group-theoretic effect—imposing the attaching loop as a normal relation.
Check Your Understanding
For each question, identify the group before attachment, the element represented by the attaching loop, and the normal subgroup used in the quotient.
- If a disk is attached to \(S^1\) by the degree-three map \(z\mapsto z^3\), what is the resulting fundamental group?
- For an attaching word \(b a b^{-1}\) on \(S^1\vee S^1\), which generator is killed in the quotient, and why?
- Explain why the normal closures of \(w\) and \(hwh^{-1}\) agree for any group elements \(h,w\).
- If the attaching loop represents the identity in \(\pi_1(X,x_0)\), what does the disk-attachment theorem imply about the fundamental group?
- Why is quotienting \(F(a,b)\) by the subgroup \(\langle a^2\rangle\) not the correct general procedure for attaching a disk along \(a^2\)?