What It Means to Attach a Disk
A disk can be added to a space by prescribing where each point of its boundary is to go. The boundary map may identify different boundary points with the same point of the original space; this affects the resulting quotient, so the map—not merely the two spaces being joined—determines the construction.
For \(n\geq 1\), let \(D^n=\{x\in\mathbb{R}^n:\|x\|\leq 1\}\) have the subspace topology inherited from Euclidean space, and let \(S^{n-1}=\{x\in\mathbb{R}^n:\|x\|=1\}\) have its subspace topology. Let \((X,\mathcal{T})\) be a topological space, and give \(S^{n-1}\) its topology as the boundary of \(D^n\). An attaching map is a continuous map \(f:S^{n-1}\to X\).
Here is the point-level description. An interior point of the disk, meaning a point of \(D^n\setminus S^{n-1}\), remains in a class by itself. A boundary point \(u\) is in the same class as \(f(u)\). If distinct boundary points \(u,v\) have \(f(u)=f(v)\), then they are in the same class as each other as well. Points of \(X\) that are not in the image of \(f\) remain in singleton classes. Thus the boundary is attached to \(X\); it is not discarded.
Write \(j_X:X\to X\cup_fD^n\) and \(j_D:D^n\to X\cup_fD^n\) for the maps induced by the two inclusions into the disjoint union. They satisfy \(j_D(u)=j_X(f(u))\) for every \(u\in S^{n-1}\). In general \(j_X\) need not be injective: if \(f(u)=f(v)\) for distinct boundary points, then the quotient also identifies \(u\) and \(v\), but it does not identify distinct points of \(X\).
The Disk Interior Survives the Attachment
Proof. Let \(V=D^n\setminus S^{n-1}\), the open unit disk. Its preimage under \(q\) is exactly the copy of \(V\) in the disjoint union: no point of \(X\) or of the boundary is equivalent to an interior point. This preimage is open in the disjoint union. By the definition of the quotient topology, \(q(V)=j_D(V)\) is open in the attached space. The restriction \(j_D|_V\) is injective because no two interior points are identified, and it is surjective onto \(j_D(V)\) by definition. For any open subset \(W\subseteq V\), \(W\) is open in \(D^n\), since \(V\) is open in \(D^n\). Its full preimage under \(q\) is just \(W\), so \(j_D(W)\) is open in the attached space, and therefore open in \(j_D(V)\). Thus the bijection \(j_D|_V:V\to j_D(V)\) is open and continuous, so its inverse is continuous. It is a homeomorphism. \(\square\)
This proposition describes the new part of the construction without requiring the attaching map to be one-to-one. Only the boundary is subject to identifications; the open disk is unchanged as a topological space.
A Useful Test for Showing Two Quotients Are Homeomorphic
In the examples, we will use the following elementary fact. A space is compact if every open cover has a finite subcover. A space is Hausdorff if every pair of distinct points has disjoint open neighborhoods.
Proof. Let \(g:K\to Y\) be a continuous bijection, where \(K\) is compact and \(Y\) is Hausdorff. To show that \(g^{-1}\) is continuous, it is enough to show that \(g\) maps closed sets to closed sets. If \(C\) is closed in \(K\), then \(C\) is compact: any open cover of \(C\), together with \(K\setminus C\), covers \(K\), so a finite subcover of \(K\) gives a finite subcover of \(C\). A compact subset \(L\) of a Hausdorff space \(Y\) is closed. Indeed, for \(y\notin L\), choose for each \(z\in L\) disjoint open neighborhoods \(U_z\) of \(z\) and \(V_z\) of \(y\). Finitely many \(U_z\) cover \(L\), by compactness; the intersection of the corresponding finitely many \(V_z\) is an open neighborhood of \(y\) disjoint from \(L\). Hence \(Y\setminus L\) is open. Apply this to \(L=g(C)\), which is compact by continuity. Thus \(g(C)\) is closed in \(Y\), and \(g^{-1}\) is continuous. Therefore \(g\) is a homeomorphism. \(\square\)
Worked Constructions
Worked Example: An Interval with Its Endpoints Identified
Take \(X\) to be a one-point space and attach \(D^1=[-1,1]\) by sending both points of \(S^0=\{-1,1\}\) to that one point. The quotient is \(D^1\) with \(-1\) and \(1\) identified. Equivalently, using \([0,1]\) in place of \(D^1\), this is the interval with its endpoints identified.
Define \(g:[0,1]\to S^1\) by \(g(t)=(\cos(2\pi t),\sin(2\pi t))\), where \(S^1\) has the subspace topology from \(\mathbb{R}^2\). This map is continuous. Its values at the endpoints agree: \(g(0)=g(1)=(1,0)\). Moreover, \(g(s)=g(t)\) exactly when \(s-t\) is an integer; for \(s,t\in[0,1]\), this means either \(s=t\) or \(\{s,t\}=\{0,1\}\). Thus \(g\) is constant on the quotient classes and distinguishes all distinct classes.
By the definition of the quotient topology, \(g\) induces a continuous bijection from \([0,1]/(0\sim1)\) to \(S^1\). The interval is compact, so its quotient is compact; \(S^1\), as a subspace of the Hausdorff space \(\mathbb{R}^2\), is Hausdorff. The lemma gives a homeomorphism \([0,1]/(0\sim1)\cong S^1\).
Check: the only two distinct points of \([0,1]\) with the same image are \(0\) and \(1\), since \(s-t\) must be an integer in \([-1,1]\). These are precisely the points identified in the quotient.
Worked Example: A Disk with Its Boundary Collapsed
Let \(X\) be a one-point space and let \(f:S^{n-1}\to X\) be the constant map. Attaching \(D^n\) along \(f\) identifies every point of the boundary to one point and makes no identifications among interior points. We show that the result is homeomorphic to \(S^n\), with its subspace topology from \(\mathbb{R}^{n+1}\).
For \(x\in D^n\), set \(r=\|x\|\). Define \(G:D^n\to S^n\) by \[ G(x)= \begin{cases} \left(\dfrac{\sin(\pi r)}{r}x,\ \cos(\pi r)\right),&r>0,\\ (0,\ldots,0,1),&r=0. \end{cases} \] For \(r>0\), the squared length of the first \(n\) coordinates is \(\sin^2(\pi r)\), and the square of the last coordinate is \(\cos^2(\pi r)\). Their sum is \(1\), so \(G(x)\in S^n\). At \(r=0\), the displayed value also lies in \(S^n\). Continuity at \(0\) follows because the norm of the first \(n\) coordinates is \(|\sin(\pi r)|\to0\), while the last coordinate \(\cos(\pi r)\to1\). Elsewhere the formula is continuous.
When \(r=1\), \(G(x)=(0,\ldots,0,-1)\), independent of the boundary point \(x\). For \(0\leq r<1\), the last coordinate is \(\cos(\pi r)\), which decreases from \(1\) to \(-1\) as \(r\) increases from \(0\) to \(1\). Hence each point of \(S^n\) other than \((0,\ldots,0,-1)\) determines a unique \(r\in[0,1)\). If \(r>0\), its first \(n\) coordinates determine the direction \(x/r\); for \(r=0\), the only point is \((0,\ldots,0,1)\). Therefore \(G\) is onto and its fibers are exactly the boundary \(S^{n-1}\) as one class and singleton classes for interior points.
Consequently \(G\) induces a continuous bijection \(D^n/S^{n-1}\to S^n\). The disk is compact by the Heine–Borel theorem, so its quotient is compact. The sphere is Hausdorff as a subspace of Euclidean space. The lemma proves that \(D^n/S^{n-1}\cong S^n\).
Worked Example: Attaching a Disk to a Circle by the Identity
Let \(X=S^1\), and attach \(D^2\) using the identity boundary map \(f:S^1\to S^1\). Here both circles are the unit circle in \(\mathbb{R}^2\), so a boundary point \(u\in\partial D^2\) is identified with the same point \(u\in X\). We show that \(S^1\cup_fD^2\cong D^2\).
Define \(H:S^1\bigsqcup D^2\to D^2\) to be the usual inclusion on each component. The two definitions agree on identified points: for \(u\in S^1=\partial D^2\), both send \(u\) to \(u\in D^2\). Thus \(H\) induces a continuous map \(\widetilde H:S^1\cup_fD^2\to D^2\). Every interior point of \(D^2\) has its own class, and each boundary point has one class consisting of that point in \(X\) and the corresponding point in \(\partial D^2\). These classes map to distinct points of \(D^2\), and every point of \(D^2\) is reached. Hence \(\widetilde H\) is bijective.
The disjoint union of the two compact spaces \(S^1\) and \(D^2\) is compact, so its quotient is compact. The target \(D^2\) is Hausdorff as a subspace of \(\mathbb{R}^2\). The lemma now shows that \(\widetilde H\) is a homeomorphism. In this example the original circle becomes the boundary of the disk that has been attached.
A Common Mistake: Forgetting the Identifications Forced by the Map
A tempting but incorrect description is: “each boundary point is identified only with its image, and different boundary points always remain distinct.” The second clause fails whenever the attaching map is not injective. If \(f(u)=f(v)\), then both \(u\) and \(v\) are identified with the same point of \(X\), so they belong to the same equivalence class.
For a concrete correction, take \(X\) to be a one-point space and \(f:S^1\to X\) constant. The mistaken description would leave distinct boundary points separate and would not describe the quotient formed by this attaching map. The actual relation identifies every \(u\in S^1\) with the unique point of \(X\); hence all boundary points lie in a single class. The calculation above shows that the resulting space is \(D^2/S^1\cong S^2\), not a disk with its boundary still present.
The choice of attaching map is therefore essential. It specifies how the boundary is placed in \(X\), while the quotient topology specifies the topology of the resulting space. When checking an example, first list the equivalence classes dictated by the map; only then identify the resulting space.
Check Your Understanding
For each question, state the equivalence classes before deciding what the quotient looks like.
- If \(f:S^1\to X\) is injective, can two distinct boundary points of the attached disk be equivalent? Explain.
- What is obtained by attaching \(D^1=[-1,1]\) to a point when both endpoints are sent to that point?
- In the constant-map construction \(D^n/S^{n-1}\), which points form the nonsingleton equivalence class?
- Why does the proposition about the disk interior not require the attaching map to be injective?
- When attaching \(D^2\) to \(S^1\) by the identity boundary map, which points of the disjoint union represent a point of \(\partial D^2\) in the quotient?