Identifying Basepoints
A wedge sum joins spaces at chosen points without identifying any of their other points. This construction is useful when a space is assembled from several pieces that meet at one distinguished point: each piece retains its own topology, while the common point records where the pieces are joined. The quotient topology is essential here, particularly when deciding what counts as a neighborhood of the shared point.
A pointed space is a topological space \(X\), with topology \(\mathcal{T}_X\), together with a chosen point \(x_0\in X\), called its basepoint. For each index \(i\) in a nonempty set \(J\), let \((X_i,x_i)\) be a pointed space. The disjoint union \(\bigsqcup_{i\in J}X_i\) has the disjoint-union topology: a subset is open exactly when its intersection with every summand \(X_i\) is open in \(X_i\).
Write \(q:\bigsqcup_{i\in J}X_i\to\bigvee_{i\in J}X_i\) for the quotient map, and \(\iota_i:X_i\to\bigsqcup_{k\in J}X_k\) for the inclusion of a summand. The canonical map from \(X_i\) into the wedge is \(j_i=q\circ\iota_i\). Every non-basepoint \(x\in X_i\) gives a point \(j_i(x)\) belonging only to that summand's image; all the points \(j_i(x_i)\) are the single wedge point.
Reading the Quotient Topology
Proof. By definition of the quotient topology, \(O\) is open if and only if \(q^{-1}(O)\) is open in the disjoint union. The intersection of \(q^{-1}(O)\) with the \(i\)-th summand is exactly \(\iota_i(j_i^{-1}(O))\). The definition of the disjoint-union topology says that \(q^{-1}(O)\) is open if and only if each of these intersections is open in its summand. Identifying the summand with \(X_i\) gives precisely the stated condition. \(\square\)
This test also describes neighborhoods of the wedge point. If \(O\) contains the wedge point, then every set \(j_i^{-1}(O)\) contains \(x_i\); for \(O\) to be open, each such slice must be an open neighborhood of \(x_i\) in its own summand. There is no requirement that these neighborhoods have the same shape or size. If \(O\) omits the wedge point, then every slice omits its summand's basepoint, and the slices still must be open. The agreement about whether the basepoint is included is automatic, since it is one point of the wedge, not a separate point in each summand.
Worked Example: Two Circles Joined at One Point
Give each copy of \(S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}\) its subspace topology from the Euclidean plane, and choose \(p=(1,0)\) as basepoint in both copies. The wedge \(S^1\vee S^1\) has one point \(*\) representing the two copies of \(p\). Every other point can be written \((i,z)\), where \(i\in\{1,2\}\) indicates the circle and \(z\in S^1\setminus\{p\}\).
For any subset \(O\) of this wedge, its two slices are \(U_i=j_i^{-1}(O)\subseteq S^1\). The open-set test gives the complete answer: \(O\) is open exactly when \(U_1\) and \(U_2\) are open in \(S^1\). If \(*\in O\), then \(p\in U_1\) and \(p\in U_2\); if \(*\notin O\), then \(p\notin U_1\) and \(p\notin U_2\).
For a concrete open neighborhood of \(*\), let \(U_1=U_2=\{(u,v)\in S^1:u>0\}\). The coordinate function \((u,v)\mapsto u\) is continuous, and \((0,\infty)\) is open in \(\mathbb{R}\), so each slice is open in \(S^1\). Both contain \(p\), since its first coordinate is \(1>0\). Therefore \(O=j_1(U_1)\cup j_2(U_2)\) is open in the wedge and contains \(*\). Its intersection with either circle's image is exactly that circle's right semicircle, including the common point. This verifies the neighborhood test directly.
The two embedded circles intersect in \(\{*\}\), not in two copies of \(p\). Away from \(*\), points on different circles remain distinct.
The Summands Keep Their Topology
Proof. First, \(j_i\) is continuous: for every open \(O\) in the wedge, the open-set test says \(j_i^{-1}(O)\) is open in \(X_i\). The map is injective because the equivalence relation identifies distinct points only when they are basepoints from different summands. It remains to show that its inverse from \(j_i(X_i)\), with the subspace topology, to \(X_i\) is continuous. It suffices to show that \(j_i(U)\) is open in \(j_i(X_i)\) whenever \(U\) is open in \(X_i\).
Let \(U\subseteq X_i\) be open. If \(x_i\notin U\), form \(V\subseteq\bigsqcup_{k\in J}X_k\) whose slice in \(X_i\) is \(U\) and whose slice in every other summand is empty. This set is open and saturated: it contains no basepoint, so it contains either all or none of each equivalence class. If \(x_i\in U\), instead take the slice in \(X_i\) to be \(U\) and the slice in every other summand \(X_k\) to be all of \(X_k\). This is open and contains every basepoint, so it is saturated. In either case, \(q^{-1}(q(V))=V\), and the quotient definition therefore makes \(q(V)\) open. In both cases \(q(V)\cap j_i(X_i)=j_i(U)\), so \(j_i(U)\) is open in the subspace \(j_i(X_i)\). Thus \(j_i\) is a homeomorphism onto its image. Finally, two different summands have no equivalent points in common except their basepoints, and all those basepoints map to the wedge point. Hence their images intersect exactly there. Finiteness is not needed for this embedding argument: in the basepoint-containing case, use the whole space \(X_k\) in every other summand. \(\square\)
The proof shows why a quotient construction need not distort the topology of each individual piece. It does, however, change how the pieces fit together at the shared point. The open-set test, rather than a visual picture of intersecting curves or surfaces, determines that local behavior.
Worked Example: A Circle Joined to a Sphere
Take \(S^1\) with its subspace topology from \(\mathbb{R}^2\) and basepoint \(p=(1,0)\). Take \(S^2=\{(a,b,c)\in\mathbb{R}^3:a^2+b^2+c^2=1\}\), with its subspace topology from \(\mathbb{R}^3\), and basepoint \(n=(0,0,1)\). In \(S^1\vee S^2\), the point \(p\) and the north pole \(n\) become one point \(*\); no other points are identified.
Consider the subset whose circle slice is \(U_1=\{(u,v)\in S^1:u>0\}\) and whose sphere slice is \(U_2=\{(a,b,c)\in S^2:c>0\}\). The first set is open because it is the preimage of \((0,\infty)\) under the first-coordinate function; the second is open because it is the preimage of \((0,\infty)\) under the third-coordinate function. They contain their respective basepoints: \(1>0\) for \(p\)'s first coordinate, and \(1>0\) for \(n\)'s third coordinate. The open-set test therefore says their union in the wedge is an open neighborhood of \(*\).
The canonical-inclusions theorem checks the result another way: the circle retains its original subspace topology, the sphere retains its original subspace topology, and their images intersect only at \(*\). In particular, the sphere slice \(c>0\) does not have to resemble the circle slice \(u>0\); each only has to be open in its own space and contain its own basepoint.
Worked Example: Three Copies of a Sphere
Let \(m=3\) and \(n\ge1\). Give each copy of \(S^n=\{z\in\mathbb{R}^{n+1}:\|z\|=1\}\) the subspace topology from Euclidean \(\mathbb{R}^{n+1}\), and choose \(e=(1,0,\ldots,0)\) as basepoint in each. The underlying set of \(S^n\vee S^n\vee S^n\) can be written as \(\{*\}\cup\{(i,z):i\in\{1,2,3\},\ z\in S^n\setminus\{e\}\}\). Thus each sphere contributes all its non-basepoint elements, while its basepoint contributes no separately tagged point.
Let \(U_i=\{z\in S^n:z_1>1/2\}\) in each of the three copies, where \(z_1\) is the first coordinate. Each \(U_i\) is open as the preimage of \((1/2,\infty)\) under the continuous first-coordinate function. Each contains \(e\), since \(e_1=1\) and \(1>1/2\). Therefore the set consisting of \(*\) and all tagged points \((i,z)\) with \(z\in U_i\setminus\{e\}\) is open in the wedge. Its preimage under \(q\) is exactly the disjoint union of the three sets \(U_i\), which is open in the disjoint union and is saturated.
For each \(i\), the canonical image of the \(i\)-th sphere intersects each of the other two images only in \(\{*\}\). The canonical-inclusions theorem also confirms that the topology on each image is exactly that of \(S^n\), including at its basepoint. This construction works for \(n=1\), giving three circles, and for \(n=2\), giving three 2-spheres; the underlying set description and open-set test are the same in both cases.
One-Point Summands and Infinite Families
Proof. Let \(P=\{p\}\) be the one-point space with basepoint \(p\), and let \(W\) be a wedge of at least one pointed space. In \(W\vee P\), the point \(p\) is identified with the basepoint of \(W\), and \(P\) contributes no other points. The map from \(W\) to \(W\vee P\) induced by the quotient is therefore a bijection. For a subset \(O\subseteq W\vee P\), its slice in \(P\) is either empty or all of \(P\); both are open. By the open-set test, \(O\) is open exactly when its slice in \(W\) is open. Thus the bijection and its inverse preserve open sets, so it is a homeomorphism. \(\square\)
The index set \(J\) need not be finite in either the wedge definition or the open-set test. For an infinite family, an open neighborhood of the wedge point still has to contain an open neighborhood of every basepoint. The choices may vary from summand to summand; no single neighborhood size is required.
Worked Example: A Neighborhood in a Countable Wedge of Circles
Take copies \(X_i=S^1\), indexed by \(i\in\mathbb{N}\), each with the standard subspace topology from \(\mathbb{R}^2\) and basepoint \(p=(1,0)\). In each copy let \(U_i=\{(u,v)\in S^1:u>0\}\). Each \(U_i\) is open and contains \(p\). The set \(O\) consisting of the wedge point and all tagged points \((i,z)\) with \(z\in U_i\setminus\{p\}\) has \(j_i^{-1}(O)=U_i\) for every \(i\). The open-set test applies to every index, so \(O\) is open in the countable wedge. This verifies the condition for infinitely many summands directly; the embedding proof likewise uses the whole \(X_k\) in every other summand when needed.
A Common Mistake at the Wedge Point
Common pitfall. One might try to declare a subset open when its slice is open in one circle, while allowing its slice in the other circle to be arbitrary. For example, in \(S^1\vee S^1\), take the entire first circle together with only the shared point from the second circle. This set contains \(*\). Its first slice is \(S^1\), which is open, but its second slice is \(\{p\}\), which is not open in \(S^1\).
To verify the last claim, every open neighborhood of \(p=(1,0)\) in the subspace topology contains points of \(S^1\) other than \(p\): an ambient open ball around \(p\) contains points of the circle sufficiently close to \(p\). Hence \(\{p\}\) is not a neighborhood of \(p\), and in particular is not open. The proposed set is therefore not open in the wedge: its quotient preimage has a non-open slice in the second summand. The correction is to check every summand. Any open set containing the wedge point must restrict to an open neighborhood of every basepoint.
Wedge sums provide a precise way to build spaces from pointed pieces and to keep track of the topology at their junction. The quotient definition handles the identification, the open-set test handles neighborhoods, and the canonical inclusions guarantee that no summand's own topology is lost. These facts are the basic checks to make whenever a construction joins several pointed spaces at one point.
Check Your Understanding
Use the quotient definition and the open-set test to answer the following.
- For \(X_1\vee X_2\), describe the preimage of a subset containing the wedge point. What must be true of its two slices for the subset to be open?
- Why is \(j_i\) injective even though all basepoints are identified? Which points are identified by the equivalence relation?
- In \(S^1\vee S^1\), is the set consisting of the full first circle and only the wedge point from the second circle open? Justify using the open-set test.
- For three pointed spaces, what is the intersection of the images of the first and third summands? Explain why.
- Does the proof that each canonical inclusion is an embedding require the family of summands to be finite? Identify the open saturated set used when the chosen open subset contains its basepoint.