Attaching a Space Along a Map
Gluing two spaces is not just a matter of deciding which points to identify: the resulting topology is part of the construction. An attaching map specifies which points are joined, while the quotient topology specifies when a set of joined points is open. The map need not be injective, so one point of the space being attached may join several points of the original space.
Let \((X,\mathcal{T}_X)\) be a topological space, let \(A\) be a subspace of a topological space \(Y\), and let \(f:A\to X\) be continuous. We use the disjoint union \(X\bigsqcup Y\), with the disjoint union topology: a subset is open precisely when its intersection with each component is open in that component. The disjoint union keeps the two copies separate before the identifications are made.
Here “generated by” matters. The relation must be an equivalence relation, so it includes every identification forced by reflexivity, symmetry, and transitivity. Points of \(X\) and \(Y\) are regarded as belonging to separate copies, even if their original names happen to coincide.
Equivalence Classes and the Quotient Topology
Proof. Let the relation described in the theorem be denoted by \(R\): each \(x\in X\) is related to the points of \(f^{-1}(\{x\})\subseteq A\), and all points not joined in this way are related only to themselves. These proposed classes partition \(X\bigsqcup Y\), so being in the same proposed class defines an equivalence relation. It contains every required pair \(a\sim f(a)\). Conversely, any equivalence relation containing those required pairs must put each \(a\in A\) in the class of \(f(a)\). If \(f(a)=f(b)=x\), transitivity then puts \(a\) and \(b\) in the same class as \(x\). No point of \(Y\setminus A\) is in any required pair, so the generated relation does not identify it with another point. Thus these are exactly the classes of the generated relation. \(\square\)
The quotient topology can also be tested directly on the original spaces. A subset \(S\subseteq X\bigsqcup Y\) is saturated if it is a union of equivalence classes; equivalently, it contains a whole class whenever it contains any point of that class. A subset \(U\) of the quotient has saturated preimage \(q^{-1}(U)\), and every saturated subset is the preimage of exactly one subset of the quotient.
Proof. By the definition of the quotient topology, \(U\) is open precisely when \(q^{-1}(U)\) is open in the disjoint union. By the definition of the disjoint union topology, a subset of \(X\bigsqcup Y\) is open precisely when its intersections with both components are open in those components. Applying this to \(q^{-1}(U)\) proves both directions. \(\square\)
This criterion includes an essential compatibility: the preimage must be a union of whole equivalence classes. To test openness, first include or exclude every point in each class together, and then check openness on each component.
Worked Example: A Finite Adjunction Space
Let \(X=\{p,q\}\) have topology \(\{\varnothing,\{p\},X\}\). Let \(Y=\{a,b,c\}\) have the discrete topology, let \(A=\{a,b\}\), and define \(f:A\to X\) by \(f(a)=p\) and \(f(b)=p\). The map is continuous because the only nonempty proper open set in \(X\) is \(\{p\}\), whose preimage is \(\{a,b\}\), open in \(A\).
By the class description, the quotient has three points: \(P=q(p)=q(a)=q(b)\), \(Q=q(q)\), and \(C=q(c)\). To compute its open sets, consider a saturated open subset of \(X\bigsqcup Y\). If it contains \(P\), it must contain \(p,a,b\). Its intersection with \(X\) can then be \(\{p\}\) or \(X\); its intersection with \(Y\) can be \(\{a,b\}\) or \(Y\). If it contains \(Q\), its intersection with \(X\) must be \(X\), which also forces it to contain \(P\). The point \(C\) can be included or excluded independently, since \(\{c\}\) is open in \(Y\).
Consequently the quotient topology is \(\{\varnothing,\{C\},\{P\},\{P,C\},\{P,Q\},\{P,Q,C\}\}\). For example, \(\{P,Q\}\) has preimage \(X\bigsqcup\{a,b\}\), which is open in the disjoint union; \(\{Q\}\) is not open because its preimage has intersection \(\{q\}\) with \(X\), and \(\{q\}\) is not open in \(X\).
Check: The listed sets are exactly the saturated open subsets just described, mapped to the quotient. In particular, the topology is not discrete: \(P\) is open, but \(Q\) is not.
Worked Example: Identifying the Endpoints of an Interval
Give \(Y=[0,1]\) its subspace topology from \(\mathbb{R}\), let \(A=\{0,1\}\), and let \(X=\{*\}\) have its one-point topology. Define \(f:A\to X\) by \(f(0)=f(1)=*\). This map is continuous because every subset of \(X\) has open preimage in \(A\).
The quotient has one class \(\{*,0,1\}\), which we call \(P\), and one singleton class \(q(t)\) for each \(t\in(0,1)\). For a subset \(U\) of the quotient, write \(V=\{t\in[0,1]:q(t)\in U\}\). The preimage \(q^{-1}(U)\) intersects the interval component in \(V\), and it intersects the one-point component in \(\{*\}\) exactly when \(P\in U\). The quotient topology therefore says precisely that \(U\) is open if and only if \(V\) is open in \([0,1]\); membership of \(0\) and \(1\) in \(V\) agrees automatically because \(q(0)=q(1)\).
For instance, \(q((1/3,2/3))\) is open: its interval preimage is the relatively open interval \((1/3,2/3)\), and it does not contain \(P\). The set \(P\cup q((0,1/3))\) is also open: its interval preimage is \([0,1/3)\), which is open relative to \([0,1]\), and its one-point preimage is all of \(X\). By contrast, \(P\) alone is not open, since its interval preimage \(\{0,1\}\) is not open in \([0,1]\).
Check: For each set above, the preimage in \(Y\) contains \(0\) if and only if it contains \(1\), as required by the equivalence class. The openness conclusions follow from the subspace topology on \([0,1]\), not merely from the point identifications.
Worked Example: Gluing Two Intervals at One Endpoint
Let \(X=[-1,0]\) and \(Y=[0,1]\), each with its subspace topology from \(\mathbb{R}\). Take \(A=\{0\}\subseteq Y\), and define \(f(0)=0\in X\). The quotient has one class containing the two copies of \(0\); every other point remains in a singleton class.
Define \(h:X\cup_f Y\to[-1,1]\) by sending the class of a point in \(X\) to that point and the class of a point in \(Y\) to that point. This is well-defined because the two representatives of the shared class both have value \(0\). It is a bijection: the two intervals cover \([-1,1]\), and their only overlap is the identified endpoint.
For \(U\subseteq X\cup_f Y\), the quotient criterion says that \(U\) is open exactly when \(h(U)\cap[-1,0]\) is relatively open in \([-1,0]\) and \(h(U)\cap[0,1]\) is relatively open in \([0,1]\). These two conditions are equivalent to \(h(U)\) being open in \([-1,1]\). Indeed, if \(h(U)\) is open, its intersections with the two subintervals are relatively open. Conversely, suppose both intersections are relatively open. At any point other than \(0\), relative openness on the relevant side supplies a neighborhood in \([-1,1]\) contained in \(h(U)\), after choosing it small enough not to reach \(0\). If \(0\in h(U)\), relative openness on each side supplies a positive radius on that side; the smaller of those two radii gives a neighborhood of \(0\) in \([-1,1]\) contained in \(h(U)\). Thus \(h(U)\) is open. It follows that \(h\) and its inverse preserve open sets, so this adjunction space is homeomorphic to \([-1,1]\).
Check: The class of \(0\) has exactly the two representatives \(0\in X\) and \(0\in Y\); all other classes are singletons. The openness test on the two components matches the usual subspace topology on their combined interval.
A Common Mistake and Why the Construction Matters
Common pitfall. It is incorrect to assume that each point of \(A\) is only paired with its image and that different points of \(A\) can never become identified. In the finite example, treating the instructions as unrelated pairs would suggest the overlapping groups \(\{p,a\}\) and \(\{p,b\}\). But equivalence classes must partition the disjoint union, and both groups contain \(p\). Transitivity forces \(a\sim p\sim b\), so the correct class is \(\{p,a,b\}\). The correct quotient has three points \(P,Q,C\), not four.
The adjunction construction separates two questions that are useful in topology. The attaching map answers which points are identified, including any extra identifications caused by repeated values of the map. The quotient topology answers which sets in the resulting space are open. Keeping these questions separate makes it possible to describe a glued space precisely before studying its further properties.
Check Your Understanding
For each question, identify the equivalence classes first, then use the quotient-topology criterion.
- Let \(X=\{p,q\}\) be discrete, \(Y=\{a,b\}\) be discrete, \(A=Y\), and \(f(a)=p,\ f(b)=q\). What are the quotient classes and quotient topology?
- In the interval endpoint example, is the subset consisting of \(P\) and \(q((1/4,3/4))\) open? Justify your answer using its preimage.
- If \(f:A\to X\) is injective, what is the class of a point \(x\in f(A)\)? What is the class of a point \(x\in X\setminus f(A)\)?
- In the finite example, explain why \(\{Q\}\) is not open but \(\{P,Q\}\) is open.
- For the two-interval example, describe the class of the shared endpoint and the classes of all other points.