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The Timpson Lyceum · General Topology (Point Set Topology)

Product and Quotient Space Topologies – Lecture 13 (Topology)

Product and Quotient Space Topologies – Lecture 13

Overview

This lecture continues the program of building new topologies out of old ones. It first finishes the product topology — reviewing its basis, warning about a common cross-product trap, and proving three theorems about how interiors, subspaces, and closed sets behave under products. It then introduces the quotient topology, the construction that lets us "glue" a space to itself to form new objects. We prove that a quotient map really does induce a topology, and we develop intuition through gluing examples that produce the circle, annulus, torus, Möbius band, Klein bottle, projective plane, and connected sums.

Key Concepts

Product topology. Given topological spaces (X,TX)(X,\mathcal{T}_X) and (Y,TY)(Y,\mathcal{T}_Y), the product topology on X×YX \times Y is generated by the basis B={ U×V:U∈TX,  V∈TY }.\mathcal{B} = \{\, U \times V : U \in \mathcal{T}_X,\; V \in \mathcal{T}_Y \,\}. A basis element is an open set of XX crossed with an open set of YY; every open set of X×YX \times Y is a union of such boxes.

Basis, not all open sets. Crossing two open sets always yields an open set, but the converse fails: a general open set of X×YX \times Y need not be expressible as one open box U×VU \times V. For example (A1×B1)∪(A2×B2)(A_1 \times B_1) \cup (A_2 \times B_2) is open yet cannot be written as a single product. (One may also restrict the generators to products of basis elements of XX and YY — these still generate the same product topology.)

Cross-product algebra (the trap). Cross products distribute over intersection but not over union: (A1∩A2)×(B1∩B2)=(A1×B1)∩(A2×B2),(A_1 \cap A_2)\times(B_1\cap B_2) = (A_1\times B_1)\cap(A_2\times B_2), whereas (A1∪A2)×(B1∪B2)(A_1\cup A_2)\times(B_1\cup B_2) is strictly larger than (A1×B1)∪(A2×B2)(A_1\times B_1)\cup(A_2\times B_2) — the union of products misses the "off-diagonal" corner pieces.

Interior of a product. For A⊆XA\subseteq X, B⊆YB\subseteq Y: Int⁡(A)×Int⁡(B)=Int⁡(A×B).\operatorname{Int}(A)\times\operatorname{Int}(B)=\operatorname{Int}(A\times B). The interior of a set is the union of all open subsets it contains (its largest open subset).

Subspace vs. product agree. Topologizing A⊆XA\subseteq X and B⊆YB\subseteq Y two ways gives the same topology on A×BA\times B: (1) take the subspace topology induced from the product X×YX\times Y, or (2) take the product of the subspace topologies on AA and BB. These coincide because (U×V)∩(A×B)=(U∩A)×(V∩B)(U\times V)\cap(A\times B)=(U\cap A)\times(V\cap B).

Product of closed sets is closed. If CC is closed in XX and DD is closed in YY, then C×DC\times D is closed in X×YX\times Y. The proof shows the complement is open by rewriting it as ((X∖C)×Y)∪(X×(Y∖D))\big((X\setminus C)\times Y\big)\cup\big(X\times(Y\setminus D)\big), a union of open boxes.

Quotient topology. Let (X,T)(X,\mathcal{T}) be a space, AA any set, and p:X→Ap:X\to A a surjective map (the quotient map). Define V⊆AV\subseteq A to be open exactly when its preimage is open in XX: V open in A  ⟺  p−1(V)∈T.V \text{ open in } A \iff p^{-1}(V)\in\mathcal{T}. The resulting collection TA\mathcal{T}_A is the quotient topology induced by pp. Different quotient maps on the same X,AX,A can induce different topologies. (Recall: p(x)=yp(x)=y makes yy the image of xx; p−1(y)p^{-1}(y) is the preimage, the set of all points mapping to yy — possibly several.)

Quotient by a partition / gluing. The maps of interest send XX onto a partition X∗X^* of XX (a family of disjoint subsets whose union is XX). Each point lands in the one block containing it. Identifying points this way is "gluing." Two spaces are homeomorphic (the same, topologically) when there is a continuous bijection both ways — rubber-sheet deformations (stretching, never tearing or new gluing) preserve the type. Standard gluings of the filled square I×II\times I:

Connected sum. To "add" two surfaces S1#S2S_1 \mathbin{\#} S_2: cut an open disk (a hole) out of each, then glue the two boundary circles together. Torus #\# torus yields the genus-2 (two-holed) surface. Cutting out an open set adds a new boundary edge; we always remove open sets so the boundary points survive to be glued.

Worked Example

Theorem. In the product topology, Int⁡(A)×Int⁡(B)=Int⁡(A×B)\operatorname{Int}(A)\times\operatorname{Int}(B)=\operatorname{Int}(A\times B).

Proof. Show each side contains the other.

(⊆\subseteq) Since Int⁡(A)\operatorname{Int}(A) is open in XX and Int⁡(B)\operatorname{Int}(B) is open in YY, their product is a basic open set of X×YX\times Y, and it sits inside A×BA\times B. As Int⁡(A×B)\operatorname{Int}(A\times B) is the union of all open subsets of A×BA\times B:

Int⁡(A)×Int⁡(B) is open and ⊆A×B,⟹Int⁡(A)×Int⁡(B) ⊆ Int⁡(A×B).\begin{aligned} \operatorname{Int}(A)\times\operatorname{Int}(B)\ &\text{is open and } \subseteq A\times B,\\ \Longrightarrow\quad \operatorname{Int}(A)\times\operatorname{Int}(B)\ &\subseteq\ \operatorname{Int}(A\times B). \end{aligned}

(⊇\supseteq) Take any point (x,y)∈Int⁡(A×B)(x,y)\in\operatorname{Int}(A\times B). This interior is open, so some basic box contains the point inside A×BA\times B:

∃ U∈TX, V∈TY:(x,y)∈U×V⊆A×B,so x∈U⊆A ⇒ x∈Int⁡(A),y∈V⊆B ⇒ y∈Int⁡(B),⟹(x,y)∈Int⁡(A)×Int⁡(B).\begin{aligned} \exists\,U\in\mathcal{T}_X,\ V\in\mathcal{T}_Y:\quad & (x,y)\in U\times V\subseteq A\times B,\\ \text{so } x\in U\subseteq A \ \Rightarrow\ x\in\operatorname{Int}(A),\quad & y\in V\subseteq B \ \Rightarrow\ y\in\operatorname{Int}(B),\\ \Longrightarrow\quad (x,y)\in \operatorname{Int}(A)\times\operatorname{Int}(B). \end{aligned}

Every point of Int⁡(A×B)\operatorname{Int}(A\times B) lies in Int⁡(A)×Int⁡(B)\operatorname{Int}(A)\times\operatorname{Int}(B), giving the reverse inclusion. Both inclusions hold, so the sets are equal. ■\blacksquare

Summary