From a Square to an Identification Space
To build a torus from a square, identify the left edge with the right edge, matching points at the same height, and identify the bottom edge with the top edge, matching points at the same horizontal position. The identifications are simple to state, but they generate classes that include more than the pairs listed on the edges: transitivity joins all four corners. We will describe every class and then define the map from the square to the resulting quotient space.
Let \(I=[0,1]\), and give the square \(I^2=I\times I\) the subspace topology inherited from \(\mathbb{R}^2\) with its standard topology. Write points as \((s,t)\), where \(s\) is the horizontal coordinate and \(t\) the vertical coordinate. We impose the identifications \((0,t)\sim(1,t)\) for every \(t\in I\), and \((s,0)\sim(s,1)\) for every \(s\in I\). These are the matching-direction edge pairings.
The relation \(\approx\) is an equivalence relation: equality is reflexive, symmetric, and transitive, and the only distinct points it relates are 0 and 1, which are related to each other. Applying it independently to both coordinates makes \(\sim\) an equivalence relation on \(I^2\). In particular, this definition includes the consequences of the edge pairings, not just the initial pairs.
Exactly Which Points Are Identified?
Proof. First, each prescribed edge pairing belongs to \(\sim\): the points \((0,t)\) and \((1,t)\) have equivalent first coordinates and equal second coordinates, while \((s,0)\) and \((s,1)\) have equal first coordinates and equivalent second coordinates. Since \(\sim\) is an equivalence relation, it contains the relation generated by those pairings.
For the reverse inclusion, suppose \((s,t)\sim(u,v)\). If \(s=u\), no horizontal-edge move is needed; if \(s\ne u\), then \(\{s,u\}=\{0,1\}\), and one prescribed left-right edge pairing changes \(s\) to \(u\) while leaving \(t\) fixed. After this, if \(t=v\), no vertical move is needed; if \(t\ne v\), then \(\{t,v\}=\{0,1\}\), and one prescribed bottom-top pairing changes \(t\) to \(v\) while leaving the first coordinate fixed. Hence \((s,t)\) can be changed to \((u,v)\) by the given pairings, so it belongs to the generated relation.
Finally, the only possible representatives of a coordinate \(r\in I\) are the elements of \(E(r)\). The two coordinates can be changed independently by the pairings, so every combination in \(E(s)\times E(t)\) lies in the class of \((s,t)\), and no other point does. If neither coordinate is an endpoint, both sets \(E(s)\) and \(E(t)\) are singletons. If exactly one coordinate is an endpoint, one factor has two elements and the other has one. If both coordinates are endpoints, each factor has two elements, giving four representatives. \(\square\)
Worked Example: An Edge Point and a Corner
Find the equivalence classes of \((0,\tfrac25)\) and \((0,0)\).
For \((0,\tfrac25)\), the first coordinate is an endpoint, so \(E(0)=\{0,1\}\). The second coordinate satisfies \(0<\tfrac25<1\), so \(E(\tfrac25)=\{\tfrac25\}\). Therefore \[ [(0,\tfrac25)] =E(0)\times E(\tfrac25) =\{(0,\tfrac25),(1,\tfrac25)\}. \] The height remains \(\tfrac25\); the edge pairing changes only the horizontal coordinate.
For \((0,0)\), both coordinates are endpoints. Thus \[ [(0,0)] =E(0)\times E(0) =\{(0,0),(0,1),(1,0),(1,1)\}. \] The four corners are in one class: the bottom-top pairing relates \((0,0)\) to \((0,1)\), and the left-right pairing relates \((0,0)\) to \((1,0)\) and \((0,1)\) to \((1,1)\).
Check: the first class has \(2\cdot1=2\) points, and the second has \(2\cdot2=4\) points, as the class formula predicts.
The quotient map is the function sending each point of the square to its equivalence class. Its fibers—the preimages of individual quotient points—are exactly the classes just computed. Distinct points of the square can therefore have the same image under \(q\), but two points have the same image precisely when they are identified by the edge pairings and their consequences.
Proof. The map is surjective because every element of \(I^2/{\sim}\) is, by definition, an equivalence class \([(s,t)]\) for some \((s,t)\in I^2\), and \(q(s,t)=[(s,t)]\). The fiber formula follows from the definition of an equivalence class and the theorem above.
By the quotient topology, \(W\subseteq I^2/{\sim}\) is open if and only if \(q^{-1}(W)\) is open in \(I^2\). This condition implies that \(q\) is continuous: for every open \(W\) in the quotient, its preimage is open. It also gives the defining test for a quotient map: a subset of the target is open whenever its preimage is open. The map is therefore a quotient map. \(\square\)
Saturation and Images of Subsets
A subset \(A\subseteq I^2\) is saturated if it contains every equivalence class that meets it. Equivalently, it is saturated when \(A=q^{-1}(q(A))\). For an arbitrary subset \(A\), the set \(q^{-1}(q(A))\) is its saturation: it contains \(A\), and it adds precisely the points identified with points of \(A\).
The quotient topology gives a practical openness test. A subset \(q(A)\) of the quotient is open if and only if its preimage \(q^{-1}(q(A))\)—the saturation of \(A\)—is open in the square. It is the saturation, rather than \(A\) itself, that must be tested.
Worked Example: Saturating a Segment
Let \(A=\{0\}\times[\tfrac13,\tfrac23]\), the closed vertical segment on the left edge with heights from \(\tfrac13\) to \(\tfrac23\). Determine \(q^{-1}(q(A))\).
Each point \((0,t)\in A\) has class \(\{(0,t),(1,t)\}\), since \(\tfrac13\le t\le\tfrac23\) implies \(0<t<1\). Taking the union of these classes gives \[ q^{-1}(q(A)) =\bigl(\{0\}\times[\tfrac13,\tfrac23]\bigr) \cup \bigl(\{1\}\times[\tfrac13,\tfrac23]\bigr). \] There are no additional points at the corners, because neither endpoint height 0 nor endpoint height 1 belongs to the interval \([\tfrac13,\tfrac23]\).
This saturation is not open in \(I^2\). For example, it contains \((0,\tfrac12)\), but every open neighborhood of \((0,\tfrac12)\) in the subspace topology contains points of the square with first coordinate strictly between 0 and 1. Those points are not in the displayed saturation. The quotient openness test therefore shows that \(q(A)\) is not open.
Check: the displayed set contains both representatives of every point of \(A\)’s image, so it is saturated. Neither endpoint of the chosen height interval is a corner height, so no other edge or corner representatives occur.
Worked Example: An Open Set in the Square
Let \(A=(\tfrac14,\tfrac34)\times(\tfrac14,\tfrac34)\), regarded as a subset of \(I^2\). Determine its saturation and decide whether \(q(A)\) is open.
Every point \((s,t)\in A\) has both coordinates strictly between 0 and 1. Its class is therefore the singleton \(\{(s,t)\}\). Consequently, \[ q^{-1}(q(A))=A. \] The set \(A\) is open in \(I^2\): it is the intersection of the open rectangle \((\tfrac14,\tfrac34)\times(\tfrac14,\tfrac34)\) in \(\mathbb{R}^2\) with \(I^2\). Thus its saturation is open, and the quotient openness test implies that \(q(A)\) is open in the quotient.
Check: no point of \(A\) lies on an edge, so no edge pairing adds a representative outside \(A\); the saturation is exactly \(A\), which is open in the stated subspace topology.
A Common Pitfall: Counting the Corners
A tempting but incorrect count treats the four corners as two separate pairs: the bottom-left corner is paired with the bottom-right corner, and the top-left corner with the top-right corner. This overlooks the other edge pairing. The bottom-top identification also relates the bottom-left corner to the top-left corner. Since equivalence classes must be transitive, all four corners belong to the same class.
Likewise, it is incorrect to say that every boundary point has exactly two representatives. The class formula shows why the qualification matters: an edge point away from the corners has two representatives, but a corner has four. Interior points have just one. These counts concern representatives in the square; the quotient space has one point for each entire class.
The quotient map records the construction precisely. It is surjective, identifies exactly the points required by the edge pairings, and equips the set of classes with the topology determined by preimages in the square. This makes it possible to study the glued space without treating the square’s edges as still distinct. The matching directions matter: the identifications used here are the ones specified above, and the resulting classes follow from those identifications together with equivalence and transitivity.
Check Your Understanding
Use the class formula and the quotient topology to answer the following.
- List every representative of \((\tfrac12,1)\)’s equivalence class.
- How many representatives does the class of \((1,1)\) have? List them.
- For \(A=[\tfrac14,\tfrac34]\times\{0\}\), determine \(q^{-1}(q(A))\).
- State the necessary and sufficient condition for \(q(A)\) to be open in the quotient, in terms of \(A\).
- Explain why the left-right pairing alone does not give the equivalence class of a corner under both edge pairings.