From a Square to a Torus
A square becomes a torus when each pair of opposite sides is glued together in the direction that matches corresponding coordinates. The identifications do more than join four edges: they also bring all four corners to a single point. Keeping track of these classes gives a small cell structure, from which the Euler characteristic can be read directly.
Let \(Q=[0,1]\times[0,1]\) with its subspace topology from the Euclidean plane. Identify points by the equivalence relation generated by \((0,y)\sim(1,y)\) for \(0\le y\le1\), and \((x,0)\sim(x,1)\) for \(0\le x\le1\). “Generated by” means that we also identify any points forced to be equivalent by reflexivity, symmetry, and transitivity. Write \(Q/{\sim}\) for the set of equivalence classes, and \(q:Q\to Q/{\sim}\) for the map sending a point to its class.
The quotient is a space whose points are classes, not points of the original square. An interior point of \(Q\) is identified with no other point. A point in the interior of a side is identified with the corresponding point on the opposite side. The corners have more identifications because each lies on two sides. We will find each type of class explicitly before counting cells.
Finding the Vertex and Edge Classes
Worked Example: The Four Corners
Name the corners \(A=(0,0)\), \(B=(1,0)\), \(C=(1,1)\), and \(D=(0,1)\). The bottom-side identification gives \(A\sim B\), and the right-side identification gives \(B\sim C\). The top-side identification gives \(D\sim C\), while the left-side identification gives \(A\sim D\). Thus transitivity puts all four corners in one class, \([A]=\{A,B,C,D\}\).
Check: each corner is one of these four points, and the generating identifications never identify a corner with a noncorner: changing an endpoint coordinate from \(0\) to \(1\) leaves the other coordinate unchanged. So this class contains exactly the four corners.
Worked Example: Two Paired Edges
Take the left-side point \((0,\tfrac13)\). Its paired point is \((1,\tfrac13)\), so its class is exactly \(\{(0,\tfrac13),(1,\tfrac13)\}\). Take also the bottom-side point \((\tfrac13,0)\). Its paired point is \((\tfrac13,1)\), giving the class \(\{(\tfrac13,0),(\tfrac13,1)\}\). Neither class has further points: these coordinates are not corners, and each generating identification pairs a point on one side with the corresponding point on its opposite side.
As the height varies from \(0\) to \(1\), the left and right open sides form one edge class. As the horizontal coordinate varies from \(0\) to \(1\), the bottom and top open sides form another. The four corners are excluded from these open edges because they belong to the vertex class just found.
Check: the two sample classes are distinct. A generating identification on a vertical side preserves the \(y\)-coordinate and changes only the endpoint \(x\)-coordinate; one on a horizontal side preserves the \(x\)-coordinate and changes only the endpoint \(y\)-coordinate. Neither sequence of changes takes \((0,\tfrac13)\) to \((\tfrac13,0)\).
When the boundary is traversed counterclockwise, label the bottom edge \(a\) from left to right and the right edge \(b\) from bottom to top. The top edge is traversed from right to left, opposite to its matching bottom edge, and is labelled \(a^{-1}\). The left edge is traversed from top to bottom, opposite to its matching right edge, and is labelled \(b^{-1}\). The resulting boundary word is \(aba^{-1}b^{-1}\). This records two edge classes, each paired with reversed boundary traversal; it does not mean that there are four distinct quotient edges.
Checking the Quotient Is the Product of Two Circles
The coordinate-wise construction gives a direct way to recognize the quotient. Define \(F:Q\to S^1\times S^1\) by \(F(x,y)=(e^{2\pi i x},e^{2\pi i y})\). This map is continuous. It has the same value at points identified along either pair of opposite sides, since \(e^{2\pi i\cdot0}=e^{2\pi i\cdot1}=1\). Consequently \(F\) is constant on every equivalence class and defines a map \(g:Q/{\sim}\to S^1\times S^1\) with \(F=g\circ q\).
Proof. First, \(g\) is continuous. For any open set \(V\subseteq S^1\times S^1\), the definition \(F=g\circ q\) gives \(q^{-1}(g^{-1}(V))=F^{-1}(V)\), which is open in \(Q\). By the definition of the quotient topology, \(g^{-1}(V)\) is open in \(Q/{\sim}\).
Next, \(g\) is surjective: every point of \(S^1\) can be written \(e^{2\pi i t}\) for some \(t\in[0,1]\), so every pair of points of \(S^1\) is \(F(x,y)\) for some \((x,y)\in Q\). To check injectivity, suppose \(F(x,y)=F(x',y')\). Equality of the first coordinates implies \(x-x'\) is an integer. Since \(x,x'\in[0,1]\), either \(x=x'\) or the two coordinates are \(0\) and \(1\). The second coordinates likewise are equal or are the pair \(0,1\). These changes of endpoint coordinates are exactly the identifications generated by the paired sides, so \((x,y)\sim(x',y')\). Thus \(g\) is injective.
Finally, \(Q\) is compact by the Heine–Borel theorem. Its continuous image \(Q/{\sim}\) under \(q\) is compact: an open cover of the image pulls back to an open cover of \(Q\), and a finite subcover of \(Q\) gives a finite subcover of the image. The target \(S^1\times S^1\) is Hausdorff: distinct pairs differ in at least one coordinate, and disjoint neighborhoods of those coordinates give disjoint product neighborhoods. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Indeed, a closed subset of a compact space is compact, its continuous image is compact, and a compact subset of a Hausdorff space is closed. Therefore the bijection sends closed sets to closed sets, which makes its inverse continuous. Applying this result to \(g\) proves the claim. \(\square\)
The Cell Structure and Euler Characteristic
A cell is a piece of a cell decomposition homeomorphic to an open ball of its dimension; its attaching boundary may meet other cells. In the quotient, the square interior supplies one 2-cell, the two pairs of open sides supply two 1-cells, and the four corners supply one 0-cell. The interior of the square is not identified with any other point, so it gives one open face. The paired open sides give the two open edges described above, and their endpoints all lie at the single vertex class.
Worked Example: Counting Cells and Computing \(\chi\)
Let \(c_k\) be the number of \(k\)-dimensional cells in this decomposition. The corner calculation gives \(c_0=1\); the two opposite-side pairs give \(c_1=2\); and the square interior gives \(c_2=1\). There are no cells of dimension greater than \(2\). For a finite cell decomposition, the Euler characteristic is the alternating sum of the cell counts: \(\chi=c_0-c_1+c_2\). Therefore \(\chi(T^2)=1-2+1=0\).
Check: the signed contributions are \(+1\) for the one vertex, \(-2\) for the two edges, and \(+1\) for the one face. Adding them gives \(1+(-2)+1=0\).
The count depends on applying the identifications before counting cells. The square boundary has four sides in the diagram, but in the quotient those sides make two 1-cells. Likewise, the diagram has four corners, but all four represent one 0-cell. The open square is the single 2-cell. These are the counts for the quotient cell structure, not for the unglued square.
Euler characteristic is useful here because it compresses a cell structure into a single integer. It is a check on the bookkeeping, but it does not by itself identify a surface: different spaces can share the same Euler characteristic. For this construction, the product map establishes the torus model, while the cell count independently records the topology of the paired-square presentation.
Check Your Understanding
Use the paired-square identifications and the cell counts, rather than counting pieces in the unglued diagram.
- Which two points are equivalent to \((0,\tfrac25)\), and why are there no others in its class?
- Write the boundary word when the square is traversed counterclockwise, starting along the bottom edge.
- Explain why the four corners form one vertex class.
- How many 0-cells, 1-cells, and 2-cells does the quotient have?
- Compute the Euler characteristic from those cell counts and show the alternating sum.