The Klein bottle as a square quotient

The Klein bottle is the quotient $K$ of the unit square $Q=[0,1{]}^2$ by

$(x,0)\sim (x,1),(0,y)\sim (1,1-y)$

with the quotient topology (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection). The named surface uses the convention for topological manifolds without boundary in Topological manifolds without boundary: Hausdorff, second-countable, and locally Euclidean spaces.

The four corners form one vertex class: the horizontal pairing identifies $(0,0)$ with $(0,1)$ and $(1,0)$ with $(1,1)$, while the vertical pairing identifies $(0,0)$ with $(1,1)$ and $(0,1)$ with $(1,0)$. There are two edge classes and one face. The four corner sectors join, under the paired edge germs, in the cycle $(00)-(01)-(10)-(11)-(00)$: the bottom-top pairing joins $(00)$ to $(01)$ and $(10)$ to $(11)$, while the left-right pairing joins $(00)$ to $(11)$ and $(01)$ to $(10)$. Thus a neighborhood of the vertex is a disk. Interior points already have disk neighborhoods, and the two half-disks at an edge interior join to a disk. Thus the quotient has no boundary.

The square $Q$ is compact by Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line. The quotient is compact as the continuous image of $Q$ and connected as the image of a connected square. Its edge-pairing relation is a finite union of the diagonal, closed graphs of the two affine edge maps, and finitely many vertex pairs, hence is closed in $Q\times Q$. The product $Q\times Q=[0,1{]}^4$ is compact by the same Heine–Borel theorem. If $q:Q\to K$ is the quotient map, then for closed $F\subset Q$ the saturation $q^{-1}(q(F))$ is the projection of that closed relation intersected with $F\times Q$; it is compact and therefore closed in $Q$ (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones). The quotient topology makes $q(F)$ closed. Distinct equivalence classes are disjoint compact subsets of the Hausdorff square, so they have disjoint open neighborhoods $U,V$ (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones). The open sets $K\setminus q(Q\setminus U)$ and $K\setminus q(Q\setminus V)$ separate their quotient points, proving Hausdorffness. A countable basis for $Q$ gives a countable basis for $K$ by taking, for each finite union $B$ of basis members, $K\setminus q(Q\setminus B)$: these are open, and compactness of each fiber shows they refine every quotient-open neighborhood. (Each fiber is closed in the compact square, hence compact.) The definition and this finite verification use no full choice axiom; any finite selection is covered by finite choice.

Orient the square boundary counterclockwise. The bottom and top sides form the oppositely traversed $a,a^{-1}$ pair; the right and left sides form the equally traversed $b,b$ pair. The one-polygon word is therefore $ab{a}^{-1}b$. The later example checks its orientability and Euler characteristic directly.