2.2 Gluing Spaces Together · Homology & Cohomology

A mathematician named Klein
Thought the Möbius band was divine.
Said he: ‘If you glue
The edges of two,
You’ll get a weird bottle like mine.’

— Leo Moser, a limerick

In the arcade game Asteroids, a little spaceship flies off the right-hand edge of the screen and reappears on the left; it leaves through the top and comes back through the bottom. The game never says so, but the ship lives on a doughnut. What put it there is gluing: declaring that certain points are to be regarded as the same. Take a single square, decide which edges to glue and which way round, and out come a cylinder, a Möbius band, a torus, a sphere, a Klein bottle and a projective plane. Two of them refuse to be built in our three-dimensional world without passing through themselves.

Gluing is the topological version of an idea you have met before. In §1.2 Equivalence and Quotients we glued the two ends of an interval together — as a set — by declaring $0 \sim 1,$ and called the result a circle. Now that we know what a topological space is (§2.1 Spaces and Continuity), we can finish the job: give the glued set a topology, so that it really is a circle, and then do the same in two dimensions, where far stranger things can happen.

A screen that wraps around

Fly the ship in Figure 2.2.1 (it will fly itself until you take the controls). Leave through the right edge and you reappear on the left, at the same height. Leave through the top and you come back through the bottom, at the same distance from the left. As far as the pilot is concerned, there are no edges at all. The universe is finite — its whole area is one screen — yet it has no boundary anywhere: wherever the ship is, space continues in every direction.

Flight on a glued square Arrow keys or the buttons to steer · try all three gluings

Show neighbouring copies

Edges crossed: 0
The ship is the right way round.

Figure 2.2.1 The arrows on the edges say which edges are glued, and which way round. In torus mode, opposite edges are glued straight across; the faint neighbouring copies show what the pilot would see — the same square repeated forever. In the other two modes some gluings flip the ship over, and it comes back as its own mirror image.

What shape is this universe? Picture the screen as a sheet of rubber. Gluing the top edge to the bottom edge rolls the sheet into a tube. Gluing the left edge to the right edge then bends the tube round until its two end circles meet. The result is the surface of a doughnut, the torus. The ship’s straight flight paths become loops wound around the doughnut.

There is another way to see the same thing, and the neighbouring copies in the figure show it. Instead of gluing, imagine the square repeated in every direction, like tiles on a floor, with a copy of the ship in every tile. Looking right, the pilot sees the back of their own ship one tile away. This tiled picture is often the easiest way to think about straight lines on a torus, and it is how a flat torus appears from the inside.

The other two modes of the game change the gluing instructions slightly — crossing certain edges now flips you over. We will come back to them once we have the language to say what is going on.

Gluing, made precise: the quotient topology

Half of the work is already done: §1.2 Equivalence and Quotients glued sets, and we only need to remember how. To glue, we choose an equivalence relation $\sim$ on a space $X :$ $x \sim y$ means “ $x$ and $y$ are to become the same point”. The glued set is the quotient set $X / \sim,$ whose elements are the equivalence classes $[x]$ (read “the class of $x ”:$ the pile of all points glued to $x).$ And there is the projection

q : X \to X / \sim, q (x) = [x],

which sends every point to its pile ( $q$ for “quotient”). Gluing the ends of the interval $X = [0, 1]$ means declaring $0 \sim 1$ and nothing else. Then the class $[0] = [1] = {0, 1}$ has two points, every other class has just one, and $X / \sim$ is a set that should be a circle.

But a set is not yet a space. Which subsets of $X / \sim$ should be open? There is a natural answer: look back at what the set was before the glue was applied.

In words: a set of glued points is open if, before the gluing, the same points formed an open set. The preimage does the looking back, exactly as in the definition of continuity — another reversed arrow. Figure 2.2.2 shows the rule at work on the interval; Munkres (2000, §22) develops the whole theory, with every proof.

Looking back before the glue Play or scrub the gluing · compare the two sets

interval circle

Before gluing, this set is $[0, ε) \cup (1 - ε, 1]$ : open in $[0, 1]$ , since every point of it has some room inside $[0, 1]$ . So after gluing it is open — the glued point has room on both sides.

Figure 2.2.2 A little arc around the glued point is open, because before gluing it was

[0, ε) \cup (1 - ε, 1],

two pieces each with room inside

[0, 1] .

A one-sided arc that ends at the glued point is not open: before gluing it was

[0, ε) \cup {1},

and the lonely point

1

has no room at all. The quotient topology agrees with our eyes.

Two facts make the quotient topology the right one.

Continuity of $q$ is immediate: preimages of open sets are open by definition. And we could not declare any more sets open without some preimage failing to be open. Gluing should be continuous — continuous maps are allowed to glue, as we saw in §2.1 Spaces and Continuity — but beyond that, the glued space should have as many open sets as possible, so that it does not get blurrier than it needs to be.

This is the topological version of the “well-defined” principle from §1.2 Equivalence and Quotients: a continuous function on a glued space is the same thing as a continuous function on the original space that gives glued points the same value. Functions that respect the glue pass through it unharmed.

The quotient $[0, 1] / \sim$ is compact (a continuous image of the compact interval, via $q)$ and the circle is Hausdorff (it is a metric space), so $g$ is a homeomorphism: $[0, 1] / (0 \sim 1) ≅ S^{1} .$ The same theorem, applied again and again, tells us that the surfaces we glue in this chapter really are the surfaces we draw. The failure in Figure 2.1.6 came from the half-open interval $[0, 1),$ which is not compact.

Glue spreads

In practice we rarely list the whole equivalence relation. We give a few gluing instructions — “glue this edge to that edge” — and let them generate the rest, exactly as in §1.2 Equivalence and Quotients: the equivalence relation generated by some pairs is the smallest one containing them. The important consequence is that glue spreads by transitivity. If $p$ is glued to $q$ and $q$ is glued to $r,$ then $p$ is glued to $r,$ even though no instruction mentioned them together. This is why, when we glue the edges of a square in a moment, the corners can end up glued in surprising ways.

Gluing diagrams: a square with arrows

A gluing diagram is a polygon whose edges carry letters and arrows. The rule is simple: edges with the same letter are glued, matching the arrows — tail to tail, head to head — so that a point a quarter of the way along one arrow is glued to the point a quarter of the way along the other. Edges without a letter are left free; they will form the boundary of the surface.

Here is the torus. Its bottom and top edges, both called $a,$ point to the right; its left and right edges, both called $b,$ point up.

The torus as a square

Figure 2.2.3 The square is

[0, 1] \times [0, 1],

with

(x, y)

meaning “

x

across,

y

up”. The arrows say: glue

(x, 0)

(x, 1)

and

(0, y)

(1, y) .

These are exactly the flight rules of torus mode in Figure 2.2.1.

Two pieces of notation make diagrams easy to talk about. First, the edge word: start at the bottom-left corner and walk once around the square anticlockwise. Write down each letter you pass, as $x$ if you walk along its arrow and as $x^{- 1}$ (“ $x$ inverse”) if you walk against it. For the torus: along the bottom $a$ (with the arrow), up the right side $b$ (with the arrow), back along the top $a^{- 1}$ (against it), down the left side $b^{- 1}$ (against it). The torus is $ab a^{- 1} b^{- 1} .$ The word records the whole diagram in one line.

Second, the corners. Gluing edges glues their end points, so corners get glued too — and glue spreads. In the torus, the bottom-left corner is the tail of the bottom $a,$ so it is glued to the tail of the top $a,$ the top-left corner. Both are also ends of the $b$ edges… follow the arrows and you will find that all four corners become a single point. Counting these corner classes will matter a great deal later, when we compute the Euler characteristic and homology.

The gluing workshop

Figure 2.2.4 is the heart of this chapter. Pick a gluing; the square on the left shows the diagram, its equivalence relation, its word and its corner classes. On the right, a rubber sheet with the same grid and the same coloured edges bends — never tearing — into the glued surface. Watch for the moment when edges of the same colour meet: their chevrons line up, because the arrows match.

The gluing workshop Pick a surface · play or scrub the gluing · drag the picture to turn it

Rule: $(x, 0) \sim (x, 1),$ $(0, y) \sim (1, y)$
Word: $ab a^{- 1} b^{- 1}$
Corners: 1 point
In 3D?: fits without crossing itself

The flat square

square torus

Figure 2.2.4 Six surfaces from one square. Gold edges are called

a,

teal edges

b;

dashed ivory edges are free and become the boundary. Corners of the same colour end up at the same point. The Klein bottle and the projective plane cannot be finished without the sheet passing through itself; the crossing is drawn in rose.

One pair of edges: the cylinder and the Möbius band

Glue just the left and right edges, straight across: $(0, y) \sim (1, y) .$ The square rolls into a cylinder (the product $S^{1} \times [0, 1]$ of §2.1 Spaces and Continuity). The top and bottom edges stay free and become two boundary circles.

Now glue them with a twist: $(0, y) \sim (1, 1 - y),$ so the arrow on the right points down. To match the arrows, the sheet must turn over before its ends meet — try it with a strip of paper, giving one end half a turn before taping. The result is the Möbius band. Watch the free edges: the top edge runs into the bottom edge, and the two together form a single boundary circle. A Möbius band has one edge — and, if you run a finger along its surface, only one side. (One-sidedness has an intrinsic cousin, non-orientability, which a creature living inside the band could detect without ever leaving it; it is studied properly in §2.4 Manifolds and Surfaces.)

The corners show the difference too. Both surfaces have two corner classes, but paired differently: in the cylinder the two bottom corners are glued to each other, and so are the two top corners; in the Möbius band each bottom corner is glued to the top corner diagonally opposite.

In the workshop the sheet stretches as it rolls. Topology does not mind: in the words of Poincaré quoted at the start of §0.1 The Shape of a Question, “the proportions may be grossly altered”. Real paper does mind. You cannot make a Möbius band from a square of paper without crumpling it; the strip must be long and narrow. How long it must be was an open question for almost fifty years.

Conjecture The shortest paper Möbius band

Proved in 2023 Posed by Benjamin Halpern and Charles Weaver, 1977; proved by Richard Evan Schwartz, 2023

A strip of paper of width $1$ and length $λ$ can be bent into a Möbius band, without stretching, tearing or passing through itself, only if $λ > 3 \approx 1.73,$ and the bound $3$ cannot be improved.

This is a question of geometry, not topology: every strip glued with a half twist gives the same topological Möbius band, and only lengths decide whether paper can do it.

How it was settled. Schwartz posted his proof in August 2023, and it appeared in the Annals of Mathematics in 2025. A paper band bends only along straight lines, and he found a pair of these lines forced to meet in a T shape; comparing that T with triangles gives $λ > 3 .$

Before and after. Bands that come close to the bound look more and more like a folded, flattened triangle, the triangular Möbius band, which is not a genuine band. If the paper may pass through itself, the threshold drops to $π /2,$ as Halpern and Weaver showed.

A way in. Fold one. Take a strip $3$ cm wide and $5.2$ cm long, a ratio just over $3;$ fold it twice at $60°$ so that it flattens into an equilateral triangle, and tape the ends. Then take a strip twice as long as it is wide and make a genuine, slightly puffy band.

(Schwartz 2025; Halpern & Weaver 1977) Status as of October 2026 All open problems

Both pairs: the torus and the Klein bottle

Gluing both pairs straight across gives the torus $ab a^{- 1} b^{- 1} :$ roll the square into a tube (gluing the $b$ edges), then bend the tube round (gluing the end circles $a).$ In the finished doughnut the gold edge $a$ has become a loop around the tube and the teal edge $b$ a loop around the hole, crossing at the single corner point.

Now reverse just one arrow: let the top edge point left. The rule becomes $(x, 0) \sim (1 - x, 1)$ together with $(0, y) \sim (1, y),$ and the word becomes $aba b^{- 1} .$ This is the Klein bottle $K .$ Roll the square into a tube as before. Now the two end circles must be glued, but with their directions reversed. Bending the tube round as for the torus would match them the wrong way. The only way to make the arrows agree is to bring one end inside the tube and meet the other end from within — and in our three-dimensional space, getting inside means passing through the tube’s own wall. The finished picture is the familiar bottle: stood upright, its narrow neck curls over and dips through the side wall, then opens out to join the base from inside. In the workshop, the rose circle is where the neck passes through.

Go back to Figure 2.2.1 and switch to Klein bottle mode. Fly through the top edge: you come back through the bottom, but mirror-reversed — the rose stripe that was on your left wing is now on your right. The space itself has turned you over. Fly through the top edge again and you are restored. All four corners of the Klein bottle square are glued to one point, just as for the torus.

Neighbouring edges: the sphere

A gluing diagram may also glue edges that meet at a corner. Glue the bottom edge to the left edge, and the top edge to the right edge, as in the envelope-shaped word $ab b^{- 1} a^{- 1} :$ $(x, 0) \sim (0, x)$ and $(x, 1) \sim (1, x) .$ Fold the square along the diagonal from its bottom-left to its top-right corner: the two triangles lie on top of each other, the bottom edge on the left edge and the right edge on the top. Zip the matching edges together and inflate, and you get a sphere $S^{2},$ with the zip running from pole to pole. This time the corners form three classes: the two corners on the fold stay separate (they become the poles), while the other two are glued together, halfway along the zip.

The same idea works with fewer edges. A polygon with only two edges, $a$ and then $a^{- 1}$ — picture a coin purse, or a lens shape whose two curved edges are zipped together — is also a sphere. Its word is $a a^{- 1} .$

Twisting both pairs: the projective plane

Finally, reverse both: $(x, 0) \sim (1 - x, 1)$ and $(0, y) \sim (1, 1 - y),$ with word $abab .$ Notice what these rules do: every point of the boundary is glued to the point diametrically opposite it, through the centre of the square. Round the square off into a disk and the description shrinks to a single line: a disk with each point of its rim glued to the opposite point of the rim. This is the real projective plane $RP^{2} .$ Its corners form two classes.

The projective plane has another life, which explains its name. Every line through the centre of a ball pierces the ball’s surface in two opposite points. So the set of lines through a point of space is the sphere with opposite points glued. Most lines are already accounted for by the upper half of the sphere, which they pierce exactly once; only the horizontal lines, which pierce the equator twice, still need their two points glued. That is our disk with opposite rim points glued.

The same space turns up in perspective drawing, where parallel railway lines meet at a vanishing point on the horizon. Give the plane one such “point at infinity” for every direction, and squash the whole plane into an open disk, so that the points at infinity form its rim. A line runs off to infinity in two opposite directions, yet it has only one vanishing point — so opposite rim points must be glued, and the result is $RP^{2}$ once more.

In the workshop, the disk grows into a famous surface called Boy’s surface. It crosses itself along three rose loops, which meet at a single triple point where three sheets pass through each other. Fly in projective-plane mode in Figure 2.2.1: every edge you cross flips you over.

Surface	Rule	Word	Corners	In 3D
Cylinder	$(0, y) \sim (1, y)$	—	2	yes
Möbius band	$(0, y) \sim (1, 1 - y)$	—	2	yes
Torus $T^{2}$	straight, straight	$ab a^{- 1} b^{- 1}$	1	yes
Klein bottle $K$	flip, straight	$aba b^{- 1}$	1	no
Sphere $S^{2}$	neighbours	$ab b^{- 1} a^{- 1}$	3	yes
Projective plane $RP^{2}$	flip, flip	$abab$	2	no

Why the Klein bottle will not fit in our space

The torus sits comfortably in ordinary space; the Klein bottle and the projective plane do not. Every picture of them crosses itself somewhere. Is that our fault, for not being clever enough, or is it a fact about the surfaces? To answer, we need two words for “putting a space into another space”.

The pictures in the workshop are immersions. Each little patch of the Klein bottle looks just like a patch of a smooth surface, but the neck passes through the wall. The crossing is a defect of the picture, not of the surface: an ant living on the Klein bottle would never notice it, any more than the pilot in the game notices an edge.

We will not prove the first statement here, but here is the idea. A closed surface sitting in $R^{3}$ without crossing itself always divides space into an inside and an outside. (This is the Jordan–Brouwer separation theorem, a three-dimensional relative of the fact that a loop drawn in the plane without crossing itself has an inside and an outside.) A surface with an inside and an outside has two sides: you could paint the inside red and the outside blue. But the Klein bottle and the projective plane contain Möbius bands — look at the vertical strip in the middle of the Klein bottle’s square, glued with a flip — so they have only one side, and cannot have an inside and an outside. The full proof, when it comes, is startlingly algebraic: the $Z /2$ in the Klein bottle’s homology, promised above, is exactly what forbids it a place in ordinary space (Hatcher 2002, Cor. 3.46).

The second statement has a lovely picture. A fourth dimension means a fourth number attached to every point. Show it as a colour. In Figure 2.2.5, turn on the colour for one of the Klein bottles: at the rose curve, the two sheets that cross have different colours. In four dimensions they are at different “heights” and do not meet at all, just as a bridge does not meet the road it crosses, although on a map the two lines intersect.

Surfaces that must cross themselves Pick a surface · drag to turn it · try the fourth coordinate

The neck passes through the wall along the rose circle.

Colour = a fourth coordinate

Figure 2.2.5 Two pictures of the Klein bottle and two of the projective plane, each with its double curve in rose. Boy’s surface is a smooth immersion with a single triple point; the cross-cap is simpler to describe but has two pinch points, where it fails to be smooth. For the two Klein bottles, colour can show a fourth coordinate: the sheets crossing at each rose point get different colours, and in four dimensions they never touch.

A little history Bottles, bands and Boy

Felix Klein 1849 – 1925

David Hilbert 1862 – 1943

August Möbius and Johann Listing found the one-sided band independently in 1858; Listing published first, but the band carries Möbius’s name. Felix Klein described his bottle in 1882, the year his health collapsed in a race with Poincaré, in words rather than pictures: take a piece of rubber tubing, turn it inside out, and let it pass through itself so that when the ends are bent together the outside meets the inside (Klein 1882, §23). Klein brought David Hilbert to Göttingen in 1895. In 1900 Hilbert set mathematics twenty-three problems for the new century, and he gave his doctoral student Werner Boy the task of proving that the projective plane cannot be immersed in space without singular points. In 1901 Boy found an immersion instead (Boy 1903). Our formula for Boy’s surface is much younger: it comes from Robert Bryant and Rob Kusner in the 1980s (Kusner 1987).

Wedges, cones, suspensions and collapsing

Squares are not the only things we can glue. Four other gluings come up so often in topology that they have names, and each answers a plain request: join two spaces at a point; crush part of a space to a point; cap a cylinder at one end; cap it at both. First, a way to put two spaces side by side.

The wedge of two circles, $S^{1} \lor S^{1},$ is the figure eight: two loops sharing one point. It is not homeomorphic to a circle — removing the shared point leaves two pieces, while removing any point of a circle leaves one (a cut point argument from §2.1 Spaces and Continuity). The wedge $S^{2} \lor S^{1}$ is a balloon with a loop of string attached at one point.

The most important example: take a disk $D^{2}$ and collapse its boundary circle, written $\partial D^{2}$ (the symbol $\partial,$ “boundary”, will become one of the heroes of this book). Picture a cloth bag with a drawstring around its rim: pull the string tight, so the whole rim shrinks to a point, and the bag closes up into a ball. So

D^{2} / \partial D^{2} ≅ S^{2} .

The same is true in every dimension, with $D^{n}$ the solid ball and $S^{n}$ the sphere bounding a ball in one dimension higher: $D^{n} / \partial D^{n} ≅ S^{n} .$ Collapsing the rim of a segment $[0, 1]$ (its two end points) gives a circle, the case $n = 1$ — our very first gluing. Figure 2.2.6 shows the disk and four more collapses.

Collapsing a subspace Pick an example · play or scrub the collapse · drag to turn

Collapsing the boundary circle of the disk:

D^{2} / \partial D^{2} ≅ S^{2}

as is collapsed

Figure 2.2.6 In each example the gold set

A

shrinks to a single point. Collapsing the rim of a disk gives a sphere. Collapsing the top circle of a cylinder gives a cone; collapsing both ends gives a sphere again. Collapsing the equator of a sphere pinches it into two spheres touching at a point,

S^{2} \lor S^{2} .

Collapsing one circle around the tube of a torus gives a “pinched torus”, which is a sphere with its north and south poles glued together.

The cone on a circle is an ice-cream cone, which is homeomorphic to a disk: $C S^{1} ≅ D^{2} .$ The suspension of a circle is a double cone, which is homeomorphic to a sphere: $Σ S^{1} ≅ S^{2} .$ The suspension of two points $S^{0}$ — two segments with their tops glued and their bottoms glued — is a circle. In general $Σ S^{n} ≅ S^{n + 1} :$ suspension climbs up the dimensions. Every cone, on the other hand, can be squashed down to its tip, a fact we will make precise in §2.3 Homotopy. As Allen Hatcher remarks, “Suspension becomes increasingly important the farther one goes into algebraic topology, though why this should be so is certainly not evident in advance” (Hatcher 2002, p. 9).

Three constructions Pick a construction and a space · play or scrub the gluing

cylinder cone

Figure 2.2.7 A wedge sum glues two spaces at a single point. A cone collapses the top

X \times {1}

of the cylinder over

X;

a suspension collapses the bottom as well. The cone on anything is “solid” — a disk, a triangle, a filled shape — while the suspension of a sphere is a sphere one dimension up.

Every closed surface from a single polygon

A square with four edges gave us the torus, a doughnut with one hole. What about a doughnut with two holes — the surface of a pretzel? Take an octagon and label its eight edges, going anticlockwise,

a_{1} b_{1} a_{1}^{- 1} b_{1}^{- 1} a_{2} b_{2} a_{2}^{- 1} b_{2}^{- 1} .

The word is two copies of the torus word, one after the other. Figure 2.2.8 shows why the result has two holes. Cut the octagon along a diagonal $c$ that separates the first four edges from the last four. Each half is a pentagon with word $a_{1} b_{1} a_{1}^{- 1} b_{1}^{- 1} c^{- 1}$ or $c a_{2} b_{2} a_{2}^{- 1} b_{2}^{- 1} :$ a torus diagram with an extra free edge. Gluing its $a$ and $b$ edges gives a torus with a hole cut out, the hole’s rim being $c .$ Glue the two holes together along $c,$ undoing the cut, and you have a surface with two handles: the genus-two surface $Σ_{2} .$

From an octagon to a two-holed doughnut Press play or drag the slider · drag the picture to turn it

The octagon, and its diagonal c

octagon two handles

Figure 2.2.8 The octagon

a_{1} b_{1} a_{1}^{- 1} b_{1}^{- 1} a_{2} b_{2} a_{2}^{- 1} b_{2}^{- 1}

is cut along its diagonal

c .

Each half bends round until its copy of

c

closes up into a circle, which opens out into a hole; then it rolls and bends, as the square does in Figure 2.2.4, into a torus with that hole. Last, the two holes are drawn out into a neck and glued along

c,

its waist. Each edge carries half of its arrowhead, so the arrowheads close up as the edges meet. On the finished surface

a_{i}

runs round a tube and

b_{i}

round a hole. In the quotient all eight corners of the octagon are a single point, through which all five loops pass; the picture draws that point stretched out, so that

c

is a separate circle and each torus has its own crossing of

a_{i}

with

b_{i} .

The pattern continues. A polygon with $4 g$ edges labelled $a_{1} b_{1} a_{1}^{- 1} b_{1}^{- 1} \dots a_{g} b_{g} a_{g}^{- 1} b_{g}^{- 1}$ glues up into the surface $Σ_{g}$ with $g$ handles — a doughnut with $g$ holes (Hatcher 2002, p. 5). The number $g$ is called the genus. Following the arrows shows that all $4 g$ corners become a single point, so counting corners minus edges plus faces gives

1 - 2 g + 1 = 2 - 2 g .

For the sphere ( $g = 0)$ that is $2,$ for the torus $0,$ for the pretzel $- 2 .$ Words in which a letter appears twice the same way round give one-sided surfaces: $aa$ (a lens whose two curved edges are glued with a twist, another picture of $RP^{2}),$ or $aabb,$ which turns out to be another picture of the Klein bottle (Hatcher 2002, pp. 51–52). A famous theorem, the classification of surfaces, says that every connected closed surface is homeomorphic to exactly one of these: a sphere, a surface $Σ_{g}$ with $g$ handles, or a one-sided surface built from projective planes. We will meet it properly in §2.4 Manifolds and Surfaces. For now, savour the fact that a single word written around a single polygon is enough to describe any of them.

Exercises

Exercise Count the corners

For each word, label the corners of the polygon and find which are glued together: (a) the torus $ab a^{- 1} b^{- 1};$ (b) the projective plane $abab;$ (c) the hexagon $ab c a^{- 1} b^{- 1} c^{- 1} .$ Then compute corners − edges + faces for each.

Hint

Each edge glues its tail to the tail and its head to the head of its partner. Write down the pairs of corners this produces, then let the glue spread.

Solution

(a) One class (all four corners), and $1 - 2 + 1 = 0 .$ (b) Two classes: the bottom-left corner is glued to the top-right, and the bottom-right to the top-left; $2 - 2 + 1 = 1 .$ (c) Number the corners $0, 1, \dots, 5$ anticlockwise, with edge $a$ running from corner $0$ to corner $1 .$ The second $a$ runs from corner $4$ to corner $3,$ so $0 \sim 4$ and $1 \sim 3;$ the $b ’$ s give $1 \sim 5$ and $2 \sim 4;$ the $c ’$ s give $2 \sim 0$ and $3 \sim 5 .$ The classes are ${0, 2, 4}$ and ${1, 3, 5} :$ two corners, three edges, one face, $2 - 3 + 1 = 0 .$ (The hexagon with opposite sides glued is another picture of the torus.)

Exercise Which surface?

Identify the space obtained from the square by: (a) gluing only the bottom edge to the top edge, with a flip, $(x, 0) \sim (1 - x, 1);$ (b) collapsing the whole boundary of the square to a single point; (c) gluing the bottom edge to the top edge straight across, and then collapsing the left edge to a point and the right edge to another point.

Solution

(a) A Möbius band: one pair of edges glued with a twist. (It is the band of the workshop turned on its side.) (b) A sphere: the square is homeomorphic to a disk, and collapsing the rim of a disk gives $D^{2} / \partial D^{2} ≅ S^{2} .$ (c) Gluing the bottom to the top gives a cylinder whose two boundary circles come from the left and right edges; collapsing each of those circles to a point gives the suspension of a circle, a sphere.

Exercise Cutting a Möbius band

Use the square diagram $(0, y) \sim (1, 1 - y)$ to predict what happens when a Möbius band is cut along its centre line $y = \frac{1}{2} .$ How many pieces? How many boundary circles?

Hint

Cutting along $y = \frac{1}{2}$ splits the square into a bottom half and a top half. Where does the gluing rule send the left edge of the bottom half?

Solution

The bottom half $[0, 1] \times [0, \frac{1}{2}]$ has its left edge $(0, y),$ $y \leq \frac{1}{2},$ glued to $(1, 1 - y),$ which lies on the right edge of the top half. Likewise the top half’s left edge is glued to the bottom half’s right edge. So the two halves join end to end into one long strip, glued at both ends without a net flip: a single piece, a band with two boundary circles — topologically a cylinder. (In space it comes out with two full twists, which is why it surprises people.)

Exercise Unglued corners

Using the table in this chapter, show that no homeomorphism can turn the sphere’s square into the torus’s square edge by edge and corner by corner. Then explain why this alone does not prove that the sphere and torus are different surfaces.

Solution

The sphere’s diagram has three corner classes and the torus’s has one, so no matching of diagrams sends corners to corners. But a surface can be drawn by many different diagrams — the torus is also a hexagon with two corner classes (the first exercise). Corner counts belong to the diagram, not the surface. What is a property of the surface is the combination corners − edges + faces ( $2$ versus $0),$ but proving that this does not depend on the diagram is real work, done in §2.6 The Euler Characteristic and, more powerfully, by homology.

Exercise A Klein bottle is two Möbius bands

In the Klein bottle square, with $(x, 0) \sim (1 - x, 1)$ and $(0, y) \sim (1, y),$ cut along the vertical lines $x = \frac{1}{4}$ and $x = \frac{3}{4} .$ Show that each of the two resulting pieces is a Möbius band, and that the two bands are glued along their boundary circles.

Hint

The middle strip is $[\frac{1}{4}, \frac{3}{4}] \times [0, 1] .$ Where does the rule $(x, 0) \sim (1 - x, 1)$ send it? The two outer strips are joined by $(0, y) \sim (1, y) .$

Solution

The middle strip’s bottom edge $(x, 0),$ $\frac{1}{4} \leq x \leq \frac{3}{4},$ is glued to $(1 - x, 1),$ and $1 - x$ is also between $\frac{1}{4}$ and $\frac{3}{4} :$ the strip’s top is glued to its own bottom with a flip, which is a Möbius band. The outer strips $[0, \frac{1}{4}]$ and $[\frac{3}{4}, 1]$ are joined by $(0, y) \sim (1, y)$ into one strip; its bottom is glued to its top by $x \mapsto 1 - x,$ which swaps the two parts — again with a flip, so another Möbius band. The cut lines $x = \frac{1}{4}$ and $x = \frac{3}{4}$ are glued end to end (the top of one to the bottom of the other) into a single circle, which is the boundary of both bands. So $K$ is two Möbius bands sewn together along their edges — the “weird bottle” of the limerick at the top of this chapter.

Exercise Cut and paste: two words, one bottle

Cut the Klein bottle square $aba b^{- 1}$ along the diagonal $c$ that runs from its bottom-left to its top-right corner. Turn one of the two triangles over, and glue the triangles back together along their $a$ edges. Show that the new square has the word $aabb,$ after renaming its letters — so the two words describe the same surface.

Hint

Write each triangle’s word anticlockwise, as for the square. Gluing a polygon with word $x P$ to one with word $x^{- 1} Q$ along $x$ gives a polygon with word $P Q .$ Turning a polygon over reverses its word: $x y z$ becomes $z^{- 1} y^{- 1} x^{- 1} .$

Solution

With the arrows of Figure 2.2.4 (bottom $a$ pointing right, top $a$ pointing left, both $b ’$ s pointing up), the lower triangle reads $a b c^{- 1}$ and the upper one $c a b^{- 1} .$ Both contain $a$ the same way round, so before gluing we turn the upper triangle over: its word becomes $b a^{- 1} c^{- 1},$ or, starting from $a^{- 1},$ $a^{- 1} c^{- 1} b .$ Now glue $a (b c^{- 1})$ to $a^{- 1} (c^{- 1} b)$ along $a :$ the new square reads $b c^{- 1} c^{- 1} b,$ which, starting from the last letter, is $b b c^{- 1} c^{- 1} .$ Reverse the arrow on $c$ and call it $d :$ the word is $bb dd,$ which is $aabb$ with the letters renamed. Cutting and regluing never changes the surface, only its picture — the same moves drive the proof of the classification of surfaces in §2.4 Manifolds and Surfaces.

Summary

Recap

To glue is to pass to a quotient $X / \sim .$ Its topology looks back before the glue: $U$ is open exactly when $q^{- 1} (U)$ is open. Continuous functions on $X / \sim$ are continuous functions on $X$ that respect the glue.
A gluing diagram labels edges with letters and arrows; its word reads them anticlockwise, and glue spreads to the corners.
One square gives the cylinder, Möbius band, torus $ab a^{- 1} b^{- 1},$ Klein bottle $aba b^{- 1},$ sphere $ab b^{- 1} a^{- 1}$ and projective plane $abab$ — a disk with opposite rim points glued.
The Klein bottle and $RP^{2}$ are one-sided, so they cannot be embedded in $R^{3};$ they can only be immersed, crossing themselves. In $R^{4}$ they fit.
Wedge sums join spaces at a point; collapsing $A$ gives $X / A,$ with $D^{2} / \partial D^{2} ≅ S^{2};$ cones are solid and suspensions raise dimension: $Σ S^{n} ≅ S^{n + 1} .$
A $4 g$ -gon with word $a_{1} b_{1} a_{1}^{- 1} b_{1}^{- 1} \dots a_{g} b_{g} a_{g}^{- 1} b_{g}^{- 1}$ gives the surface with $g$ holes, and corners − edges + faces $= 2 - 2 g .$

Gluing Spaces Together

A screen that wraps around

Gluing, made precise: the quotient topology

Glue spreads

Gluing diagrams: a square with arrows

The gluing workshop

One pair of edges: the cylinder and the Möbius band

Both pairs: the torus and the Klein bottle

Neighbouring edges: the sphere

Twisting both pairs: the projective plane

Why the Klein bottle will not fit in our space

Wedges, cones, suspensions and collapsing

Every closed surface from a single polygon

Exercises

Summary

Further reading

References