2.3 Homotopy · Homology & Cohomology

One of the main ideas of algebraic topology is to consider two spaces to be equivalent if they have ‘the same shape’ in a sense that is much broader than homeomorphism.

— Allen Hatcher, Algebraic Topology (2002), Chapter 0

A rubber band is wrapped once around a flagpole. You may slide it up and down, twist it, stretch it into strange shapes — but as long as you are not allowed to lift it over the top, you will never get it off. The pole is a hole in the space the band lives in, and the band can feel it. To turn that feeling into mathematics, we let maps move instead of spaces: continuously, like the frames of a film. Films of maps will let us call two spaces the same when one can be squashed onto the other, and they will hand us our first algebraic invariant of a space — a group whose elements are loops.

In §2.1 Spaces and Continuity we called two spaces the same when there is a homeomorphism between them: a continuous bijection with a continuous inverse, a perfect rubber-sheet correspondence. That notion is exact, but for counting holes it is too fussy. A thick ring and a thin circle are not homeomorphic (one is two-dimensional, the other one-dimensional), yet anyone can see that they have “the same hole”. A solid disk and a single point are not homeomorphic either, yet neither has any hole at all. Homotopy is the coarser, more forgiving notion of sameness that sees holes and nothing else.

Paths: journeys inside a space

Everything in this chapter is built out of one simple object, which you met in §2.1 Spaces and Continuity: the path. We write $I = [0, 1]$ for the unit interval, the set of real numbers between $0$ and $1$ inclusive. Think of $I$ as a clock: $0$ is the moment we set off and $1$ the moment we arrive.

Read $γ$ (“gamma”) as a traveller’s itinerary: at each moment $s \in I$ it tells you where the traveller is, $γ (s) \in X .$ Continuity says the traveller never teleports. Two features of this definition will matter again and again.

A path is the whole journey, timing included, not just the trail it leaves. Walking along a road at a steady pace and walking along the same road while stopping for lunch halfway are different paths with the same trail (the same image $γ (I)).$ Re-timing a path without changing its trail is called reparametrising it.
A path may cross itself, double back, or stand still. The constant path $c_{x} (s) = x$ for all $s,$ which never leaves the point $x,$ is a perfectly good path. So is a path that runs around a circle five times.

A space is path-connected when any two of its points can be joined by a path. The plane is path-connected; so is the plane with a point removed (walk around the missing point); two disjoint disks are not.

Homotopy: a movie of maps

Here is the question that drives the chapter. In the plane with a hole punched in it, take two paths from a point $x_{0}$ to a point $x_{1} .$ Can one path be continuously slid onto the other, keeping its ends pinned down, without ever passing through the hole? Try it below: the gold path $γ_{0}$ stays put; drag the teal bead to reshape $γ_{1},$ then drag the white bead along its track to run the movie that slides one onto the other.

Paths around a hole Drag the teal bead to reshape γ₁ · drag the white bead to run the movie

Not homotopic The paths pass on opposite sides of the puncture, so every movie from one to the other has to sweep across it. The straight-line movie hits it at $t \approx 0.50$ .

Figure 2.3.1 The straight-line movie

H (s, t) = (1 - t) γ_{0} (s) + t γ_{1} (s) :

every point of

γ_{0}

slides along a straight track (the faint lines) to the matching point of

γ_{1},

and the dashed curves are frames of the movie. When both paths pass on the same side of the puncture, the movie never touches it. When they pass on opposite sides, the frames are forced across the puncture (the pink dashed frame), and no other movie can do better: the two paths are not homotopic.

Let us describe the sliding precisely. At each moment of the film $t \in I$ we have a whole path, which we may call $H_{t};$ at $t = 0$ it is $γ_{0},$ at $t = 1$ it is $γ_{1},$ and in between it changes gradually. A path is a function of one time variable $s$ (where the traveller is during the trip), and now the trip itself evolves in a second time variable $t$ (which frame of the film we are watching). Packaging the two together gives a single function of two variables, $H (s, t) = H_{t} (s) .$ The continuity of this function in both variables at once is what makes the film smooth: no frame jumps, and no frame tears.

Nothing here is special to paths. Any continuous map can be the subject of such a film.

Let us read every symbol. $X \times I$ is the product of the space $X$ with the time interval: its points are pairs $(x, t)$ — “the point $x,$ at time $t ”.$ The homotopy $H$ takes such a pair and says where $x$ has been sent at that moment. The two equations pin down the first and last frames: at time $0$ the film shows $f,$ at time $1$ it shows $g .$ The symbol $≃$ is read “is homotopic to”.

For paths we usually want more: the ends should stay pinned down for the whole film, otherwise every path could be reeled in to a single point by sliding its endpoints together.

Homotopy is an equivalence relation

“Is homotopic to” behaves like a sameness relation, in the precise sense of §1.2 Equivalence and Quotients: it is an equivalence relation. Each of the three properties is an operation on films.

Reflexive ( $f ≃ f):$ the still film $H (x, t) = f (x),$ which shows $f$ in every frame.
Symmetric (if $f ≃ g$ then $g ≃ f):$ play the film backwards, $H ˉ (x, t) = H (x, 1 - t) .$
Transitive (if $f ≃ g$ and $g ≃ h$ then $f ≃ h):$ show the first film and then the second, each at double speed. If $H$ runs from $f$ to $g$ and $K$ from $g$ to $h,$ the joined film is $H (x, 2 t)$ for $t \leq \frac{1}{2}$ and $K (x, 2 t - 1)$ for $t \geq \frac{1}{2} .$ At $t = \frac{1}{2}$ both formulas give $g (x),$ so the two halves meet without a jump.

The equivalence classes are called homotopy classes; the class of $f$ is written $[f] .$ The same three films, with ends held still, show that path homotopy is an equivalence relation too.

The same shape in a broader sense

Now we use films to compare spaces. Hatcher opens his book, straight after the sentence in our epigraph, with the alphabet (Hatcher 2002, pp. 1–2). Write each capital letter in two ways: as a thin skeleton of curves, and as a fat, inflated version of itself. The thin letter sits inside the fat one, and the fat one can be squashed onto it by sliding each point straight in towards the skeleton.

Thick letters, thin letters Tap a letter to shrink it · sort them

Figure 2.3.2 Each thick letter shrinks onto its gold skeleton without tearing. Sorted by homotopy type, the letters fall into just three families: those that can be shrunk to a point, those with one loop (like a circle

S^{1}),

and

B,

with two loops (like the figure eight

S^{1} \lor S^{1}).

Homeomorphism would tell many of these apart; homotopy only sees the loops.

What exactly is the relationship between a fat letter $X$ and its thin skeleton $A ?$ There are two natural maps between them. One is the inclusion $i : A \to X,$ which regards each point of the thin letter as a point of the fat one. The other is the squash $r : X \to A,$ which sends every point of the fat letter to where it ends up after the shrinking. Neither is a homeomorphism. But each undoes the other up to homotopy:

squashing a point of the thin letter does nothing (it is already on the skeleton), so $r \circ i = id_{A}$ exactly;
including after squashing, $i \circ r : X \to X,$ is not the identity of the fat letter — it crushes everything onto the skeleton — but the shrinking film itself is a homotopy from $id_{X}$ to $i \circ r .$

That is the pattern we make into a definition.

Compare this with a homeomorphism, where we demand $g \circ f = id_{X}$ and $f \circ g = id_{Y}$ on the nose. A homotopy equivalence only asks the round trips to be deformable to the identity. Every homeomorphism is a homotopy equivalence (take the still films), but the converse fails wildly, as the letters show. With the same three film operations as before one checks that $≃$ is an equivalence relation on spaces (one of the exercises does the transitive step).

Space	is homotopy equivalent to	but not homeomorphic to it, because
a solid disk $D^{2},$ or all of $R^{n}$	a point	a point is a single point: there is not even a bijection
a thick ring (annulus)	the circle $S^{1}$	removing two points disconnects $S^{1}$ but not the ring
the plane minus a point $R^{2} ∖ {0}$	the circle $S^{1}$	$S^{1}$ is compact, the punctured plane is not
the Möbius band	the circle $S^{1}$	removing two points disconnects $S^{1}$ but not the band
the torus with one point removed	the figure eight $S^{1} \lor S^{1}$	removing its crossing point disconnects the figure eight; no single point disconnects the punctured torus
the letter $A$	the letter $O$	$A$ has points where three strokes meet

Deformation retractions

All the examples in the table above are of one convenient kind: the bigger space can be squashed onto a subspace while the subspace itself stays perfectly still.

Explore four of them below. Pick a space, then play or scrub through time: you are watching the frames $f_{t}$ of the film, and the faint grid lines are painted on the space so that you can see every point travel. The inset shows the same film drawn flat.

Deformation retractions Choose a space · play or scrub time · drag to rotate

t = 0 t = 1

at time t: $slide outward from the puncture$

Figure 2.3.3 A disk shrinks to its centre; an annulus is squeezed onto its middle circle; a Möbius band is squeezed onto its gold core circle (watch its single teal edge, which runs round twice, collapse onto the core); and a torus with one puncture is pushed outward from the puncture until only the gold circle

a

and the teal circle

b

remain. In the last one, the pink rim of the puncture ends up running along

a,

then

b,

then

a

backwards, then

b

backwards: the loop

ab a^{- 1} b^{- 1},

which will be important later.

Contractible spaces, and why homotopy is the right lens

The disk in Figure 2.3.3 shrank all the way to its centre, and the letters C and T of Figure 2.3.2 can go as far: squeeze the skeleton of a C along itself and it ends as a single point. The simplest homotopy type of all is that of a point, and spaces of that type get a name.

$R^{n},$ every disk and ball, every convex set, every tree (a graph without cycles) and the letters $C, E, T, X$ are contractible. A contractible space is always path-connected: if $H$ contracts $X$ to the point $x_{0},$ then for each $x$ the map $t \mapsto H (x, t)$ is a path from $x$ to $x_{0} .$ So any two points can be joined through $x_{0} .$

This is the moment to say why homotopy is the right lens for this book. We are looking for invariants: numbers, or groups, attached to spaces in such a way that equal spaces get equal labels. The finer the notion of “equal”, the more labels are allowed and the harder they are to compute. Homotopy is the coarse, generous notion — and invariants that do not change under homotopy equivalence have a superpower.

Loops, and how to compose them

Fix a point $x_{0}$ of a space $X,$ and call it the basepoint: a home from which all our journeys start and to which they return.

Two loops at the same basepoint can be done one after the other: first take trip $α,$ then trip $β .$ But our trips must fit into the single time interval $I,$ so each is run at double speed.

At $s = \frac{1}{2}$ the first formula gives $α (1) = x_{0}$ and the second gives $β (0) = x_{0},$ so the two halves join up. (The same formulas concatenate any two paths, provided the first ends where the second starts.) The order matters: in $α \cdot β$ we do $α$ first.

Is concatenation associative? Not quite, if we are strict. In $(α \cdot β) \cdot γ$ the trip $α$ is done at quadruple speed during the first quarter of the time, $β$ during the second quarter and $γ$ at double speed during the second half; in $α \cdot (β \cdot γ)$ it is $α$ that gets the first half. Same trails, same order, different timetables. The difference is only one of timing, and timing can be changed continuously.

Timetables Drag the point in the square · run the trip

At stage t = 0.40 the trip runs α until s = 0.35, β until 0.60, then γ. Only the timing changes from t = 0 to t = 1.

Figure 2.3.4 The square of a homotopy

H,

with trip time

s

running across and movie time

t

running up, and the space, three loops at

x_{0} .

Each point

(s, t)

of the square is a position

H (s, t)

in the space: the white bead. Sliding the two breakpoints from

\frac{1}{4}, \frac{1}{2}

\frac{1}{2}, \frac{3}{4}

turns

(α \cdot β) \cdot γ

into

α \cdot (β \cdot γ) .

Choose “Inverses” to see

α \cdot α ˉ

shrink to the constant loop by turning back earlier and earlier.

Three facts make loops, up to homotopy, behave like the elements of a group:

Associativity up to homotopy: $(α \cdot β) \cdot γ ≃ α \cdot (β \cdot γ),$ by sliding the breakpoints, as in the figure.
The constant loop does nothing, up to homotopy: $c \cdot α ≃ α ≃ α \cdot c .$ The loop $c \cdot α$ waits at home for half the time and then rushes round $α;$ shrinking the waiting time to zero is a homotopy.
A loop and its reverse cancel, up to homotopy: $α \cdot α ˉ ≃ c .$ Go out along $α$ and come straight back; at stage $t$ of the film, turn back after only the fraction $1 - t$ of the way. At $t = 1$ you never leave home.

The fundamental group

We are ready for the chapter’s main construction. Up to homotopy, loops compose associatively, have an identity and have inverses — so homotopy classes of loops form a group, in the sense of §1.3 Groups.

Read $π_{1}$ as “pi one”; the “1” is there because there are higher versions $π_{2}, π_{3}, \dots$ built from spheres instead of loops, which we will only mention. Remember that a homotopy of loops keeps the basepoint fixed for the whole film: the loop may wriggle as it likes, but it is always anchored at home.

One point needs checking before this is a definition at all. The product is defined by choosing loops from the two classes; what if we had chosen others? This is the “is it well-defined?” question from §0.2 How to Read Mathematics. Suppose $α ≃ α^{'}$ by a film $F$ and $β ≃ β^{'}$ by a film $G .$ Then running $F$ and $G$ side by side — $F$ in the first half of each frame, $G$ in the second — is a film from $α \cdot β$ to $α^{'} \cdot β^{'} .$ So the class of the product depends only on the classes we started with, and the three facts of the previous section become the group axioms.

The fundamental group is an invariant in exactly the sense we wanted. A continuous map $f : X \to Y$ carries loops at $x_{0}$ to loops at $f (x_{0})$ and homotopies to homotopies, so it gives a homomorphism $f_{*} : π_{1} (X, x_{0}) \to π_{1} (Y, f (x_{0})),$ $[α] \mapsto [f \circ α] .$ With a little more work one shows:

Every convex set is simply connected: the straight-line film $H (s, t) = (1 - t) α (s) + t x_{0}$ shrinks any loop $α$ to $x_{0},$ and the basepoint never moves because $α (0) = α (1) = x_{0} .$ By homotopy invariance, so is every contractible space. Now compare a sphere with a torus.

Loops on a sphere and on a torus Pull the loops tight · drag to rotate

Figure 2.3.5 Pull the loops tight. On the sphere

S^{2}

the gold loop slides over the top and shrinks to a point. On the torus

T^{2}

the gold loop round the tube and the teal loop round the hole straighten into clean circles — the teal one slides to the inner equator, where it is shortest — and then they are stuck: shrinking further would mean leaving the surface.

Proof

Take a loop on $S^{2},$ and suppose first that it misses some point $p$ of the sphere. Removing a point from a sphere leaves something homeomorphic to the plane (stereographic projection from $p$ flattens the rest of the sphere onto $R^{2},$ as we will see in §2.4 Manifolds and Surfaces), and in the plane the straight-line film shrinks any loop. So our loop shrinks inside $S^{2} ∖ {p},$ hence inside $S^{2} .$

The honest difficulty is the first step. There are continuous loops that pass through every point of the sphere — “space-filling curves”, relatives of the famous curves of Peano and Hilbert — so “pick a point the loop misses” is not always possible. The fix is to first replace the loop by a homotopic one that does miss a point: chop it into short arcs, each lying in a small cap of the sphere, and replace each arc by the shortest great-circle arc with the same ends (a homotopy inside the cap). The new loop is made of finitely many great-circle arcs, and those cannot cover the whole sphere. Hatcher dodges the difficulty differently: he chops the loop into pieces that each miss the north pole or the south pole, and shrinks them one at a time (Hatcher 2002, Prop. 1.14).

Winding numbers: the fundamental group of the circle

Now for the torus’s stuck loops, and the rubber band on the flagpole. The simplest space with a hole is the circle $S^{1},$ or equivalently (by the deformation retraction above) the punctured plane $R^{2} ∖ {0} .$ A loop in the punctured plane can wind around the hole some whole number of times. Let us measure it.

Imagine standing at the hole and watching a traveller go round the loop, always turning to face them. Count the total angle you turn through — counter-clockwise turns counting as positive, clockwise as negative — and divide by a full turn, $2 π .$ When the traveller gets home you are facing the same way you started, so the total must be a whole number of turns. That whole number is the winding number of the loop.

Winding number Reshape the loop · drag the puncture · drag the teal probe along the loop

angle swept so far +0.61 turns winding number $1$

Figure 2.3.6 As the probe runs round the loop, the dashed ray from the puncture turns; the teal spiral records the total angle so far, and the graph plots it in turns. This graph is the loop lifted to the real line. It always ends on a whole number — the winding number. Deform the loop however you like: the end value cannot change unless you drag the loop across the puncture, where it jumps.

The graph on the right deserves a name. The angle of a point of the circle is only defined up to adding whole turns: “a quarter turn” and “one and a quarter turns” are the same direction. Following the traveller continuously picks out one consistent value at each moment, a function $α ~ : I \to R$ (read “alpha tilde”) with no jumps. It is called the lift of the loop: picture the real line wound into an endless spiral staircase above the circle, one storey per turn, and $α ~$ as the route of someone walking up and down the staircase while their shadow traces the loop.

Proof

Here is the shape of the argument; Hatcher’s proof of Theorem 1.7 fills in every detail. There are three steps.

Lifts exist and are unique. Every path in the circle has exactly one lift to the spiral staircase starting at a chosen storey. Every homotopy of paths lifts too, frame by frame, and continuously.
The winding number is a homotopy invariant. Lift a homotopy of loops. During the film the lifted endpoint can only move continuously, yet at every moment it must sit on a whole number (the loop ends at home). A continuously varying whole number is constant. So homotopic loops have equal winding numbers — and the winding number is defined on $π_{1} (S^{1}) .$
It is a bijective homomorphism. Lifting $α \cdot β$ means climbing the stairs for $α$ and then for $β,$ so winding numbers add: the map is a homomorphism. It is onto, because $ω_{n}$ has winding number $n .$ And it is one-to-one, because a loop with winding number $n$ can be slid onto $ω_{n} :$ straighten its lift into a straight ramp from $0$ to $n$ — the staircase is a line, so the straight-line film works there — and project the film back down. In particular only the class of the constant loop $ω_{0}$ goes to $0 .$

Three dividends arrive at once. First, the rubber band on the flagpole really is stuck: its winding number is $1,$ the shrunken band would have winding number $0,$ and winding numbers do not change under homotopy. Second, the question we left open has its answer: the circle is not contractible, because a contractible space has trivial fundamental group and $π_{1} (S^{1}) ≅ Z$ is not trivial. Third, since the punctured plane deformation retracts onto the circle, $π_{1} (R^{2} ∖ {0}) ≅ Z$ as well.

When order matters: the figure eight

Every fundamental group so far has been abelian: the order of composing loops did not matter. The figure eight changes that, and in doing so reveals what homology will keep and what it will throw away.

Let $X = S^{1} \lor S^{1}$ be two circles joined at one point, which we take as the basepoint $x_{0} .$ Let $a$ be the loop once round the left circle and $b$ once round the right one, both counter-clockwise, and write $a^{- 1}, b^{- 1}$ for their reverses. Any loop can be pulled tight into a sequence of these four moves, recorded as a word such as $a b a^{- 1} b^{- 1}$ or $a^{3} b^{- 2} a$ (read left to right: first $a$ three times, and so on). Some words visibly describe the same loop: $a a^{- 1}$ goes round and straight back, and shrinks to nothing. Cancelling every such adjacent pair until none is left produces a reduced word.

That different reduced words give non-homotopic loops is the hard part; it is usually proved with van Kampen’s theorem or with covering spaces (Hatcher 2002, Example 1.21, §1.3), and we shall take it on trust. The figure below shows the idea behind the covering-space proof. Build your own loops from the four letters and watch three pictures at once.

Loops on the figure eight Build a word from the letter keys · tap a letter to remove it · pull it tight

the loop on

S^{1} \lor S^{1}

π_{1}

: a walk on the tree of words

H_{1}

: a walk on the grid

Z^{2}

word

in $π_{1}$: $ab$ already reduced
in $H_{1}$: $a + b$ only the net counts survive

Figure 2.3.7 First the loop, one petal per letter, numbered in order (dashed petals cancel). Then the same trip as a walk on an infinite tree, one direction per letter — this tree is the “staircase” of the figure eight, and the walk’s end point is the reduced word. Last, the walk on the grid

Z^{2},

which only remembers how many steps went each way. Compare

ab

with

ba :

different ends on the tree, the same end on the grid.

Try $ab$ and then $ba .$ As loops, they go round the two circles in opposite orders; as reduced words they are different, so they are not homotopic: $ab \neq = ba$ in $π_{1} .$ The fundamental group of the figure eight is not abelian. Now try the commutator $ab a^{- 1} b^{- 1} .$ Each circle is traversed once in each direction, so every “net count” is zero, and the grid walk returns home. Yet the word is already reduced, the tree walk ends four steps away, and the loop cannot be shrunk.

Here, finally, is the punctured torus again. It deformation retracts onto the figure eight, so its fundamental group is free on $a$ and $b$ too — and the rim of the puncture, which the film of the earlier figure carried onto $ab a^{- 1} b^{- 1},$ is a loop that cannot be shrunk. Fill the puncture back in and the square reappears to shrink it: the torus’s $ab = ba$ is precisely the effect of the one missing disk.

Loops around a knot

Order-sensitive loops answer a question from §2.1 Spaces and Continuity: why can a trefoil not be untied? Homotopy itself is no help with the knot. As a map from the circle into space, every knot is homotopic to a round circle, because a homotopy may pass the string through itself; Figure 2.1.8 did exactly that. The knot is remembered by the space around it, the complement $R^{3} ∖ K .$ Moving the knot about in space, without passing the string through itself, carries its complement along by a homeomorphism, so the fundamental group of the complement cannot change. If two knots have different groups, neither can be moved onto the other.

The loops in the complement are lassos. Put the basepoint high above the knot. For each arc of the drawing (a stretch of string from one undercrossing to the next) there is a lasso that drops down, goes once round that arc, and climbs back. Call the trefoil’s three lassos $x,$ $y$ and $z .$ A lasso can slide along its arc freely; but sliding it under a crossing changes it, because on the far side it also has to go round the strand on top. Each crossing therefore ties three lassos together by a relation such as $z = x y x^{- 1} .$ The lassos generate the whole group, and these relations are all there is: this is the Wirtinger presentation of the knot group (Rolfsen 1976, Ch. 3), and like the figure eight’s it is a fact we take on trust.

Lassos around a trefoil Choose a crossing to see its relation · then send the lassos to the triangle and tap a lasso to change its flip

Slide the lasso $z$ along its strand, under this crossing: it arrives as the lasso $x$ , but on the way it has had to go round the strand on top, whose lasso is $y$ . So

x = y^{- 1} z y

All three: $x = y^{- 1} z y, z = x^{- 1} y x, y = z^{- 1} x z$

Figure 2.3.8 Three arcs, three lassos, and one relation at each crossing. In the second view each lasso is sent to one of the three flips of the triangle from §1.3 Groups, and each crossing is checked:

f_{a} f_{b} f_{a}

must be the flip of the arc coming out. With three different flips every check passes, and since flips do not commute, neither do the lassos.

The test is the one in the figure. Send $x, y, z$ to the three flips $f_{1}, f_{2}, f_{3}$ of the triangle. A relation $z = x y x^{- 1}$ becomes $f_{c} = f_{a} f_{b} f_{a},$ and conjugating one flip by another always gives the third flip (or the same one, if $a = b);$ so every relation holds, and the assignment respects the whole group. But $f_{1} f_{2} \neq = f_{2} f_{1},$ so $x y \neq = y x :$ the trefoil’s group is not abelian. For an unknotted circle the complement’s group is $Z,$ generated by a single lasso, and is abelian. The groups differ, so no amount of moving the trefoil will untie it. Knot theory, the subject of §3.7 Knots and Links, is full of arguments like this one: to tell two knots apart, find something computable that sees the difference.

Forgetting the order: a preview of homology

The grid on the right of the figure is a picture of what happens if we insist that loops commute, declaring $ab = ba$ for all elements. A word then only remembers its exponent sums: the total number of $a ’$ s (counting $a^{- 1}$ as $- 1)$ and the total number of $b ’$ s. The non-abelian free group collapses to the abelian group $Z^{2},$ written additively as $m a + n b .$ This process — forcing a group to become commutative in the most economical way — is called abelianization. (Precisely, one divides the group by the subgroup generated by all commutators; for free groups the result is always “count each letter”.)

There is a second way to see what homology gives up, and it is a picture. Shrinking a loop turns out to be the same as filling it in.

So the fundamental group asks of every loop: is it the rim of a disk? That is the meaning of “hole” in most popular accounts — something that stops a loop from shrinking. Homology will ask a more generous question: is the loop the rim of some surface, a disk or anything else, handles and all? The punctured torus shows the difference. The rim of its puncture, the commutator $ab a^{- 1} b^{- 1},$ cannot shrink, yet it is the entire edge of a surface — the rest of the torus, a torus with a hole in it. So homology counts it as zero, just as the grid does. On a closed two-holed surface the same happens to the “waist” that separates the two handles: it is the rim of either half, but of no disk. (André Henriques draws exactly this curve in his Oxford lecture on chains and cycles; see Further reading.)

Why trade the richer group for the poorer one? Because the fundamental group is hard. Its words can be arbitrarily complicated, and there are spaces built from finitely many cells for which no computer program can decide whether two words describe the same loop: the problem is provably unsolvable (Boone 1958), (Hatcher 2002, Cor. 1.28). The higher homotopy groups $π_{n}$ are wilder still: even for the 2-sphere, $π_{n} (S^{2})$ is non-zero for infinitely many $n$ (Hatcher 2002, p. 98), and nobody knows them all. Homology gives up the order of loops, and in exchange becomes computable — by the linear algebra of §1.5 Linear Algebra — in every dimension at once, detecting the sphere’s hollow along the way. Hatcher’s way of putting the bridge: abelianizing frees loops from their basepoint, so that “loops become cycles” (Hatcher 2002, p. 99). Cycles are where Part III begins.

How hard is the fundamental group? Hard enough that a famous question about the simplest possible group is still open. The knot group above was given by generators (the lassos) and relations (one per crossing), a description called a presentation. Some presentations describe the group with one element in disguise, and recognising them is the problem.

Conjecture The Andrews–Curtis conjecture

Open since 1965 Posed by James Andrews and Morton Curtis, 1965

If a presentation with as many relations as generators describes the trivial group, it can be turned into $⟨ x_{1}, \dots, x_{n} ∣ x_{1}, \dots, x_{n} ⟩$ by three kinds of moves: multiplying one relation by another, inverting a relation, and conjugating a relation.

Every presentation gives a space built from one point, one loop for each generator and one disk for each relation, whose fundamental group is the group presented. When there are as many relations as generators and the group is trivial, that space has the homology of a point, and in fact it can be shrunk to a point. The conjecture asks whether the shrinking can always be done in a particularly simple, controlled way, one step at a time.

What is known. Most experts suspect it is false, but no counterexample is known. The best-known candidates are the presentations $⟨ x, y ∣ x y x = y x y, x^{n} = y^{n + 1} ⟩$ of Akbulut and Kirby (1985). In 2024 a team led by Sergei Gukov used computer search, including reinforcement learning, to dispose of many candidates, and showed that the case $n = 3$ can be reduced if extra generators are allowed; without them it is still open. Some presentations that do reduce need astronomically many moves, so blind search alone cannot settle the question.

A way in. Make $⟨ x, y ∣ x y x = y x y, x^{n} = y^{n + 1} ⟩$ commutative: the first relation forces $x = y,$ and then the second forces $x = 1 .$ So the group is trivial even after forgetting the order, and homology sees no obstruction at all. Then reduce the small example $⟨ x, y ∣ x y^{2}, y ⟩$ to $⟨ x, y ∣ x, y ⟩$ with the three moves.

(Andrews & Curtis 1965; Akbulut & Kirby 1985; Shehper et al. 2024) Status as of October 2026 All open problems

Exercises

Exercise The alphabet up to homotopy

Sort the 26 capital letters, written in a plain sans-serif font as thin curves, by homotopy type. Then find two letters that are homotopy equivalent but not homeomorphic.

Hint

Count the independent loops. For homeomorphism, count the pieces left when you remove a well-chosen point.

Solution

Contractible (no loops): C E F G H I J K L M N S T U V W X Y Z. One loop, $≃ S^{1} :$ A D O P Q R. Two loops, $≃ S^{1} \lor S^{1} :$ B. (Fonts matter: in some fonts the tail of Q only touches the circle, in others it crosses it; the loop count does not change, but the branch points do. The answer depends on the shapes, not the names.) For a pair: $T$ and $X$ are both contractible, but removing the crossing point of $X$ leaves four pieces, while no point of $T$ does that — its junction leaves three. A homeomorphism would carry the special point of one to a point of the other with the same property, so they are not homeomorphic.

Exercise Two nails

The plane with two points removed, $P = R^{2} ∖ {p, q},$ deformation retracts onto a figure eight, so $π_{1} (P)$ is free on the loops $a$ (round $p)$ and $b$ (round $q).$ (a) Compute the winding numbers of the commutator $ab a^{- 1} b^{- 1}$ around $p$ and around $q .$ (b) Explain why it is nevertheless not homotopic to a constant loop in $P,$ but becomes homotopic to one if $q$ is filled back in.

Solution

(a) Around $p,$ only the $a ’$ s wind: $+ 1 - 1 = 0 .$ Around $q,$ only the $b ’$ s: $+ 1 - 1 = 0 .$ (b) In $π_{1} (P)$ the word $ab a^{- 1} b^{- 1}$ is reduced and non-empty, so it is not the identity: winding numbers, being abelian, are blind to it. If $q$ is filled in, the space becomes $R^{2} ∖ {p},$ whose fundamental group is $Z,$ generated by $a;$ the loop $b$ now shrinks to the constant loop, so the word becomes $a a^{- 1},$ which cancels. This is the picture-hanging puzzle, and a first example of a loop that homology cannot see but homotopy can.

Summary

Recap

A homotopy $H : X \times I \to Y$ is a continuous film of maps from $f$ to $g;$ “homotopic” ( $≃)$ is an equivalence relation. For paths we keep the ends fixed.
Spaces are homotopy equivalent ( $X ≃ Y)$ when maps both ways compose to something homotopic to the identity. This is much coarser than homeomorphism: it forgets thickness and dimension and keeps the pattern of holes.
A deformation retraction squashes a space onto a subspace that stays still: the annulus and punctured plane onto a circle, the Möbius band onto its core, the punctured torus onto a figure eight, $R^{n}$ onto a point (so $R^{n}$ is contractible).
Homotopy classes of loops at a basepoint form the fundamental group $π_{1} (X, x_{0});$ composition is “one loop then the other”, inverses run backwards. It is a homotopy invariant.
$π_{1} (S^{1}) ≅ Z$ via the winding number; $π_{1} (T^{2}) ≅ Z^{2};$ $π_{1} (S^{2}) = 0,$ yet $S^{2}$ is not contractible — loops cannot see the sphere’s hollow.
$π_{1} (S^{1} \lor S^{1})$ is free on $a, b$ and non-abelian: $ab \neq = ba .$ Abelianizing keeps only the net counts and gives $Z^{2}$ — a preview of $H_{1} (X) ≅ π_{1} (X)^{ab} .$
A loop shrinks exactly when it is the rim of a disk. Homology will be content with the rim of any surface, which is why the commutator, the rim of a punctured torus, counts as zero there.

Paths: journeys inside a space

Homotopy: a movie of maps

Homotopy is an equivalence relation

The same shape in a broader sense

Deformation retractions

Contractible spaces, and why homotopy is the right lens

Loops, and how to compose them

The fundamental group

Winding numbers: the fundamental group of the circle

When order matters: the figure eight

Loops around a knot

Forgetting the order: a preview of homology

Exercises

Summary

Further reading

References