The group a space has at a point

The first rung on this ladder counted loops in a ring and found that the count is all there is: two loops can be slid onto one another exactly when they wind the same number of times. It stopped short of saying what kind of object the collection of loops is, and the answer is that it is a group — but only after two pieces of care that the counting picture hides completely.

The first is that loops have to start somewhere. The second is that the group operation is defined on classes of loops rather than on loops, and it is not obviously well defined even there.

Why a starting point is needed at all #

To multiply two loops, run one and then the other. That only makes sense if the first ends where the second begins, so a collection of loops closes under the operation exactly when they all begin and end at one point. Fix a point $x_0$ — the base point — and consider the loops that start and finish there.

Without a base point there is no operation. Two loops in different parts of a space cannot be run one after the other, and even two loops that happen to meet cannot be composed canonically, because which meeting point is chosen changes the result.

This is more restrictive than it first appears. The set of loops at $x_0$ is not the set of all loops; it is the loops that pass through one particular place. Whether that is a serious restriction is the question the next two sections answer.

The operation is not defined on loops #

Composition of loops is not associative and has no identity, which sounds fatal.

Take three loops $f$, $g$ and $h$, each traversed over the unit interval. In $(fg)h$, the first loop occupies the first quarter of the time, the second the next quarter, and the third the second half. In $f(gh)$, the first occupies the first half. As functions these are different maps, so the product is not associative on the nose.

And $f$ composed with the constant loop at $x_0$ is not $f$: it is $f$ run twice as fast followed by standing still, which as a function is not the same as $f$.

Both problems disappear once loops are replaced by their classes, and the reason is the same in each case: reparametrising a loop — running it at a different speed — is a homotopy, so a loop and any reparametrisation of it are in the same class. The associativity failure and the identity failure are both reparametrisations, and both are invisible at the level of classes.

This is worth dwelling on, because it is the pattern the whole subject runs on. The object with the good algebraic properties is not the object one first writes down; it is a quotient of it, and the quotient is chosen precisely to make the properties true. Two loops are the same when one slides onto the other was introduced as a way of counting; it turns out to be the thing that makes the algebra work.

And the operation must respect the classes #

Even on classes, one thing needs checking: if $f$ is slid to $f'$ and $g$ to $g'$, is $fg$ slid to $f'g'$?

It is, and the proof is a picture rather than an argument. A homotopy from $f$ to $f'$ is a continuous family of loops indexed by time; likewise for $g$. Running the two families at the same time gives a family of composed loops, which is a homotopy from $fg$ to $f'g'$. Nothing else is needed.

With that, the classes form a group: associative, with the constant loop as identity, and with the reverse loop as inverse — since running a loop and then running it backwards is homotopic to standing still, by pulling the turning point back along the loop.

That group is $\pi_1(X, x_0)$, the fundamental group of $X$ at $x_0$.

Moving the base point #

Now the question the base point raises. Choose a different point $x_1$. Is $\pi_1(X, x_1)$ the same group?

If there is a path $\gamma$ from $x_0$ to $x_1$, then yes, and the isomorphism is explicit: send the class of a loop $f$ at $x_0$ to the class of $\bar\gamma f \gamma$ — run the path backwards, run the loop, run the path forwards — which is a loop at $x_1$. That map respects composition, because the two copies of $\gamma$ in the middle of a product cancel.

If there is no path, the two groups have nothing to do with each other. The fundamental group sees only the path-component of the base point, which is why the whole subject is usually set up for path-connected spaces and why the base point is dropped from the notation there.

What the choice of path costs #

The isomorphism above depends on $\gamma$, and two paths from $x_0$ to $x_1$ generally give different isomorphisms.

Take two paths $\gamma$ and $\delta$. The loop $\gamma\bar\delta$ is a loop at $x_0$, of some class $c$. Comparing the two isomorphisms shows they differ by conjugation by $c$: one sends $[f]$ where the other sends $c[f]c^{-1}$.

So the group at $x_1$ is isomorphic to the group at $x_0$, canonically up to conjugation and no better. Two consequences follow, and they are the practical content of the whole rung.

Any quantity defined on the fundamental group that is invariant under conjugation — the order of an element, whether the group is abelian, the number of conjugacy classes, the isomorphism type of the group itself — is a property of the space, with no base point in it. Anything not invariant under conjugation is not.

And when the group is abelian, conjugation is trivial, so the isomorphism is canonical after all. That is why the ring gives no hint of any of this: its group is $\mathbb{Z}$, conjugation does nothing, and moving the base point is entirely free.

A worked check: the ring’s group really is the whole numbers #

The claim that $\pi_1$ of the ring is $\mathbb{Z}$ has two halves and only one of them is the counting picture.

The winding number is a homomorphism. Running one loop and then another adds the angles turned through, so the winding of a product is the sum of the windings. That is the map $\pi_1 \to \mathbb{Z}$, and it is well defined on classes because winding is unchanged by sliding — which is the first rung’s content.

It is onto, because a loop going $k$ times round has winding $k$ for every whole $k$, and such loops exist.

And it is one-to-one, which is the half the pictures do not establish. A loop of winding zero has to be shown contractible, and the argument is the lift: pass to the covering space, where the loop becomes a path whose ends agree exactly when the winding is zero, and a path in a simply-connected space can always be pulled to a point. Pushing that contraction back down contracts the original.

That third step is where the real work is, and it is where the drawings stop. Every figure on this rung and the last shows loops with different windings and invites the conclusion that different windings mean different classes — which is the easy direction. The hard direction is that the same winding means the same class, and the evidence for it is not a picture of two loops; it is a covering space argument that the next rung sets up.

It is worth being clear that this is the standard shape of a computation of a fundamental group. Producing elements is easy; proving there are no others needs a structure the space is compared against, and the covering space is that structure.

Where it does bite #

To see conjugation doing something, the group has to be non-commutative, and the smallest example on this site is the plane with two points removed.

The group there is free on two generators. The commutator $aba^{-1}b^{-1}$ has zero winding about each hole and is not trivial, which is the first thing on this ladder that winding numbers cannot detect. And now conjugation matters: $a$ and $bab^{-1}$ are different elements of the group, so a loop going round the first hole and the same loop carried to another base point by a path that goes round the second hole are genuinely different.

What survives is the conjugacy class, and that is what a base-point-free statement can mention. Asking whether a specific loop in a space is trivial is a well-posed question; asking whether it equals a specific element of the fundamental group is not, until a path to the base point has been fixed.

Setting those two words side by side is the fastest way to see what non-commutativity means concretely. Both loops use each hole twice. One has non-zero winding about each hole and is caught by the counting invariant; the other has zero winding about both and is not, and is nevertheless not contractible. The difference between them is the order of the letters and nothing else, and a commutative group is by definition one that cannot tell the difference.

That is also the sharpest available statement of why the ring is a poor laboratory. It has one hole, its group has one generator, and a group with one generator is commutative whatever else is true of it. Everything on this rung that is about order rather than about counting needs at least two holes to exist at all.

What it costs to work this way #

Three honest costs, in decreasing order of how often they matter.

A functor needs base points and maps do not always preserve them. A continuous map $X \to Y$ induces a homomorphism of fundamental groups only if it sends the base point to the base point, and a homotopy between two such maps induces the same homomorphism only if it moves the base point along a path — whose class then enters the comparison as a conjugation, exactly as above. Working with based spaces everywhere is the standard fix and it is a genuine tax on the notation.

The group is enormous and usually not computable by staring. The group of a knot complement is finitely presented and deciding whether two presentations give the same group is undecidable in general. So having the group does not automatically mean having an answer, which is the same gap that a presentation of any group has.

A presentation is not an algorithm. Even where the group is known — free on two generators for the twice-punctured plane, say — deciding whether two given words are the same element requires a normal form, and normal forms exist for free groups and for many others but not in general. The word problem is undecidable for finitely presented groups, and fundamental groups of reasonable spaces realise every finitely presented group. So every undecidability of group theory is an undecidability about spaces, transmitted by this construction.

And it forgets a great deal. All the higher-dimensional structure of a space is invisible to it: the sphere has a trivial fundamental group and is not a point. That is not a defect — it is the throwing away that makes the invariant computable — but it is worth stating that a trivial group is very far from proving a space is trivial.

What the pictures cannot show #

Every loop drawn is a smooth curve and the theory is about continuous maps. A continuous loop can be a space-filling monster, and nothing about the definitions excludes it. Smooth drawings are a convenience; the theorems are proved for the wild ones too, and it takes real work to know that the smooth ones represent every class.

A homotopy is a family and each figure is a member. Sliding one loop onto another is a continuous family indexed by time, of which a figure can show two or three stages. That is a picture of the endpoints of a proof rather than of the proof.

A group is not a picture and the figures draw its elements rather than it. Six loops in a ring are six elements of an infinite group, and every statement about the group as a whole — that it is infinite cyclic, that it is commutative, that its only automorphisms are the identity and negation — is about the collection rather than about any member. There is no drawing of a group here, only drawings of things in one.

And the base point’s cost cannot be drawn on a ring, which is what the two figures above admit. Exhibiting conjugation doing something requires a non-commutative group, and every non-commutative example on this site lives on a surface with two holes, where the drawings become hard to read for a different reason.

Where the definition came from #

Poincaré introduced the group in 1895, in the first of his papers on analysis situs, and his motivation was not the one modern treatments give.

He was classifying three-dimensional manifolds and needed an invariant finer than the Betti numbers, which are the counts of holes in each dimension. The Betti numbers cannot distinguish certain three-manifolds that are plainly different, and Poincaré’s response was to attach to a space something with more structure than a number: a group, whose non-commutativity would carry the extra information. He produced examples with identical Betti numbers and different fundamental groups within the same series of papers, which settled the question he had asked.

The base point is nowhere in his original treatment, and the modern insistence on it is a later tidying. Poincaré works with loops and their compositions without worrying about where they start, which is fine for the arguments he makes and produces exactly the conjugation ambiguity described above the moment anyone tries to make it precise. The care taken in the sections above is the price of stating in one sentence something he stated correctly and informally.

That trajectory — an object introduced to settle one classification question, then made precise decades later, then found to be the beginning of a subject — is common enough to be worth expecting. The precision is added by people who want to prove theorems about the invariant rather than with it, and the tax it imposes falls on the notation rather than on the ideas.

Where the ladder goes next #

The next rung finds the group’s subgroups drawn as spaces: every cover of a space corresponds to a subgroup, with the number of sheets equal to the index. After that, the group is computed rather than described — by cutting the space into pieces — and the last rung asks what happens when loops are replaced by spheres, where the arithmetic becomes commutative for a reason nobody finds obvious.

Sideways, the group’s elements as words in generators is a Cayley graph, the quotient-to-make-the-algebra-work move is the same one that makes a curve a circle, and the invariance under sliding is the same continuity argument as a winding number’s integrality.

What is worth carrying away #

An arbitrary choice that changes the answer by a symmetry is not really arbitrary; it is a statement about which questions are well posed.

The base point is a choice, and everything about the fundamental group survives it up to conjugation. So the well-posed questions about a space are the conjugation-invariant ones, and that class is large enough to include almost everything anybody wants to ask. The residue — that a specific element cannot be named without a path — is not a defect of the theory but a fact about the space.

The habit worth taking is to ask, of any invariant defined with a choice, what group acts on the choices. Whatever that group fixes is the real invariant, and whatever it moves was never a property of the object.