Pico's Math Blog!

Examples of Spaces

We'll use a few examples of spaces in this post.

In the top-left corner in the image above is the circle, denoted S¹. It consists of all points a fixed distance away from a given point in 2D space (usually, that distance is taken to be 1 and the given point is the origin). Though the circle is embedded in 2D space, it is itself one dimensional.

In the top-right corner, there is the sphere, denoted S². It consists of all points in 3D space a fixed distance away from a given point. The points enclosed by the sphere are not part of the sphere, only those on the surface are, so the sphere forms a 2D surface.

In the bottom-left corner, there is the torus, denoted T². A torus is the surface of a doughnut so, like the sphere, the torus forms a 2D surface.

Last, in the bottom-right corner, there is the figure eight. The figure eight consists of two circles joined together at one point. Just like the circle, the figure eight is a one-dimensional space. The usual notation for the figure eight is S¹ ∨ S¹, but I'll use 8.

These four spaces will be our main examples in the rest of the post.

Paths

In some space X, we can have a path from a point x to a point y. If p is a path from x to y, then we denote that with a squiggle arrow. So, p: x ↝ y means that p is a path from x to y. Note that paths have a direction: a path from x to y is not the same as a path from y to x (unless x and y are the same point). Going from school to your grandma is not the same as going from your grandma's home to school.

A path is a continuous function from the closed interval [0,1] to the given topological space.

In the picture above, we have four points. From left-to-right, call them a, b, c and d. We also have three paths. We say the red path is p, the blue one is q and the green one is r. So, we have p: a ↝ b, q: b ↝ c and r: c ↝ d.

A path must remain within the given space. For example, if the given space is a sphere, then your path is not allowed to ever go off the sphere. And when on a circle, you can never go into the middle of the disc it encloses, as only the boundary of that disk is actually part of the circle.

We can see, in the picture above, that the end-point of p is equal to the start-point of q. So, we can concatenate the two paths into one path going from a directly to c. We denote this concatenation by p;q, pronounced "p, then q". Different people use different notations, but I like to use the semicolon to denote concatenation. Another notation I like to use is q·p ("q follows p"), but I think writing the paths in reverse order might be confusing.

We can also concatenate all three paths, p, q and r, into one path. If we do this, then we get a new path p;q;r ("p, then q, then r") from the point a to the point d.

Concatenating three paths isn't a new operation from concatenating two paths: we can first concatenate p and q into the path p;q, and then concatenate that path with r to obtain the path (p;q);r = p;q;r, and we only ever concatenate two paths at once.

We could also have gotten the path p;q;r from concatenating the paths p, q and r in a different order. We could have first concatenated q and r to obtain a path q;r: b ↝ d, and then concatenate p with that to obtain p;(q;r) = p;q;r.

Concatenating three (or more) paths in different ways always results in the same path. This principle is called associativity. In symbols, we write this principle as (p;q);r = p;(q;r).

We can also reverse paths. If p is a path from a to b, then p' is a path from b to a. This path is the result of following the path p backwards. Reversing a path twice is the same as doing nothing, so that p'' = p (we say the operation of reversing paths is an involution).

We can also reverse concatenated paths. For paths p: a ↝ b and q: b ↝ c, we have that p;q is a path from a to c, and so (p;q)' is a path from c to a. Following p, then q, but all in reverse is the same as first following q in reverse, then p in reverse. So, we have that (p;q)' = q';p'.

Last, every point has an identity path. For a point a, we have a path from a to a that is just staying at a. This path is called the identity or constant path on a, and is denoted 1ₐ or 1_a.

Doing nothing, then following a path p, is the same as just following the path p. So, we have 1ₐ;p = p. Conversely, first following the path p, then doing nothing at b, is also the same as just following p. So, we have p;1_b = p.

Consider the path p;p': a ↝ a, where we first follow the path p, and then go back along the same path. This seems like a different path from 1ₐ. Along p;p', we do visit the point b, but along 1ₐ, we not. However, we still say the two paths are the same or, more precisely, that they are homotopic: we can continuously deform one path into the other while remaining in the given space.

For paths p,q: a ↝ b in a space X, a path homotopy from p to q is a continuous map H: [0,1] × [0,1] -> X satisfying H(0,x) = p(x), H(1,x) = q(x), H(t,0) = a and H(t,1) = b. There is a more general definition of a homotopy between continuous maps, which we won't go over in this blog-post.

Below is a gif of a homotopy

The homotopy (purple) is a continuous deformation between the red path and the blue path. Both paths go from the bottom-left point to the top-right point. Every path of the homotopy remains in the given space and has the same start and end point, so it is a valid homotopy. Would the space in the gif be punctured in the middle, then the homotopy wouldn't be valid: it would go through the puncture, which is not part of the space.

As long as there is a homotopy from one path to another, then we regard the two paths as the same path. Technically, the identities (p;q);r = p;(q;r) and 1ₐ;p = p = p;1_b have also been only up to homotopy, due to the different "speeds" at which you would be following the paths.

Considering paths up to homotopy, we thus have the equations p;p' = 1ₐ and p';p = 1_b. So, we say that the operation of reversing a path is an inverse with respect to concatenation.

Path-Connected Components

Let X be a space and let a and b be points in X. We could ask if there is a path p: a ↝ b between the two given points. For example, let X be the space of two circles that are disjoint from eachother. If a and b are points on the same circle, then they are connected by a path. However, if a and b are points on different circles, then they are not connected by a path.

For a given point a, the subspace of all points that have a path to/from a is called the path-connected component of a. In the example above, if a is on one of the two circles, then its path-connected component is the circle it is on. We denote the path-connected component of a by C(a).

For a space X, we write π₀(X) for the set of all path-connected components of X. In the example above, π₀(X) has two elements: each circle is one. We sometimes call π₀(X) the zeroth homotopy group, even though it is not a group.

We say a space is path-connected iff π₀(X) has exactly one element. Depending on convention, this term is sometimes defined as π₀(X) having at most one element, so allowing the space X to be empty.

If you know some topology, then you're probably familiar with what it means for a space to be connected: we say X is connected iff it has exactly two clopen sets (the empty set and X itself). It turns out, connectedness is a different notion from path-connectedness. As an exercise, try proving the lexicographic square is connected but not path-connected. What are the path-connected components of the lexicographic square?

Fundamental Group

Let X be a space and let a be a point in X. A loop on a is a path from a to a. For example, if X is a circle, then we have a loop on a which goes around the circle counterclockwise once. Recall paths have a direction, so this loop is different from the loop that goes around the circle clockwise once.

If we fix a point a, then we can consider the set of all loops on a. As all these loops have the same start and end point, we can concatenate these loops (and their inversions) any way we like. By the properties we discussed in the previous chapter, these loops form a group:

We have set of loops (up to homotopy)

For any two loops p and q, there is a loop p;q

For any three loops p, q and r, we have (p;q);r = p;(q;r)

There is a loop 1ₐ for which p;1ₐ = 1ₐ;p = p for any loop p

For every loop p, there is a loop p' for which p;p' = p';p = 1ₐ

To be precise, a group consists of some set (in this case a set of loops), a way to combine elements of that set (in this case concatenation), and must satisfy the last three equations above.

We call this group of loops the fundamental group of X at a and we denote it π₁(X,a). The fundamental group is also known as the first fundamental group. If X is path-connected, then we get (basically) the same group at any point. In this case, we sometimes omit the base-point a in the notation and simply write π₁(X).

For example, consider the circle S¹ and take some point a in S¹, e.g. the point on the right. We have a loop w on a that goes around the circle anticlockwise once. We can compose this loop with itself to obtain a loop w;w that goes around the circle twice, or we could invert it to obtain a loop w' that goes around clockwise once, etc. More generally, for any integer k, we have a loop wₖ in the circle defined as follows:

If k is positive, then wₖ = w;...;w is a loop that goes around the circle anticlockwise k times;

If k = 0, then w₀ = 1ₐ does not go around the circle;

If k is negative, then wₖ = w';...;w' is a loop that goes around the circle clockwise -k times.

One can see that, for example, w₁;w₋₂ = w;w';w', which goes around the circle anticlockwise once, then clockwise twice, is the same as w₋₁ = w', which goes around the circle clockwise once. More generally, for any two integers s and t, we have wₛ;wₜ = wₛ₊ₜ.

One might be tempted to say that the fundamental group of the circle is the integers with addition, in the sense that every loop wₖ has a corresponding integer k, where that correspondence is so that loop concatenation corresponds with integer addition. However, one has first to show that the family of loops wₖ are in fact all loops on the circle, and second that no two wₖ are the same (i.e. that wₛ and wₜ cannot continuously be deformed into one another for distinct integers s and t). We'll do this in the next few chapters.

As another example, consider the sphere S² and some point a on S². Given a loop p on a, we can always deform p into the constant loop 1ₐ on a. If we take some point b not on the loop p, then we can deform p by pushing everything away from b, towards a. Technically, this doesn't work for space-filling curves, but those can also be deformed into the constant path.

So, the fundamental group π₁(S²) is trivial: there are no loops besides the constant loop. We say a path-connected space with a trivial fundamental group is simply connected.

Given two points a and b in a space X and a path p: a ↝ b between them, we can identify loops on a with loops on b. For any loop q on a, we have that p';q;p is a loop on b (we first go from b to a, follow the loop q, then go back from a to b). Conversely, for any loop r on b, we have that p;r;p' is a loop on a for the same reason.

Someone who has studies group theory might recognize these operations as conjugation, although technically we are not conjugating elements of the same group. The loop p';q;p is the right-conjugate of q by p, and p;r;p' is the left-conjugate of r by p. Left- and right-conjugation are inverses of eachother: first left-conjugating a loop r on b by the path p, then right-conjugating, is equivalent to doing nothing. I.e. p';(p;r;p');p = p';p;r;p';p = r (the p and p' cancel out). Similarly, applying right-conjugation by p to a loop q on a, then applying left-conjugation, is equivalent to doing nothing.

The action of conjugation preserves the structure of the group. I.e. concatenating loops q₁ and q₂ on a corresponds to concatenating their conjugation by p. The conjugate of q₁;q₂ by p is p';q₁;q₂;p, and the concatenation of conjugates of q₁ and q₂ by p is (p';q₁;p);(p';q₂;p). The p and p' in the middle cancel out, and we're left with p';q₁;p;p';q₂;p = p';q₁;q₂;p, which we saw is the conjugate of q₁;q₂ by p.

Do note that the action of conjugation does usually depend on the specific path p we choose. If p₁ and p₂ are both paths from a to b and q is a loop on a, it may be that p₁';q;p₁ is not the same loop on b as p₂';q;p₂. For example, let X = 8 be the figure eight. Let a be the point in the middle, where the two circles meet, and let b the the point on the left of the left circle. Let p₁: a ↝ b be the path from a to b that goes through the top of the left circle, and let p₂: a ↝ b be the path that goes through the bottom. Let q: a ↝ a be the loop on a that goes around the right circle clockwise once. Then, p₁';q;p₁ is the loop on b that goes through the top of the left circle, goes around the right circle clockwise once, and then goes back through the top of the left circle. The loop p₂';q;p₂ is analogous, but goes through the bottom of the left circle instead. There is no way to deform the loop p₁';q;p₁ into the loop p₂';q;p₂.

For groups G and H, a group homomorphism φ: G -> H is a function preserving the group operation. I.e. φ(ab) = φ(a)φ(b). A group isomorphism is a bijective group homomorphism, or a group homomorphism that has an inverse homomorphism. For a path p: a ↝ b, we have a group isomorphism Φₚ: π₁(X,a) -> π₁(X,b) that maps a loop x on a to the conjugated loop p';x;p on b. For someone who knows category, try proving Φ is a functor from the fundamental groupoid Π₁(X) of X (defined in the obvious way) to the category of groups. That is, show that Φ_1ₐ is the identity on π₁(X,a) and show that, for paths p: a ↝ b and q: b ↝ c, we have Φ_(p;q) = Φ_q ∘ Φ_p.

Covering Spaces

A covering space of a space X is a space Y that "locally looks like" X. More precisely, each point X has one or more corresponding points in Y so that, for any point a in X and any corresponding point b in Y, there is a small part of X around a that looks the same as a small part of Y around b.

For spaces X and Y, a covering map is a continuous surjection π: Y ↠ X for which, for all x in X, there is an open neighbourhood U of x for which its preimage π⁻¹(U) under π is homeomorphic to a disjoint union of copies of U. Equivalently, there is an open cover Σ of X for which, for every U in Σ, π⁻¹(U) is a disjoint union ⨆{i∈I} V_i of open sets V_i homeomorphic to U. We call the V_i in Y the sheets of Y.

In the gif above, our space X is the circle (bottom). Three covering spaces of the circle are shown.

The left covering space is two copies of a circle. This covering space is called a trivial covering space, as it is some number of copies of the original space.

In the middle, we have a double loop. Every point on the circle has two corresponding points in the double loop: one in the inner loop, and one in the outer loop. The gif shows one of these two corresponding point at each moment in time.

Last, on the right, we have an infinite spiral. Every point on the circle has an infinite number of corresponding points in the infinite spiral: one on each level. The gif shows half of the visible covering points at each moment in time.

The first two covering spaces are double covers of the original space: every point in the original space has two corresponding points in the covering spaces. The last cover is an infinite cover, as every point in the original space has infinitely many corresponding points in the covering space.

Generally, we say a covering space Y of a space X is an n-sheeted cover if every point in X has n corresponding points in Y. The term "sheeted" may be confusing, but this refers to the fact that every small "patch" in X has a n corresponding "patches", called sheets, in Y. A 2-sheeted cover is a double cover, a 3-sheeted cover is a triple cover, etc.

Not every covering needs to be an n-sheeted cover for some n. For example, let X be the space of two disjoint circles. We can cover this with a space Y, which covers the first circle with just a single circle, and the second circle with two copies of that circle. Then, every point in the first circle has one corresponding point in the covering space, but every point in the second circle has two corresponding points in the covering space. So, the covering is neither 1-sheeted nor 2-sheeted.

This problem disappears, however, if the space we cover is path-connected. In this case, every point in X has the same number of points that cover it. Only X needs to be path-connected, the space Y may be disconnected.

For a point a in X, the collection of all points in Y corresponding to a is called the fibre of a. So, a covering is n-sheeted if every fibre has exactly n elements. We denote the fibre of a point a by Yₐ or Y_a.

Let π: Y ↠ X be a covering space and define the function n on X as follows: n(x) is the number of elements of the fibre Yₓ of x. Show that n is locally constant. I.e. show that n is continuous as a map from X to the set ℕ ∪ {∞} with the discrete topology.

Path-Lifting Property

Let X be a space and let Y be a covering space of X. Let a and b in points in X, let c be a point in Y in the fibre of a and let d be a point in Y in the fibre of b. Suppose we have a path p: c ↝ d in Y. Then, p has a corresponding path q: a ↝ b in X: every point on the path p has a corresponding point on the path q in X. We denote this path q by π ∘ p. This new path π ∘ p is called the projection of p onto X.

For example, let X be the circle and let Y be the infinite spiral. Let a and b be both the point on the right of the circle, let c be some point in the fibre of a and let d be the point one level below c. Then, the path p: c ↝ d that walks from c down to d corresponds to the loop w: a ↝ b that goes around the circle counterclockwise once (w is a loop as a = b are the same point).

We can also reverse this correspondence of paths. For a path p: a ↝ b in X and a point c in Y in the fibre of a, we can always find some point d in the fibre of b and a path p̃: c ↝ d in the covering space Y that projects onto p. We can do this because Y "locally looks like" X: we can start at c, we look at how p moves close to a in X, copy that in Y, then see how p moves close to where we were then, copy that in Y, and constantly take tiny steps in Y copying those in X, until we are at the end of the path. We call the path p̃ a lift of the path p. Note that the end-point of a lift is uniquely determined by its start-point and the path we're lifting.

Let π: Y ↠ X be a covering space, let p: [0,1] -> X be a path in X and let c be a point in the fibre of p(0). Show that there is a unique path p̃: [0,1] -> Y in Y for which p̃(0) = c and π ∘ p̃ = p.

For example, let X = S¹ be the circle, let a be the point on the right and let w be the loop on a that goes around the circle anticlockwise once. We consider the three covering spaces in the gif. In the first covering space, if we let c be the point on the right of the top circle, then the lift w̃ of w starting at c is a loop on c that goes around the top circle anticlockwise once. In the second covering space, if we let c be the point on the right on top, then the lift w̃ of w starting at c is a path that goes around the inner loop once and ends up at the bottom point. If we let c be the other point in the fibre of a, i.e. the point on the right on the bottom, then w̃ goes around the outer loop once and ends up at the top point on the right. Last, in the third covering space, if we let c be any point in the fibre of a, then w̃ is the path that goes down from c one level and ends up at the point directly below c.

As the end-point of a lift is uniquely determined by its start-point and the path we're lifting, we can define an action of paths in X onto points in Y. For a path p: a ↝ b in X and an element c of the fibre Y_a of a, define c·p ("c, apply p") to be the end-point of the lift p̃ of p starting at c. Then, c·p is a point in the fibre Y_b of b.

For a paths p: a ↝ b and q: b ↝ c in X, let d be a point in the fibre of a. Then, d·p is a point in the fibre of b, so we can apply q to it to obtain a point (d·p)·q in the fibre of c. However, this is equivalent to applying the concatenated path p;q to the point d directly, obtaining the point d·(p;q). So, we have (d·p)·q = d·(p;q). Further, applying the constant path 1ₐ to any point d results in d.

(unimportant note) If we read c·p as applying the path p in X to the point c in Y, it might make more sense to reverse the notation and write p·c instead ("p applied to c"). If we do this, then we get q·(p·c) = (p;q)·c, note that p and q swap here. If we want that the paths p and q do not swap in the notation, then it makes more sense to write concatenation in reverse: i.e. we write q·p for the path q after the path p, and we get q·(p·c) = (q·p)·c. What notation you like to use comes down to personal preference. In this blog, I'll stick to using p;q to denote the concatenation of p and q, and c·p for applying a path p in X to a point c in Y.

We can restrict to looking at the loops of some point a. As loops on a both start at end at a, they send elements of the fibre of a to the fibre of a. As we saw above, applying paths in X to points in Y respects concatenation of loops, so we say we have an action of the fundamental group π₁(X,a) (the group of loops, in case you forgot) on the fibre Yₐ of a.

c·1ₐ = c

(c·p)·q = c·(p;q)

The covering map π: Y -> X induces a functor Π₁(X) -> Set that maps a point x in X to its fibre Yₓ and maps a path p: x ↝ y to a function Y_x -> Y_y. This is how we categorically represent the action of all paths in X on a covering space.

If we have the circle with the infinite spiral as covering space, then each loop wₖ acts on Yₐ by moving each point up or down. If k is positive, then wₖ moves every point down k steps. If k is negative, then wₖ moves every point up k steps. From this, we may conclude the loops wₖ are indeed all distinct: each loop wₖ acts differently on Yₐ (moving all points down one step is not the same as moving them down two steps, or three, or up one step, etc). It's not possible for homotopic paths to act differently on the same point, as we can also lift the homotopy to show the two path lifts are homotopic and, in particular, must have the same end-point.

So, we are halfway done with computing the fundamental group of the circle. Now, we only need to show the loops wₖ are the only loops in the circle. We'll do this in a few chapters by showing the infinite spiral is a universal covering of the circle.

Functoriality of π₁

Let X be a path-connected space, let a be some point in X, let Y be a path-connected covering space of X and let b be some point in Y in the fibre of a. Every loop on b projects down to some loop on a. However, a loop on a needs to only lift up to a path between elements of the fibre of a. Only some loops on a lift to another loop on b.

For a loop p on a, the following are equivalent:

p lifts to a loop on b;

p is the projection of a loop on b.

If p satisfies the first condition, then it is the projection of that lift. If it satisfies the second, then lifting p will give the loop we originally projected.

Recall that π₁(X,a) is the group of all loops on X. If we let H be the set of all loops on a that are projections of loops on b, then H is a subgroup of π₁(X,a):

For loops p and q in H, we have that p;q is again in H

For a loop p in H, we have that p' is again in H

The constant loop 1ₐ is in H

The lift of p;q in Y is simply the lift of p, concatenated with the lift of q, both of which are loops and therefore concatenate to a loop. Similarly, the lift of p' to Y is the lift of p, reversed, which is still a loop. Last, 1_a is simply the projection of 1_b onto X, and 1_b is a loop on b.

For example, let X be the circle and let Y be the double loop. The loop w in X that goes around the circle anticlockwise once lifts to a path w̃ in Y that goes from the point on the right on top to the point on the right on the bottom. However, this lift w̃ is not a loop: its start-point and end-point are different. If we consider the loop w₂ in X that goes around the circle anticlockwise twice, then its lift w̃₂ is a loop: it goes from the point on the top on the right, around the inner loop to the bottom, and then around the outer loop to go back to the top. In general, a loop p in the circle lifts to a loop in Y if and only if p goes around the circle (clockwise or anticlockwise) an even number of times. So, the subgroup H consists of loops that go around the circle an even number of times.

If we instead let Y be the triple loop (a path-connected triple covering of the circle), then H would consist of all loops that go around the circle some number of times that is a multiple of three. If Y is the quadruple loop, then H consists of loops that go around the circle a multiple of four times. If Y would be the infinite spiral, then the subgroup H consists of only the constant loop.

We denote the subgroup H of all loops on a that are a projection of a loop on b as π⁎(π₁(Y,b)). Usually, the star is written in the subscript, but no such unicode character exists. The first π denotes the projection map of Y onto X.

Let X be a space and let S be a subspace of X. For example, X can be the sphere and S the equator of the sphere (so S is a circle in the sphere). As S is a subspace of X, every point in S is also a point in X and every path in S is also a path in X. Assume S and X are path-connected and let a be a point in S. We can consider the group of π₁(S,a) of loops on a within the space S. It might seem this is a subgroup of the group π₁(X,a) of loops on a within the space X. However, it doesn't need to be, for a bit of a subtle reason.

The space X is the blue disk. The subspace S is the white circle within the disk. We have a point a at the bottom of the circle and a point b at the top. Consider the two paths p,q: a ↝ b, where p goes around the left side of the circle from a to b, and q goes around the right side. The gif above shows a homotopy from p to q. However, this homotopy is in X. We have that the two paths p and q are homotopic as paths in X. However, this homotopy goes outside of the circle S. The paths p and q are not homotopic as paths in S.

The same happens with loops. Consider (in the example above) the loop on a that goes around the circle anticlockwise once. This loop, in S, is distinct from the constant loop on a. However, in X, we can contract this loop down to the constant loop. The blue disk X is simply connected, while the white circle S is not. As S and X disagree on which loops are the same and which are different, we have that π₁(S,a) is not a subgroup of π₁(X,a).

However, every loop in S is still a loop in X, even if they don't agree on which loops are equal. What we have here is a group homomorphism i⁎ from π₁(S,a) to π₁(X,a). Here, i denotes the inclusion map of S in X. The group homomorphism i⁎ maps a loop p in S to the loop p in X. We denote the loop p in X as i⁎(p).

The homomorphism i⁎ might forget some information: some loops in S get mapped to the constant loop in X. In the example of the disk and the circle above, the loop on a that goes around the circle anticlockwise once gets mapped to the identity loop in X (as they are homotopic in X), even though that loop is not homotopic to the identity loop in S. We can measure the information that a group homomorphism forgets with its kernel. The kernel of the map i⁎, denoted ker(i⁎), is the set of all loops in S that get mapped to the constant loop in X. In the disk-circle example, as X is simply connected, the kernel of i⁎ consists of all loops in S.

For a bit more of a non-trivial example, let S be the figure eight and let X be the figure eight with the left circle filled in. Let a be the point in the middle of the figure eight, where its two circles meet. The loop p that goes around the right circle of the figure eight is not homotopic to the constant loop in S. This loop p is also not homotopic to the constant loop in X, so p is not in the kernel of i⁎. Now, consider the loop q that goes around the left circle once. This loop is not homotopic to the constant loop in S but, as in X, the left circle is filled in, we can contract q down to the constant loop in X. So, q is in the kernel of i⁎. Last, consider the loop p;q;p', which first goes around the right loop, then around the left loop, and then around the right loop backwards. The loops q and p;q;p' are different loops in S (try coming up with a covering space of the figure eight that shows this). In X, we can contract the loop q within p;q;p' to the constant loop, so we're left with p;p'. Now, as p and its inverse are right next to eachother, we can contract p;p' to the constant loop. So, the loop p;q;p' is in the kernel of i⁎.

If we concatenate two loops in the kernel or reverse a loop in the kernel, we always get another loop in the kernel. So the kernel ker(i⁎) is a subgroup of π₁(S,a). However, as we saw with p;q;p' in the example above, we actually have that ker(i⁎) is a special kind of subgroup of π₁(S,a). For any loop q in the kernel, and any loop p in π₁(S,a) possibly outside the kernel, the conjugate p;q;p' of q by p is again in the kernel. So, as ker(i⁎) is closed under conjugating by elements outside of it, we say it is a normal subgroup of π₁(S,a).

One might ask if knowing what loops get mapped to the constant loop is enough to know what information i⁎ forgets as we go from π₁(S,a) to π₁(X,a). What if we have two different loops p and q in π₁(S,a) that are homotopic in X, but are not homotopic to the constant loop? Do we know p and q are homotopic in X by only looking at the kernel ker(i⁎)? Yes, we do! We can concatenate p with the inverse of q to obtain a loop p;q'. If p and q are homotopic in X, then p;q' is homotopic to p;p' in X, which is homotopic to the constant loop. So, p and q get mapped to the same loop in π₁(X,a) (i.e. are homotopic in X) exactly when p;q' is in the kernel of i⁎.

Although π₁(S,a) isn't itself a subgroup of π₁(X,a), we can still ask what loops in X are also loops in S. We say the set of loops in π₁(X,a) that are (homotopic in X to) a loop in S is the image of i⁎. We denote this image by im(i⁎) or i⁎(π₁(S,a)). The image of i⁎ is a subgroup of π₁(X,a), as we can concatenate and reverse loops in the image while still remaining in the image.

A pointed space is a space X along with a point a in X, called the base-point. A base-point preserving map from a pointed space (X,a) to a pointed space (Y,b) is a continuous map f: X -> Y for which f(a) = b. I'll denote the category of pointed spaces and base-point preserving maps as Top*.

Let (X,a) and (Y,b) be pointed spaces and let f: X -> Y be a base-point preserving map. The map f induces a group homomorphism f⁎: π₁(X,a) -> π₁(Y,a) that maps a loop p: [0,1] -> X on a to the loop f ∘ p: [0,1] -> Y on b. Show that f⁎ is a well-defined homomorphism, i.e. that it maps homotopic paths to homotopic paths and that f⁎(p;q) = f⁎(p);f⁎(q). Further, show that π₁ along with the action of turning f into f⁎ is a functor from Top* to Grp. I.e. show, where id_X: (X,a) -> (X,a) is the identity map, that (id_X)⁎ = id_π₁(X,a) and show, for pointed spaces (X,a), (Y,b) and (Z,c) and base-point preserving maps f: X -> Y and g: Y -> Z, that (g ∘ f)⁎ = g⁎ ∘ f⁎. Last, for a covering map π: Y ↠ X, show that π⁎ is an injection.

Universal Covering

Let X be a path-connected space. A universal covering space of X is a simply connected covering space X̂. I.e. it is a path-connected covering space with a trivial fundamental group.

Not every space has a universal covering. However, if it exists, then it is unique. I'll refer to spaces that have a universal covering as good spaces from now on, this terminology is not standard. Note that all good spaces must be path-connected.

An example of a path-connected space that is not good (i.e. is bad) is the Hawaiian earrings, pictured below.

The Hawaiian earrings consist of smaller and smaller circles all wedged together at a point.

The four example spaces from the beginning of the post, i.e. the circle, sphere, torus and figure eight, are all good spaces.

A space X is semi-locally simply connected (slsc) iff every point x has a neighbourhood basis of opens U for which i⁎: π₁(U,x) -> π₁(X,x) is trivial (maps everything to the identity element), where i is the inclusion map of U in X. I.e. every loop on x in U is homotopic in X with the constant loop. An example of a slsc space that is not locally simply connected is the cone on Hawaiian earrings.

If X is slsc, connected and locally path-connected, then it has a universal covering. If we fix some point a in X, then we can consider the set of all paths from a. We identify paths if they are homotopic, and we define a function from this set of paths to X by mapping a path to its end point. With the right topology, this set of paths from a up to homotopy is a universal covering of X.

There are a few reasons why a universal covering space is interesting. First is that it covers every other path-connected covering space of X. For example, consider the circle S¹ with its universal covering Ŝ¹, the infinite spiral. We have that the double loop is a path-connected covering space of S¹, and the infinite spiral Ŝ¹ covers the double loop where points on the inner loop correspond to points in the even levels of Ŝ¹, and points on the outer loop correspond to points in the odd levels of Ŝ¹ (the gif at the beginning of the chapter Covering Spaces shows, for each point of the double loop, all its corresponding points in the infinite spiral).

The second reason is that loops on a given point a in the space X have a one-to-one correspondence with points in the fibre of a in the universal covering space of X.

Let X be a good space and let Y be a path-connected covering space of X. We'd like to know why the universal covering of X covers Y. Fix some point a in X, some point b in Y in the fibre of x, and some point c in the universal covering X̂ in the fibre of x. Let z be a point in X̂ that projects down to some point x in X. As X̂ is path-connected, there is some path p: c ↝ z in X̂. This path projects down to some path q: a ↝ x which, in turn, lifts up to some path q̃: b ↝ y in Y. We'd like to define y as the projection of the point z in X̂ (thus making X̂ a covering of Y), but there is a problem: how do we know the point y does not depend on the path p: c ↝ z in X̂ we chose? I.e. can the procedure above for different paths p₁,p₂: c ↝ z in X̂ ever result in different points y₁,y₂ in Y? This is where the assumption that the universal covering X̂ is simply connected comes in: the path p₁;p₂' is homotopic with the constant path (by definition of simply connected), and therefore p₁ and p₂ are homotopic to eachother (you can probably see this intuitively, but proving it formally could also be fun). Homotopic paths project down to homotopic paths and lift up to homotopic paths, so the end-points y₁ and y₂ of the lifts of the projections must be the same. So, which path p: c ↝ z in X̂ we choose indeed doesn't matter.

As for the second claim, that loops on a point in X have a one-to-one correspondence to points in the fibre of a in the covering space X̂, let X be a good space. Fix some point a in X and a point b in X̂ in the fibre of a. Then, every loop p on a acts on the point b giving us a point b·p (the end-point of the lift of p, starting at b). As X̂ is path-connected, every point in the fibre of a is of the form b·p: for a point z in the fibre of a, we can find a path q: b ↝ z in X̂, which then projects down to some loop p on a for which b·p = c. So, every point in the fibre of a has a corresponding loop on a. However, this is true for any path-connected covering space of X. The claim is that this corresponding loop on a is unique. I.e. for loops p and q on a, if b·p = b·q, then the loops p and q are homotopic in X. To prove this, we use the fact that X̂ is simply connected. Let z = b·p = b·q. Then, we have lifts p̃,q̃: b ↝ z of the paths p̃ and q̃. As X is simply connected, the paths p̃ and q̃ are homotopic to eachother. Homotopic paths project down to homotopic paths, so p and q are homotopic, as desired.

The infinite spiral is a simply connected covering space of the circle. Let a be the point in the circle on the right. The loops wₖ on a in the circle act on the fibre of a in the infinite spiral by sending elements up or down some number of steps. As the loops wₖ are enough to send any point of the fibre to any other point, we know, by the one-to-one correspondence of points and loops, that the loops wₖ are all loops in the circle. So, we may conclude the fundamental group of the circle is the integers with addition.

Covering Spaces are G-Sets

In this chapter (and only this chapter), we won't require covering maps to be surjective. To emphasize this, I'll write π: Y -> X instead of π: Y ↠ X to denote covering spaces. This might seem a bit weird, but this will make the category Cov(X) introduced in the next paragraph nicer.

Let X be a space and let π: Y -> X and ρ: Z -> X be covering spaces of X. A homomorphism of covering spaces from Y to Z is a continuous map f: Y -> Z respecting the covering maps, i.e. for which ρ ∘ f = π. As compositions of homomorphisms of covering spaces are homomorphisms of covering spaces, we have a category Cov(X) of covering spaces of X. What are the initial and terminal object of this category? What are the coproducts? Assume X is simply connected, show that Cov(X) is equivalent to Set, the category of sets.

For a space X, a point a in X and a covering space Y of X, we have that the fundamental group π₁(X,a) acts on the fibre Yₐ of a: for a loop p on a and a point b in the fibre, b·p is the point in the fibre that is the end-point of the lift of p that starts at b.

More precisely, for a group G, where we can combine elements p and q of G into a new element p;q, and for some set A, an action of G on A has the following data and conditions:

For an element a of A and p of G, we have that a·p is an element of A

Where 1 is the identity element of G, a·1 = a

For an element a of A and elements p,q of G, we have (a·p)·q = a·(p;q)

In the example at the beginning of this chapter, we have that G is the fundamental group π₁(X,a) and the set A is the fibre Yₐ. However, we can also define actions more abstractly.

For example, consider the fundamental group π₁(S¹,a) of the circle, consisting of loops wₖ for integers k. We can define an action of this group on the rationals ℚ as follows: for a rational number b and a loop wₖ on a, we set b·wₖ = b + k. We can see that b·1ₐ = b·w₀ = b + 0 = b and (b·wₛ)·wₜ = b + s + t = b·wₛ₊ₜ = b·(wₛ;wₜ), so this is a well-defined action.

Is there some covering space Y of the circle for which π₁(S¹,a) acts the same way on Yₐ as it does on ℚ with the action described above? I.e. where the points in the fibre of a correspond with rational numbers, and loops act on those points (by path-lifting) in the same way as they act on their corresponding rational numbers with the action described above?

Yes! In fact, for any good space X, every action of its fundamental group on any set can be represented as a covering space. For a group G, we call a set A along with an action of G on A a G-set, hence the chapter name Covering Spaces are G-Sets. In this chapter, we'll look at connections between covering spaces of a good space and their corresponding G-sets. In the first half of this chapter, we'll look at how to construct this covering space. In the second half, we'll compare G-sets with their covering spaces.

Let G be a group. We define a category G-Set of G-sets. Objects are G-sets (A,α) with a set A and an action α: G ↷ A (we'll take α to be a right-action, as that aligns with the notation we've been using up to this point). A morphism from (A,α) to (B,β) is a function f: A -> B for which, for all a in A and g in G, we have f(a·g) = f(a)·g.

Let X be a space, let a be a point in X and let G = π₁(X,a) be the fundamental group of X at a. We define a functor F: Cov(X) -> G-Set from covering spaces of X to G-sets. For a covering space π: Y -> X, we set F(Y) = (A,α) where A = Yₐ and G acts on A in the usual way. Let π: Y -> X and ρ: Z -> X be covering spaces and let f: Y -> Z be a homomorphism of covering spaces. We set F(f) = f|Yₐ.

Let G be a group and let A be a set G acts on. We can partition A into different orbits, where two elements are in the same orbit if there is an element of G that sends one to the other. I.e. for a and b in A, we have that a and b are in the same orbit iff there is some g in G for which a·g = b. For example, for a covering space Y of a path-connected space X, the orbits of the action of the fundamental group π₁(X,a) on the fibre Yₐ correspond to the path-connected components of Y. In the action of π₁(S¹,a) on ℚ described above, for every rational number 0 ≤ b < 1, we have an orbit consisting of copies of a shifted by integers (i.e. b + k for integers k).

We'll first look at how to construct covering spaces for specific kinds of actions: transitive actions. An action of a group G on a set A is said to be transitive iff it has exactly one orbit, i.e. any element can be send to any other element. If G is the fundamental group of some good space, we'd like to find a path-connected covering space representing that transitive action.

Another important notion is that of a free action: an action of a group G on a set A is free iff every non-identity element of G sends every element of A to a different element of A. For a good space X, the action of the fundamental group π₁(X,a) on the fibre X̂ₐ in the universal covering is free. However, the action of the fundamental group of the circle on the fibre in the double loop is not free, as the loop w₂ sends every element of that fibre to itself.

There is essentially only one way a group can act freely and transitively. If X is a good space and G = π₁(X,a) is its fundamental group, then the action of G on X̂ₐ is the free and transitive action of G. For a set A, if G acts only transitively on A, then there is some measure in which G fails to acts freely. Fix some point e of A and consider the subgroup H of G consisting of all elements of G that send e to itself. In this case, we define a quotient space Y of the universal covering X̂, which is a space resulting from gluing certain points together. Fix some point b in X̂ in the fibre of a. Let x and y be points in X̂ that are in the same fibre, i.e. project down to the same point z in X. Let p: b ↝ x and q: b ↝ y be paths in X̂. These paths project down to paths r,s: a ↝ z in X. Together, these paths define a loop r;s' on a. We glue the points x and y together in the quotient space Y iff r;s' is in the subgroup H, i.e. if r;s' acts on the point e of A by sending it to itself. Then, in the quotient space Y, we have that the point b is glued together with every point b·p for loops p on a that act on A by sending e to itself. The quotient space Y is still a covering space of X, but now the fundamental group G acts on Yₐ in the same way as it does on A.

For example, let X be the circle and let a be the point on the right of the circle. Consider the action of π₁(S¹,a) on the two-element set A = {0,1} where b·wₖ is 0 if b+k is even and b·wₖ is 1 if b+k is odd. If we set e = 0, then the subgroup H consists of all loops wₖ for which k is even (the choice of e doesn't matter in this case). Points in the universal covering Ŝ¹ are glued together iff they are in the same fibre and the distance between them, measured in levels of the infinite spiral, is an even number. Fix some point b in the infinite spiral, let x and y be points in the same fibre and let p: b ↝ x and q: b ↝ y be paths in Ŝ¹. These paths project down to paths r and s in S¹, and r;s' = wₖ for some k that measures the distance between x and y (the sign of k indicates which point is above and which is below). x and y are glued together if k is even, i.e. if the distance between them is even. So, the resulting quotient space is the double loop, as we would expect.

Now that we know how to construct covering spaces for transitive actions, we can try to construct them for actions in general. This step is a lot less complicated. Let X be a good space, let G = π₁(X,a) be its fundamental group, let A be a set and let G act on A. Then, G acts on each orbit of A individually, and this action of G on an orbit is transitive. So, for each orbit O of A, we can construct a covering space of X for the action of G on O as described above, and we can simply take all of those covering spaces of orbits together as the covering space of X corresponding to the action of G on A. So, the covering space of X corresponding to the action of G on A has a path-connected component for each orbit O of A.

Now that we have seen how to construct covering spaces of a good space corresponding to certain actions, we'll now look at properties of actions and how they reflect in their corresponding covering spaces, and vice-versa.

We already saw a transitive action corresponds to a path-connected covering space. What's also not hard to see is that a free action corresponds to a covering space that consists of some number of copies of the universal covering space.

Another property of an action are its stabilizers. For a group G acting on a set A, the stabilizer group of a point a in A is the subgroup H of G of all elements of G that send a to itself. We denote this stabilizer subgroup as Stab_a. For a covering space Y of a path-connected space X with fundamental group G, the stabilizer of a point b in Y consists of all loops in X that lift up to a loop on b in Y, i.e. the stabilizer of a point b is the group π⁎(π₁(Y,b)), corresponding to the fundamental group π₁(Y,b). Let a be a point in X and b and c be points in Y in the fibre of a. In the chapter Fundamental Group, we saw that a path p: b ↝ c induces a one-to-one correspondence between loops on b and loops on c by conjugation, i.e. an isomorphism between the fundamental groups π₁(Y,b) and π₁(Y,c). As p is a path between elements of the same fibre, we have that p projects down to some loop q on the point a, i.e. an element of π₁(X,a). For a loop r in π⁎(π₁(Y,b)), i.e. a loop on a that lifts up to a loop on b, we have that the conjugate q';r;q is a loop in π⁎(π₁(Y,c)), i.e. it lifts up to a loop on c. Stabilizer groups of points of the same orbit are conjugate subgroups of eachother, which corresponds to the fact that paths between points in a space induce isomorphisms on the fundamental groups at those points by conjugation.

Regular Coverings

Consider the following two triple covers of the figure eight:

The point in the middle of the figure eight is shown, and so are the points in its fibres in the two covers. The left circle, and the paths corresponding to the left circle in the covers, is coloured red, and the right circle is coloured blue. The arrows indicate the direction of the paths. Let p be the loop that goes around the circle clockwise once and let q be the loop that goes around the right circle clockwise once. We'll use Y to denote the left covering space and Z to denote the right covering space.

If we label the points in the first covering space, starting from the top, going clockwise, a, b and c, then p acts on {a,b,c} by sending a to b, b to c and c to a. No point stays where it is. The loop q acts on {a,b,c} by sending every point to itself. Every loop in the figure eight acts on {a,b,c} by either rotating the points (so no point gets send to itself) or mapping each point to itself.

If we label the points in the second covering space, going from let to right, x, y and z, then p acts on {x,y,z} by mapping x to x and swapping y and z. The loop q acts on {x,y,z} by swapping x and y and sending z to itself. The path p;q;p swaps x and z and sends y to itself. Some loops in the figure eight act on {x,y,z} by fixing some, but not all, of the points.

The left covering space looks more symmetric: every point in the fibre looks like any other point in the fibre. The right covering space is not symmetric: the point x is clearly distinct from the point y, as the point x has a red loop on it, while y has not.

Consider the subgroup π⁎(π₁(Y,a)) of π₁(8). Let r be a loop in π⁎(π₁(Y,a)), i.e. a loop r for which a·r = a, and let s be any loop in π₁(8). Consider the loop s';r;s. First, s' sends the point a to one of the three points a, b or c, then r sends that point to itself, and s sends that point back to a. So, we have a·s';r;s = a. So, π⁎(π₁(Y,a)) is closed under conjugation by elements of π₁(8), i.e. the subgroup π⁎(π₁(Y,a)) is normal.

Consider the subgroup π⁎(π₁(Z,x)) of π₁(8). Let r be the loop p in π⁎(π₁(Z,x)), which sends x to itself. Let s be the loop q in π₁(8). Consider the loop s';r;s = q';p;q. First, q' sends x to y, then p sends y to z and q sends z to itself. The conjugate q';p;q of a loop fixing x sends x to z, so the subgroup π⁎(π₁(Y,a)) is not closed under conjugation by elements of π₁(8) and is therefore not normal.

The space Y, on which all loops act by either fixing all points or sending all points to a different point / is symmetric / induces normal subgroups π⁎(π₁(Y,a)), is called a regular cover of the figure eight. The space Z, on which loops act on each point differently / is not symmetric / induces subgroups π⁎(π₁(Z,x)) that are not normal, is called an irregular cover of the figure eight.

For a space X and a covering space Y of X, we write Aut(Y/X) for the automorphism group of Y in the category Cov(X) of covering spaces of X. I.e. the group of homeomorphisms of Y that respect the covering map to X.

For a path-connected space X and a covering space π: Y ↠ X, TFAE:

For every loop p in π₁(X,a), either, for all b in Yₐ, we have b·p = b or, for all b in Yₐ, we have b·p ≠ b;

The group Aut(Y/X) acts transitively on each fibre of Y;

For every b in Y, π⁎(π₁(Y,b)) is a normal subgroup of π₁(X,a).

In this case, we call Y a regular covering of X.

Last Chapter

That's all I had to say for now on algebraic topology. Maybe I'll make a second part where I talk about the Seifert-Van Kampen theorem, or simplicial homology, or higher homotopy groups.