The Fundamental Group, Part II – The Basepoint

We constructed the fundamental group by looking at the homotopy classes of loops with a fixed basepoint.  The question that naturally arises is how does the fundamental group change if we move the basepoint.  To begin let’s suppose we’re in a path connected space and we have two basepoints we’re concerned about, say and .  Since is path connected, there exists some path from to .  We can use this to convert loops into loops.

What we’ll do is take a loop, say , based at and use our path to get from to , follow the loop, then take the reverse path back to .  This is illustrated in the image below (adapted from an image in Hatcher’s book).

This new path is a loop based at .  Now, if two -loops, say and are homotopic, then so are the associated -loops: .

If we now construct a map by setting we see that the map will be well-defined since our construction of the -loops respects homotopy.  But what happens when we concatenate two -loops?

Suppose and are loops based at .  We want to see what this map does to the homotopy class of .

So our map is in fact a homomorphism.  Notice that implies , which means that our map is injective.

To show the map is surjective, let be an -loop, then is an -loop.  Applying we have , and so our map is surjective.  Together with injectivity above we have shown that in a path connected space the fundamental group of and are isomorphic.  We then simply write since the choice of basepoint is basically irrelevant.

If our space is not path connected, but and are in the same path component, the above gives us that .

An Aside on Path Connectedness

Long ago we talked about topological spaces being connected, which meant we couldn’t partition the space into two disjoint non-empty open sets.  There is a stronger notion that we’ll need when working with the fundamental group called path connectedness.  We say that a space is path connected if for every pair of points there exists a path in the space, , such that and ; we can connect every pair of points with a path.

If a space is not path connected, it can be decomposed into a collection of disjoint subsets with are path connected.  These subsets are called the path components of the space.  More specifically, path connectedness is an equivalence relation on the space and the path components are the equivalence classes.  Clearly every point is path connected to itself (reflexivity); if and are connected by path , then the reverse path connects and (symmetry); and the concatenation operation from last time gives us transitivity.

We said that path connectedness was a stronger condition than connectedness.  This means that path connectedness implies regular ol’ connectedness.  To see this, suppose that is path connected.  Let be connected by path .  Let be an open set containing , and an open set containing .  Now consider the preimages and .  Since is a path it is continuous, so these are two open subsets of [0, 1].  We must have since and .  We know that [0, 1] is connected, however.  This implies that must be connected (otherwise their preimages would be disjoint open sets covering [0, 1]).  Since and were arbitrary, we can generalize to get that is connected.

Notice that this also means that a continuous path must stay inside of one path component.

The Fundamental Group, Part I – Connecting Paths

Imagine that you have two paths, and .  If it should happen that the terminal point of is the initial point of (that is, if ), then we can basically concatenate and into a new path .  To do this formally we need to define , which means we have to reparameterize and a little bit.  We’ll define so that it behaves as on [0, 1/2] and as on [1/2, 1] (notice this will be well-defined since we’re assuming the terminal point of is the initial point of ).

We will let the asterisk denote this concatenation operation: .

In the case that and are loops with the same basepoint, we’re just gluing the two loops together.  In the image below the basepoint is represented by the red dot.  We’re taking these two loops and gluing them together so that we first wind around the first loop (the one on the left) and then around the second loop.

Note that the red dot is the same basepoint for both loops, even though in our picture it doesn’t stay aligned (just for graphical convenience).

If , and are loops (with the same basepoint) and , we can show that .  Letting be the homotopy taking to , we construct the homotopy by defining .  This always behaves as on the “second half” of the loop, and deforms to on the first half.  Of course, this also can be used to show that .

We can now use to help us form an operation on the homotopy classes of loops.  We will simply define as .  This operation is clearly associative, since our concatenation is associative. The identity element of this operation is the homotopy class of the constant function.  Keeping in mind that these are loops, the constant function is the function that maps everything to the basepoint.  Let denote this map.  Then we have that , where the last equality follows from the fact that with we traverse twice as fast as normal, then just sit at the basepoint.  This is homotopic to , and so and live in the same homotopy class.

Finally, our operation is invertible.  To define the inverse operation we will take a loop and will let denote the reverse loop: .  This reverse loop will have the same shape as the initial loop, but traverse the loop in the opposite direction.  Now notice that can be compressed into the constant map by constructing a homotopy that traverses most of , then goes ahead and starts coming back through .  This homotopy slowly reels back to the basepoint, which eventually turns it into the constant map.  Since ( being the constant basepoint map), we have that , and similarly .

So, our operation is associtiative, has an identity, and is invertible, so we have a group of the homotopy classes of loops for a fixed basepoint.  This is known as the fundamental group and is denoted , where is the space and is the basepoint.  In the next post we’ll look at what happens when the basepoint changes.

Homotopy Classes

If we have two continuous maps, and , we would like to view and as essentially being the same map if they are homotopic to one another; we want to consider and as being equivalent with respect to homotopy.  To do this we need to show that homotopy is an equivalence relation on the set of continuous maps.  This is a pretty simple thing to verify.

Suppose are all homotopic to one another with being the homotopy from to , and the homotopy from to .  To show that we can simply take the identity homotopy: where .  For reflexivity we’ll construct a new homotopy where .  This simply reverses the direction of the homotopy: instead of going from to , we start at and go to .  For transitivity we construct the homotopy with .  Then takes us to by first taking us .

This shows that homotopy is an equivalence relation.  We will refer to equivalence classes here as being homotopy classes.

Just as general homotopy is an equivalence relation, so too is path homotopy.  Since path homotopy has the additional requirement of being anchored down at the path’s endpoints, though, path homotopy is an equivalence relation on all paths with two fixed endpoints.

If a function is in the same homotopy class as a constant function, we say that the function is null-homotopic.  This means that we can squeeze the function’s graph down to a single point with a homotopy.  Notice that the only way a path can be null-homotopic (when we say homotopic and are talking about paths, we mean path homotopic) is if its initial and terminal points are the same.  Such a path is called a loop and the homotopy classes of loops is what we will actually place a group structure on to construct the fundamental group.

Path Homotopy

Over the next few posts we’re going to be discussing homotopy groups in topology.  Homotopy groups will be our window into the world of algebra, giving us the opportunity to use machinery from algebra to do interesting things with topological spaces (and vice versa).  Before trying to put an algebraic structure on a topological space though, we need to discuss paths.

A path is a mathematical way of connecting two points in space.  In the case of the Euclidean plane, for instance, there is a line segment between any two points.  There are also all sorts of curves that start at one point and end at another.  These are paths as well, provided they can be represented as continuous functions.  That is, a path in a topological space is a continuous function .  We call the points and the initial and terminal points of the path.  The choice of the interval [0, 1] is semi-arbitrary, since we could just as easily have used or a finite portion of a ray in some other space.  However, [0, 1] is convenient and it’s not very difficult to reparameterize a path that uses some other interval to “fit” in [0, 1].  For example, if we had a path , we could just take , then if we define , we’d have the same path represented as .

Now, between two points, say , there may be several paths.  We would like to know if two paths connecting these points are essentially the same thing or not.  That is, if one path can be deformed into the other.  We saw last time that homotopy gave us a way to deform functions, but we have to take a little bit of care when it comes to paths.  You see, we would like to use homotopy to bend one path into another, but with the restriction that all of the intermediate functions are paths as well.  Our homotopy should be anchored down at the path’s endpoints.

If and are paths with initial point and terminal point , we say that and are path homotopic if there exists a homotopy (recall we may think of a homotopy as a family of functions) such that for every , we have that is also a path from to .  Recalling the video from last time, we had a homotopy between two curves in the plane.  This was not an example of a path homotopy because, for one thing, the endpoints of the paths changed as we bent one curve into the other.

Modifying that example slightly, however, we can construct a path homotopy.  In this case our space is and our two paths are given by and .  Using the straight-line homotopy we then have the following.

Notice how here the points (-1, 0) and (1, 0) are the endpoints for all of the paths in the homotopy.

Of course, not all paths are homotopic to one another.  In the special case of all paths are homotopic, just as before, but if we simply remove the origin from , it’s easy to construct two non homotopic paths.

Here the blue path and the red path are simply the portions of the unit circle connecting (-1, 0) and (1, 0).  However these paths are not path homotopic as we would have to either break a path or move through the origin to deform one path into the other, just as before.  Even though these two curves are not path homotopic, they are in fact homotopic in as the following video demonstrates.

The purple path bends the blue path into the red path while avoiding the origin.  Again, this is not a path homotopy since the endpoints aren’t anchored down.

Homotopy of Functions

Given two continuous real-valued functions, it’s easy to imagine how one of these functions may be deformed into the other.  For instance, if we take the sine curve modified so that it passes through the points (-1, 0) and (0, 1), we can bend and stretch it into a curve that passes through (-1, 0) and (0, 1) but as a semicircle.  Consider the video below.

Here we’re taking the curve, grabbing the endpoints and swapping them, bending the curve as we do so.  In topology this idea of deforming one function into another is formalized by homotopy.  A homotopy between two functions, and , is a continuous map that “through time” bends one curve into the other.  If a homotopy between and exists we say that and are homotopic and write .

(It should be noted that the above deformation is with parametric curves.  That is, these are functions of the form , and not functions.  This is why having the crossing and failing the vertical line test is okay.)

Formally, a homotopy between two continuous functions is a continuous function that satisfies a few properties.  Before discussing those properties, however, let’s make some ideas and notation clear.  First we’ll just let denote the closed unit interval [0, 1], mainly just to keep typing down.  Now, when we say a continuous function from to , we of course mean continuous in the topological sense (preimages of open sets are open), but what’s an open set in ?  It’d be nice to say that it was just the Cartestian product of an open set in and an open set in , but unfortunately that’s not enough.

If you think about Euclidean plane, , it seems entirely reasonable to say that the ball (that is, the collection of all points of distance less than one from the origin) is open, but we can’t represent this ball as the product of two open sets in the real line.  If we take an open set and an open set from , their product is just going to be a collection of open rectangles in the plane.  Taking , for instance, we have the following.

If and are open sets, then is going to look something like this, since every open set can be represented as a countable collection of open intervals.  Of course, we could construct a more complicated set by changing the sizes and number of intervals making up or , but we’re still limited to sets that look something like the above.

The open ball, however, can not be represented like this.  If we took lots and lots, infinitely many in fact, open rectangles of various sizes and could put them together however we like, we could construct the open ball.  That is, the open rectangles of should form a basis for the topology.

The product topology of is precisely the topology whose basis is the collection of the open “rectangles” in .  So, even though an open set in may not be the product of an open set in and an open set in , it can be described as a union of those open rectangles (where we allow infinite unions).  It is in this sense that we mean is continuous: the preimage of an open set in is an open set in with the product topology.

We may think of the function as actually being a family of functions , for where actually means the function .  We will generally think of homotopies as deforming one function into another through time, where “time” is given by the interval [0, 1].  That is, the starting time is and the stopping time is .  At each point   associate the function .  So, the homotopy is giving us a family of functions, one for each moment in time.  The condition of continuity basically means that if is small, then the change from to is small.

Now, earlier we said that this function has to satisfy a few properties to be a homotopy taking to .  The first condition was continuity, and the second is simply that and .  This means that when we start, we have the function and when we finish we have the function .  So, a homotopy gives us a way to take a function and bend it around however we like provided that we don’t break the curve and that we eventually stop bending the function when we get to the function .

In the special case of the valued functions we can actually show that all continuous maps are homotopic to one another.  What we will do is add little pieces of to little pieces of , where the portion of that we’re using shrinks, and the portion of grows, as goes from 0 to 1.  We then define the homotopy  .  It’s easy to see that this satisfies the conditions for a homotopy as when we have , and when we have .  Since this is composition of continuous functions (addition, substraction, multiplication, and the maps and ), this is also a continuous function.  This is known as the straight-line homotopy since we imagine taking the curve and the curve and drawing lines between associated points (i.e., drawing a line between f(s) and g(s) for each s), then the homotopy just moves to along those lines.  This result applies to convex subspaces of as well.

At this point you may wonder about how to find two functions that are not homotopic.  We’ve just seen that all continuous functions are homotopic to one another, so we need to consider some other space.  Suppose we just take and remove the origin; .  If is the unit circle, and is a line segment connecting the points (2, -1) and (2, 1), then we can’t deform the circle into this line segment in our space.

Intuitively it’s easy to see that in order to deform the circle like this we’d either have to tear the circle, which would make the deformation discontinuous, or the circle would have to pass through the origin as it’s deformed.  However, the origin isn’t in our space, so there can be no intermediate function in our homotopy that passes through the origin.  This means the circle and the line segment can’t be homotopic.

The Hausdorff Property

One of the important tools in topology are properties that are invariant between spaces.  These are properties of topological spaces that remain unchanged under certain “operations” that we may perform on the spaces.  We’ve already seen one such property in connectedness, and now we’re going to discuss another one: the Hausdorff property.  We say that a space has the Hausdorff property if for every pair of points in the set, there are disjoint open sets that contain the points.  For example, in the case of the real line with the standard topology, we can pick two distinct points, and , and find two open intervals, and , such that and , but .  If and , for instance, we may take and .

Proposition: If is a metric space, then the topology induced by the metric has the Hausdorff property.

Proof:  Let and denote the distance between and by .  Clearly and are disjoint open sets containing and , respectively.

At first glance, it would seem that a topological space which does not posess the Hausdorff property is a strange or esoteric abstract space, but there is an easy way to extend the real line, with its standard topology, into a non-Hausdorff space.  We simply add a second zero to real line, which we’ll call Z.  The open sets containing Z are the same as the open sets containing zero, except that zero is replaced by Z.  For example, is such an open interval.  Now, note that this space (called the real line with a double point) is not Hausdorff:  Let and be any two open sets with and .  By the way we’ve defined a topology on this space, must contain an open interval containing Z.  This open interval is precisely the same as some open interval containing zero (with zero replaced by Z).  Any open interval containing zero must have a non-empty intersection with this open set, however.  Since must contain some open interval containing zero, and must have a non-empty intersection.  This means the space can not be Hausdorff.

Continuous Functions from the Reals to the Rationals

Using the ideas of connectedness previously discussed, we can show that a continuous function from to must be constant.  We will do this by showing that if we have such a continuous function, we’d have a surjection between a connected space and a disconnected space, which we showed in the last post can’t happen.

Suppose is a non-constant map; we’re not making any assumptions about continuity or surjectivity, just that the image of the map has at least two distinct points.  Let be two distinct points in the image of ; .  Now note that is a surjection from the reals to its image, .  Assume, without loss of generality, .  Since these points are distinct, there must be an irrational number in , call it , between them.  Let and .  Notice that these are open in the subspace topology of as is open in and , and likewise for .  These two sets are clearly disjoint and cover and are non-empty as .  This means that is a disconnected space.  We know that is a surjection to a disconnected space.  This is impossible if is continuous, so we have if is non-constant it is not continuous, which by contrapositive means that if is continuous, it must be constant.

(This is a modified version of the proof in Crossley, with some gaps filled in.)

Connected Spaces

Crossley motivates the idea of connectivity of a topological space by posing the question of whether or not there can be a continuous surjection from to the set with the discrete topology.  If such a surjection existed, then and would be open subsets of .  Also note that as is partitioned by these sets, must be partitioned by and ; i.e., with .  We further require that neither nor may be empty as is a surjection.  If we can show that no such and can exist, then we will have shown there is no continuous surjection from to .

Let and and assume, wlog, that and consider the interval : one endpoint is in and the other is in .  Split the interval into two halves of equal size, and .  One of these intervals will lie entirely in one set, while the other will have an endpoint in each set.  This follows from the fact that and are open sets that partition .  Call the interval with an endpoint in each of the sets .  Notice if we split into two closed intervals of equal length, we will again have an interval with an endpoint in each of and .  Call this interval .  In general, we will let denote the interval obtained if we split into two halves and pick the half with an endpoint in and the other in .  Notice is a decreasing sequence of closed intervals.  Furthermore, because of the way we constructed the , their measure is decreasing to zero.  These sets must have a non-empty intersection, but because the measure is shrinking to zero, this intersection must consist of a single point.  This point, then, must be in both and .  This contradicts the fact that and are non-empty disjoint sets, however.  This means we can not find open sets and that are disjoint, both non-empty, and whose union is . So we conclude that there can be no continuous surjection from to .

In general a topological space that does have two disjoint, non-empty open sets that partition the space is called disconnected, while a space where no such sets exist is called connected.  We say that a subset of a topological space is disconnected if it’s a disconnected topological space when endowed with the subspace topology.  The mentioned above is a disconnected space.  Another example is with the subspace topology from .  Note that this is not simply the discrete topology on , as it was with .  (The set {0} will not be open since any open subset of will contain an open interval, so if we take some open set of reals and intersect it with the rationals, we’ll get all of the rationals in the intervals making up the open set.)  Letting and gives us two non-empty open sets whose intersection is empty, but whose union is all of .  Similarly, is disconnected; let and .

In fact, if we have any connected space , there can be no continuous surjection to .  The proof of this follows the same outline for above.  We can also show that if is any disconnected space there is a continuous surjection from to : we just split into two disjoint, non-empty open sets that partition (by disconnectedness), map everything in one set to 1 and everything in the other set to -1.  This tells us that if is connected and is disconnected, there is no continuous surjection from to .  If there were, we’d just compose that surjection with the surjection from to then we’d have a map from to .

Now we’ll show that any continuous function from to must be constant.  Suppose that is continuous and let be a point in the image of .  Let and .  Clearly and are disjoint sets whose union is   Since is continuous, both and must be open.  However, is a connected space, so one of and must be empty.  We know that is not empty, since we specifically chose it to be the preimage of a point in the image of .  This means that , so and everything maps to under , and so must be constant.

The Discrete, Trivial, and Subspace Topologies

Given a set , there are always at least two topologies that can be placed on : the discrete topology and the trivial topology.  The discrete topology is the topology where every set is open; the collection of open sets in the topology is simply the collection of all sets, .  If is a map between topological spaces where has the discrete topology, then will be continuous, regardless of how it maps the elements of to .  The trivial topology is the bare minimum for a topology: the empty set and the entire set .  For any map , then, will be continuous as and .  Regardless of the topology on , those two sets will be open, so is continuous.

Suppose now that is a topological space where is any topology.  If we can place a topology on by using the topology of .  Let be a collection of subsets of where ( is an open subset of ) if for some .  We call the subspace topology of .  For instance if we take with the standard topology, we can place a topology on using this subspace topology.  In such a case, sets like are considered open in this topology since , even though this is not considered an open subset of .

If and are topological spaces and is continuous, we’d like to know when is continuous when restricted to a subspace of .  We can show that if define as the image of under , and assign the subspace topology in , then is continuous.  Let be an open subset of .  Then for some open subset of .  We know .  By definition and is an open subset of by the continuity of .  That is, is an open subset of with the subspace topology, so the restriction of to is continuous.

In the case of , the subspace topology inherited from is the same as the discrete topology on .  Let .  Then is an open set with the subspace topology as .  This gives that every singleton set is an open set and since any set will be a union of singletons, any set will be open.