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Topology/The fundamental group

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The basic idea of the fundamental groupEdit


An easy way to approach the concept of fundamental group is to start with a concrete example. Let us consider the 2-sphere   and the surface of the torus.

Let us start thinking about two types of loops of the torus (paths starting and ending at the same point). It seems that a path around the "arm" of the torus is substantially different from a “local” simple loop: one cannot be deformed into the other. On the other hand, in the sphere it seems that all the loops can be deformed into any other loop. The set of "types of loops" in the two spaces is different: the torus seems to have a richer set of "types of loops" than the spherical surface. This type of approach constitutes the base of the definition of fundamental group and explains essential differences between different kinds of topological spaces. The fundamental group makes this idea mathematically rigorous.

Definition of fundamental groupEdit

Definition: Let   be a topological space and let p and q be points in X. Then two paths   and   are considered equivalent if there is a homotopy   such that H(x,a) is a path from p to q for any  . It is easily verified that this is an equivalence relation.

Definition: Define the composition of paths   from x to y and then   from y to z to be simply the same adjunction of paths as we had in the section on path connectedness:


We shall denote the composition of two paths f(x) and g(x) as f(x)*g(x).

Definition: Let   to be a path. Define the inverse path (not to be confused with the inverse function) as  , the path in the opposite direction.

Definition: Let X be a topological space, and let p be a point in X. Then define   to be the constant path   where f(x)=p.

Now consider the set of equivalence groups of paths. Define the composition of two equivalence groups to be the equivalence group of the composition of any two paths. Define the inverse of an equivalence group to be the equivalence group of the inverse of any within the equivalence group. Define   to be the equivalence group containing  .

We can easily check that these operations are well-defined.

Now, in a fundamental group, we will work with loops. Therefore, we define the equivalence, composition and inverse of loops to be the same as the definition as that of paths, and the composition and inverse of the equivalence classes also to be the same.

Definition: The set of equivalence classes of loops at the base point  , is a group under the operation of adjoining paths. This group is called fundamental group of   at the base point  .

In order to demonstrate that this is a group, we need to prove:

1) associtivity:  ;

2) identity:  ;

3) inverse:  .

1) It is quite obvious that when you have a path from a to b, and then b to c, and finally c to d, then the adjunction of the paths from a to b and b to d is the same as the path you get when you adjoin the paths from a to c and then c to d.

In fact, an explicit homotopy on   and   can be given by the following formula:



2)   is the identity. One can easily verify that the product of this constant loop with another loop is homotopic to the original loop.

3) The inverse of the equivalence relation as defined before serves as the inverse within the group. The fact that the composition of the two paths f(x) and   reduces to the constant path can easily be verified with the following homotopy:



Dependence on the Base PointEdit

We now have our fundamental group, but it would be of interest to see how the fundamental group depends on the base point, since, as we have defined it, the fundamental group depends on the base point. However, due to the very important theorem that in any path-connected topological space, all of its fundamental groups are isomorphic, we are able to speak of the fundamental group of the topological space for any path-connected topological space.


Let´s take   in the same path-component of  . In this case, it´s possible to find a relation between   and  . Let   be a path from   to   and   from   back to  . The map   defined by  is an isomorphism. Thus if   is path-connected, the group   is, up to isomorphism, independent of the choice of basepoint  .

When all fundamental groups of a topological space are isomorphic, the notation   is abbreviated to  .

Definition: A topological space   is called simply-connected if it is path-connected and has trivial fundamental group.

The fundamental group of Edit

This section is dedicated to the calculation of the fundamental group of   that we can consider contained in the complex topological space. One more time we can start with a visual approach.


It's easy to imagine that a loop around the circle is not homotopic to the trivial loop. It's also easy to imagine that a loop that gives two complete turns is not homotopic to one that gives only one. Simple intuition seems to be that fundamental group of   is related with number of turns. However, the rigorous calculation of   involves some difficulties.

We define  . It's possible to demonstrate the following results:

Lemma 1: Let   be a path. Then there exists   such that  . Moreover, if  , then   is unique, called a lift of  .


Lemma 2: Let   be a homotopy on paths with start point  . Let  . Then exists only one homotopy   on paths with start point   such that  .

Note: These lemmas allow to guarantee that homotopic loops have homotopic lifts.

For more information see Wikipedia.

Theorem: .

Proof: Let   be a loop with base   and  . Let   and define


The good definition of this application comes from the fact that homotopic loops defined in   to   have homotopic lifts. We have  .So,   for some  . Therefore  .

1)   is surjective. For   we define the loop  . We then have   and  ;

2)   is injective. Let  . Then,  . We then have that   is a homotopy on   and  , or either  ;

3)   is a homomorphism. We want to demonstrate that  . Let´s consider


  is an integer and  .We then have,  and


We can note that all the loops are homotopic to   for some  , or either, all the loops, up to homotopy, consist of giving a certain number of turns.

We can think this demonstration through the following scheme:



Covering spaces and the fundamental groupEdit

One of the most useful tools in studying fundamental groups is that of a covering space. Intuitively speaking, a covering space of a given space   is one which 'looks like' a disjoint union of copies of   in a small enough neighborhood of any point of  , but not necessarily globally.

This section will define covering spaces formally, state important lifting theorems for covering spaces, and then show what the consequences are for fundamental groups.

Definition: Suppose   is a topological space. A space   is called a covering space for   if we are given a continuous map   with the following property: for any  , there exists an open neighborhood   such that

(i)   is a disjoint union  of open subsets of  ;

(ii) the restriction of   to any of these open subsets   is a homeomorphism from   to U.

Unsurprisingly, we call   a covering map.

Example: In fact, we've already seen an example of a covering space. In the calculation of   above, we implicitly made us of the fact that the real line   is a covering space for  . The map   is the covering map. How can we check this? Well, recall that   iff the difference   is an integer. So, suppose we're given a point  . Let   - that is, the set consisting of the whole circle except for the point antipodal to  . Then a little thought shows that if  , we have  . In other words, the preimage of   consists of the whole real line except for a 'hole' at each point  .

It's clear (draw a picture!) that this set is a disjoint union of subintervals, and one can check that the exponential function maps each subinterval homeomorphically onto  . So we do have a covering map. Neat!  

Homotopy LiftingEdit

Now we come to a theorem which looks a bit esoteric at first, but in fact allows us to do much with covering spaces.

Theorem (Homotopy Lifting): Suppose   is a covering map for a space  . Let   be a map from the unit  -cube to  , and   a homotopy of   to another map  . Suppose (for the last time!) that   is a map satisfying  . Then there exists a unique map   satisfying the following:

(i)  ;

(ii) .  

The proof is quite technical, but straightforward, and so is omitted. Any introductory book on algebraic topology should give it --- see, for example, Armstrong, "Basic Topology" (Springer). At first sight this is pretty daunting, so let's take a concrete case to make it easier to digest. Suppose   --- then   is just a point, and hence   is just a function selecting a particular point of  . Hence   can be identified with its image, a point  . Now a homotopy from   to another map is (recalling the definition of homotopy) just a map   such that  ; hence, nothing more than a path in   starting at  . What does the theorem tell us about covering maps  ?

It says (check it!) that if   is a point such that  , and   is a path in   starting at  , then there is a unique path   in   starting at   such that  . In fancier (and looser) terminology, we say that a path in   has a unique lift to  , once the starting point of the lift has been chosen.

On reflection, this result --- sometimes known as the path lifting theorem --- is not so surprising. Think about a covering space as a 'folded-over' version of the base space  , as in Fig XXXX. If we look at a small open set  , its preimage in   is a disjoint union of open sets each homeomorphic to it. If we just concentrate on the portion of   lying inside   for now, it's clear that for each of the disjoint sets  , there is a unique path in   which maps onto   via the covering map  . So to specify a lift, we simply need to choose which of the sets   it lives in (and this is equivalent to choosing a point in the preimage of   as above). Now the whole path   can be split up as a finite 'chain' of short paths living inside 'small' open sets like   (check this!), so finite induction shows that the whole lift is uniquely determined in this way.

Covering Spaces and  Edit

Now we come to the connection between covering spaces and the fundamental group, which is of major significance.

Theorem: Given a covering space  , the map   induces a map   which is an injective (i.e. 1-1) group homomorphism.

Proof (Sketch): First, consider a path   in  : it's a continuous map  , and so we can compose it with the covering map   to get a path   in  . So we have a map

  paths in   paths in  .

We want to show this can be used to define a map

  homotopy classes of loops in   based at   homotopy classes of paths in   based at  .

This sounds complicated, but in fact isn't at all: the idea is, given a homotopy   between two paths   and   in  , the composition   is a homotopy in   between their images   and  . (If this still seems opaque, be sure to check the details.) Also, loops based at   clearly map to loops based at  .

So, we have our map   as desired, mapping   to   We still need to show (a) that it's a group homomorphism, and (b) that it's injective.

(a) is pretty easy. To prove it, choose two elements of  . These are homotopy classes of based loops, so we can choose loops to represent them. What we need to see is that if we concatenate these loops, and then look at the image of this concatenation in  , the result is homotopic to the loop we get if we first map each of the loops via   and then concatenate them. Convince yourself that this is so.

(b) is more tricky. To prove it, we must show that the kernel of the homomorphism   described above consists just of the identity element of  . So, suppose we have a path   representing an element   in the kernel: so   is the identity of  . By definition of  , this means that   is homotopic in   to the constant path at  . So suppose   is such a homotopy: the trick is to use the homotopy lifting theorem (above) to 'lift'   to  , a homotopy in   from   to the constant path at  . (Again, one should check the details of this!) Since such a homotopy   exists, this shows that the homotopy class   is the identity element of  . So the only element in the kernel of   is the identity element of  , so   is injective, as required.  

Let's think about the significance of this result for a moment. An injective homomorphism of groups   is essentially the same as a subgroup  , so the first thing the theorem tells us is that there's a significant restriction on possible covering spaces for a given space  :   can't be a covering space for   unless   is a subgroup of  . Right off the bat, this rules out whole classes of maps from being covering maps. There's no covering map from the circle   to the real line  , for example, since the fundamental group of the circle is isomorphic to the integers, as we saw above, while that of the real line is trivial (why?). Similarly, there's no covering map from the torus   to the two-dimensional sphere   --- the latter has trivial fundamental group, while that of the former is   (the direct sum of two copies of the integers). And so on, ad infinitum.

This latter example is particularly nice, I think, in that it shows how looking at algebraic invariants of spaces (in this case, fundamental groups), rather than our geometric 'mental picture' of the spaces themselves, vastly simplifies arguments about the existence or form of particular kinds of maps between spaces. Can you give me a simple geometric argument to show that there's no possible way to 'wrap'   around the torus so that every point is covered an equal number of times? Can you do the same for the three-dimensional sphere  ? For  ? (If so, I humbly salute you.)

Example: Now let's look at a specific covering space, and see what the homomorphism   we talked about above actually is in a concrete case. Think about the circle   as the unit circle in the complex plane:  . Then we can define a continuous map   by  .

I claim that   is a covering map. To see this, imagine a point   of   (with  ). It isn't hard to see that there are exactly two points   such that p(z') = z; moreover, if we look at a 'small enough' circular arc around  , its preimage under   will consist of two disjoint circular arcs, each containing one of the two preimages of  , and each mapped homeomorphically onto our original arc by  . (Check these details!)

So,   is a covering map, and so the above theorem tells us that   is a group homomorphism from the fundamental group of the circle to itself. In symbols, we have  . But what is it? To answer this, consider a path   in   that winds once around the origin. As we saw in the previous section, the equivalence class of such a path is mapped to the element   under the isomorphism  . Now, to work out  , we look at the equivalence class of the path   in   (this is just the defintion of  ). It's easy to check (do it!) that   is a path is   winding twice around the origin, and so its equivalence class is  . So we have  . Looking at   as the group of integers, we have  . So   is just the doubling map!

Of course, it's all well and good introducing new concepts like covering spaces, but this doesn't achieve a lot unless our new concepts prove useful in some way. Covering spaces have indeed proved useful in many ways, but hopefully the following example will suffice to illustrate this point:

Theorem (Nielsen-Schreier): Any subgroup of a free group is free.

Proof (Sketch): Consult Wikipedia for a rigorous definition of 'free group': roughly speaking, it is a group in which no non-trivial combination of elements equals the identity. Now, the strategy of proof is along the following lines:

1) Given a free group  , find a graph   with  . (Note that a graph is a topological space consisting of a discrete set of points to which are attached a family of line segments. Again see Wikipedia for a rigorous definition.)

2) Show that for a space   and a subgroup  , there exists a covering space   for   with  .

3) Show that any covering space for a graph is itself a graph.

4) Show that the fundamental group of a graph is a free group.

The Fundamental Theorem of AlgebraEdit

Theorem: Let f be a non-constant polynomial with coefficients in the complex numbers  . Then there exists a root of that polynomial in  . Phrased in the language of algebra, the set of complex numbers   are algebraically closed.


Suppose that   has no roots in  . Without loss of generality, we may assume   is monic (if not, then make an appropriate change of variables), and thus we write  . Given   with no roots, define a function   by   It is readily checked that   is a well defined loop in   for all choices of  . Given any  , we may construct a path homotopy   from   to   by  . But   is the constant loop at  , so   is null-homotopic for all  .

We now show that, for a particular choice of  , that   is homotopic to the loop  ,  , where   is the degree of  . Since   is a generator for  ,   is path homotopic to  , and we already know that   is null-homotopic, this will imply that  , and thus that   is a constant polynomial.

To this end, fix  , and let   be the circle in the complex plane of radius  . For all   and all  , we have  

This implies that for all  , the polynomial   has no roots on  , for if it did, this would imply that  , contrary to the above (strict) inequality. Now define   by  . It is easy to check that   is a path homotopy from   to  , which we have shown is null-homotopic. Thus  , and the proof is complete.

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