Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider some space $X$. Does the fundamental group tell us information about the equivalence between two paths $f,g: I \to X$? So there exists a homotopy $h: I \times I \to X$ such that $h(s,0) = f(s)$, $h(s,1) = g(s)$, $h(0,t) = x$ and $h(1,t) = y$?

In other words, starting with two paths, how/why do we introduce the fundamental group?

share|improve this question
6  
Have you tried reading a textbook on the subject? There are great expositions! The end result will be much, much better than even the best answer you might get here. –  Mariano Suárez-Alvarez Jun 17 '11 at 2:18
    
I'm not sure I understand your question. First of all, by equivalent do you mean homotopic, as your second question seems to imply? –  Becca Winarski Jun 17 '11 at 2:21
    
@Becca: Yes. (more characters) –  Damien Jun 17 '11 at 2:24
    
Oh, maybe a better question for me to ask: do you mean homotopic relative to the boundary of the paths where the boundary is the endpoints x and y? –  Becca Winarski Jun 17 '11 at 2:29
add comment

1 Answer

To answer the letter (but not the spirit) of your question, if $X$ is path connected, then any two maps $I\to X$ are homotopic because $I$ is contractible (homotopy equivalent to a point), and up to homotopy, a map from a point to $X$ is determined by the path component it lands in.

When dealing with the fundamental group, we want to take maps $S^1\to X$ and homotopies relative to a basepoint. One way to do this is to to take maps $f:I\to X$ such that $f(0)=f(1)=x$ where $\phantom{}x\in X$ is a basepoint, and to consider only homotopies which preserve the endpoints.


A better answer, if I correctly interpret your question to be "Why do we care about fundamental groups" is the following.

We don't introduce the fundamental group starting with two paths, we introduce the fundamental group because we want to have some sort of algebraic invariant associated to a space that is easy to work with and which tells us something useful or interesting about the space.

If $X$ is a space, and $x\in X$ is a point, we define $\pi_1(X,x)$ to be the loops in $X$ based at $x$ up to homotopy. There is a natural group structure (see any textbook on the subject, Mariano is right that even a great exposition here would pale in comparison to what a textbook can provide), and so we have for every (based) space a corresponding group, and the correspondence has a few nice properties:

  1. If $f:X\to Y$, then we have an induced map $f_*:\pi_1(X,x)\to \pi_1(Y,f(x))$.
  2. This correspondence respects composition, in the sense that $(fg)_*=f_* g_*$.
  3. This correspondence respects the identity map, in the sense that $(\operatorname{id}_X)_*=\operatorname{id}_{\pi_1(X,x)}$.
  4. If $f,g:X\to Y$ are homotopic, then $f_*=g_*$. (This implies that if two spaces are homotopy equivalent, they have the same fundamental group).

If you know any category theory, this is all contained in the statement $\pi_1$ is a covariant functor from the category of topological spaces (with morphisms homotopy classes of continuous maps) to the category of groups.

There are various ways to actually compute fundamental groups, for example covering space theory or Van Kampen's theorem, and it allows you to prove interesting theorems, such as the two dimensional version of the Brower fixed point theorem: Every continuous map from the unit ball to itself must have a fixed point. A sketch of the proof is as follows:

If we had a map $B\to B$ that didn't have a fixed point, then identifying the circle $S^1$ as the boundary of $B$, we could find a deformation retract $B\to S^1$. However, $\pi_1(S^1)=\mathbb Z$, and $\pi_1(B)={0}$ because $B$ is contractible. Therefore, if $S^1$ was a deformation retract of $B$, then we could find maps $\mathbb Z \to 0 \to \mathbb Z$, the composition of which was the identity map. This is impossible.

Here, we have taken properties of topological maps and turned them into properties of algebraic maps, and because algebraic maps are much easier to deal with, we were able to say something useful. In particular, we were able to show that certain maps did NOT exist, which in the absence of algebraic invariants or very simple geometric criterea (e.g. the intermediate value theorem) is a very hard task. In particular, algebraic invariants such at $\pi_1$ are absolutely essential to understanding spaces too complicated to visualize or for dealing with situations where things don't hold on the nose but only up to homotopy.


To link this back up with your original question, the fundamental group is a useful way to show that two maps are NOT homotopic, by showing that they induce different maps on fundamental groups. However, it is better to think about what the fundamental group does to spaces before you think about what it does to maps, as determining the induced group homomorphism can sometimes be tricky.

share|improve this answer
    
I tend to think that one should keep these things going together, i.e. viewing $\pi_1$ as a functor from spaces with base point to groups. This confirms the idea of modelling and analogy, as given by a functor. But of course to evaluate the morphism you first have to give its domain and codomain.A simple example is to evaluate what $z \mapsto z^n$ on $S^1$ does to the fundamental group of $S^1$. –  Ronnie Brown Apr 27 '12 at 16:42
    
@RonnieBrown: I agree that if you've seen category theory or the fundamental group before, treating it as a functor on based spaces is better than the ad hoc description I've given above (although I personally like unbased spaces and the fundamental groupoid). However, when just starting out, talking about based spaces seems a little odd, as it is it solving a problem that hasn't yet made itself entirely clear. –  Aaron Apr 27 '12 at 19:16
    
You make a subtle point, since actually you need not one base point but many, according to the geometry of the situation. To calculate the fundamental group of the circle, you need two base points! Further, spaces are not given as manna from heaven, but are presented by data in some way, and invariants should reflect the structure of that data. My work on higher order van Kampen Theorems has involved filtered spaces, i.e. spaces with an increasing sequence of subspaces, and I can't make it work without that. –  Ronnie Brown Apr 29 '12 at 21:28
    
Actually I like to argue that that the intuitive idea is often that of groupoid rather than group. One does not describe a set of railway stations and connections in terms of return journeys, and passing from one set of return journeys to another at a different station! –  Ronnie Brown Oct 8 '12 at 21:03
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.