Difference between revisions of "Fundamental group"

m (add category)
(re-wrote article a bit)
Line 1: Line 1:
Perhaps the simplest object of study in algebraic topology is the [[fundamental group]]. Let <math>X</math> be a [[path-connected]] topological space, and let <math>x\in X</math> be any point. Now consider all possible "loops" on <math>X</math> that start and end at <math>x</math>, i.e. all [[continuous function]]s <math>f:[0,1]\to X</math> with <math>f(0)=f(1)=x</math>. Call this collection <math>L</math>. Now define an [[equivalence relation]] <math>\sim</math> on <math>L</math> by saying that <math>p\sim q</math> if there is a continuous function <math>g:[0,1]\times[0,1]\to X</math> with <math>g(a,0)=p(a)</math>, <math>g(a,1)=q(a)</math>, and <math>g(0,b)=g(1,b)=x</math>. We call <math>g</math> a [[homotopy]]. Now define <math>\pi_1(X)=L/\sim</math>. That is, we equate any two elements of <math>L</math> which are equivalent under <math>\sim</math>.
+
Perhaps the simplest object of study in algebraic topology is the '''fundamental group'''.  
  
Unsurprisingly, the fundamental group is a group. The [[identity]] is the [[equivalence class]] containing the map <math>1:[0,1]\to X</math> given by <math>1(a)=x</math> for all <math>a\in[0,1]</math>. The [[Function/Introduction#The_Inverse_of_a_Function | inverse]] of a map <math>h</math> is the map <math>h^{-1}</math> given by <math>h^{-1}(a)=h(1-a)</math>. We can compose maps as follows: <math>g\cdot h(a)=\begin{cases} g(2a) & 0\le a\le 1/2, \\ h(2a-1) & 1/2\le a\le 1.\end{cases}</math> One can check that this is indeed [[well-defined]].
+
Let <math>(X,x_0)</math> be a [[based topological space|based]], [[path-connected]] [[topological space]] (that is, <math>X</math> is a topological space, and <math>x_0\in X</math> is some point in <math>X</math>). Now consider all possible "loops" on <math>X</math> that start and end at <math>x_0</math>, i.e. all [[continuous function]]s <math>f:[0,1]\to X</math> with <math>f(0)=f(1)=x_0</math>. Call this collection <math>\Omega(X,x_0)</math> (the '''loop space''' of <math>X</math>). Now define an [[equivalence relation]] <math>\sim</math> on <math>\Omega(X,x_0)</math> by saying that <math>f\sim g</math> if there is a (based) [[homotopy]] between <math>f</math> and <math>g</math> (that is, if there is a continuous function <math>F:[0,1]\times[0,1]\to X</math> with <math>F(a,0)=f(a)</math>, <math>F(a,1)=g(a)</math>, and <math>F(0,b)=F(1,b)=x_0</math>). Now let <math>\pi_1(X,x_0)=\Omega(X,x_0)/\sim</math> be the set of equivalence classes of <math>\Omega(X,x_0)</math> under <math>\sim</math>.
 +
 
 +
Now define a [[binary operation]] <math>\cdot</math> (called ''concatenation'') on <math>\Omega(X,x_0)</math> by
 +
<math>(g\cdot h)(a)=\begin{cases} g(2a) & 0\le a\le 1/2, \\ h(2a-1) & 1/2\le a\le 1.\end{cases}</math>
 +
One can check that if <math>f\sim f'</math> and <math>g\sim g'</math> then <math>f\cdot g\sim f'\cdot g'</math>, and so <math>\cdot</math> induces a well-defined binary operation on <math>\pi_1(X,x_0)</math>.
 +
 
 +
One can now check that the operation <math>\cdot</math> makes <math>\pi_1(X,x_0)</math> into a group. The identity element is just the constant loop <math>e(a) = x_0</math>, and the inverse of a loop <math>f</math> is just the loop <math>f</math> traversed in the opposite direction (i.e. the loop <math>\bar f(a) = f(1-a)</math>). We call <math>\pi_1(X,x_0)</math> the '''fundamental group''' of <math>X</math>.
  
 
Note that the fundamental group is not in general [[abelian group|abelian]]. For example, the fundamental group of a figure eight is the [[free group]] on two [[generator]]s, which is not abelian. However, the fundamental group of a circle is <math>{\mathbb{Z}}</math>, which is abelian.
 
Note that the fundamental group is not in general [[abelian group|abelian]]. For example, the fundamental group of a figure eight is the [[free group]] on two [[generator]]s, which is not abelian. However, the fundamental group of a circle is <math>{\mathbb{Z}}</math>, which is abelian.

Revision as of 00:43, 13 December 2009

Perhaps the simplest object of study in algebraic topology is the fundamental group.

Let $(X,x_0)$ be a based, path-connected topological space (that is, $X$ is a topological space, and $x_0\in X$ is some point in $X$). Now consider all possible "loops" on $X$ that start and end at $x_0$, i.e. all continuous functions $f:[0,1]\to X$ with $f(0)=f(1)=x_0$. Call this collection $\Omega(X,x_0)$ (the loop space of $X$). Now define an equivalence relation $\sim$ on $\Omega(X,x_0)$ by saying that $f\sim g$ if there is a (based) homotopy between $f$ and $g$ (that is, if there is a continuous function $F:[0,1]\times[0,1]\to X$ with $F(a,0)=f(a)$, $F(a,1)=g(a)$, and $F(0,b)=F(1,b)=x_0$). Now let $\pi_1(X,x_0)=\Omega(X,x_0)/\sim$ be the set of equivalence classes of $\Omega(X,x_0)$ under $\sim$.

Now define a binary operation $\cdot$ (called concatenation) on $\Omega(X,x_0)$ by $(g\cdot h)(a)=\begin{cases} g(2a) & 0\le a\le 1/2, \\ h(2a-1) & 1/2\le a\le 1.\end{cases}$ One can check that if $f\sim f'$ and $g\sim g'$ then $f\cdot g\sim f'\cdot g'$, and so $\cdot$ induces a well-defined binary operation on $\pi_1(X,x_0)$.

One can now check that the operation $\cdot$ makes $\pi_1(X,x_0)$ into a group. The identity element is just the constant loop $e(a) = x_0$, and the inverse of a loop $f$ is just the loop $f$ traversed in the opposite direction (i.e. the loop $\bar f(a) = f(1-a)$). We call $\pi_1(X,x_0)$ the fundamental group of $X$.

Note that the fundamental group is not in general abelian. For example, the fundamental group of a figure eight is the free group on two generators, which is not abelian. However, the fundamental group of a circle is ${\mathbb{Z}}$, which is abelian.

More generally, if $X$ is an h-space, then $\pi_1(X)$ is abelian, for there is a second multiplication on $\pi_1(X)$ given by $(\alpha\beta)(t) = \alpha(t)\beta(t)$, which is "compatible" with the concatenation in the following respect:

We say that two binary operations $\circ, \cdot$ on a set $S$ are compatible if, for every $a,b,c,d \in S$, \[(a \circ b) \cdot (c \circ d) = (a \cdot c) \circ (b \cdot d).\]

If $\circ,\cdot$ share the same unit $e$ (such that $a \cdot e = e \cdot a = a \circ e = e \circ a = a$) then $\cdot = \circ$ and both are abelian.