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Difference between revisions of "Fundamental group"

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If <math>\circ,\cdot</math> share the same unit <math>e</math> (such that <math>a \cdot e = e \cdot a = a \circ e = e \circ a = a</math>) then <math>\cdot = \circ</math> and both are abelian.
 
If <math>\circ,\cdot</math> share the same unit <math>e</math> (such that <math>a \cdot e = e \cdot a = a \circ e = e \circ a = a</math>) then <math>\cdot = \circ</math> and both are abelian.
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[[Category:Topology]]
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[[Category:Algebraic Topology]]

Revision as of 00:09, 13 December 2009

Perhaps the simplest object of study in algebraic topology is the fundamental group. Let $X$ be a path-connected topological space, and let $x\in X$ be any point. Now consider all possible "loops" on $X$ that start and end at $x$, i.e. all continuous functions $f:[0,1]\to X$ with $f(0)=f(1)=x$. Call this collection $L$. Now define an equivalence relation $\sim$ on $L$ by saying that $p\sim q$ if there is a continuous function $g:[0,1]\times[0,1]\to X$ with $g(a,0)=p(a)$, $g(a,1)=q(a)$, and $g(0,b)=g(1,b)=x$. We call $g$ a homotopy. Now define $\pi_1(X)=L/\sim$. That is, we equate any two elements of $L$ which are equivalent under $\sim$.

Unsurprisingly, the fundamental group is a group. The identity is the equivalence class containing the map $1:[0,1]\to X$ given by $1(a)=x$ for all $a\in[0,1]$. The inverse of a map $h$ is the map $h^{-1}$ given by $h^{-1}(a)=h(1-a)$. We can compose maps as follows: $g\cdot h(a)=\begin{cases} g(2a) & 0\le a\le 1/2, \\ h(2a-1) & 1/2\le a\le 1.\end{cases}$ One can check that this is indeed well-defined.

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.

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