Fundamental group

Revision as of 19:11, 23 January 2017 by Dli00105 (talk | contribs) (remove nonexistent category)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

Let $(X,x_0)$ be a based, topological space (that is, $X$ is a topological space, and $x_0\in X$ is some point in $X$). Note that some authors will require $X$ to be path-connected. 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.

Independence from base point

At this point, one might wonder how significant the choice of base point, $x_0$, was. As it turns out, as long as $X$ is path-connected, the choice of base point is irrelevant to the final group $\pi_1(X,x_0)$.

Indeed, pick consider any other base point $x_1$. As $X$ is path connected, we can find a path $\alpha$ from $x_0$ to $x_1$. Let $\bar\alpha(t) = \alpha(1-t)$ be the reverse path from $x_1$ to $x_0$. For any $f\in\Omega(X,x_0)$, define $\varphi_\alpha = \bar\alpha\cdot f\cdot\alpha \in\Omega(X,x_1)$ by \[\varphi_\alpha(f)(t) = (\bar\alpha\cdot f \cdot \alpha)(t) = \begin{cases} \bar\alpha(t) & 0\le t\le 1/3, \\  f(3a-1) & 1/3\le t\le 2/3,\\ \alpha(3t-2) & 2/3\le t\le 1. \end{cases}\] One can now easily check that $\varphi_\alpha$ is in fact a well-defined map $\pi_1(X,x_0)\to\pi_1(X,x_1)$, and furthermore, that it is a homomorphism. Now we may similarly define the map $\varphi_{\bar\alpha}:\pi_1(X,x_1)\to\pi_1(X,x_0)$ by $\varphi(g) = \alpha\cdot g\cdot\bar\alpha$. One can now easily verify that $\varphi_{\bar\alpha}$ is the inverse of $\varphi_\alpha$. Thus $\varphi_\alpha$ is an isomorphism, so $\pi_1(X,x_0)\cong \pi_1(X,x_1)$.

Therefore (up to isomorphism), the group $\pi_1(X,x_0)$ is independent of the choice of $x_0$. For this reason, we often just write $\pi_1(X)$ for the fundamental of $X$.

Functoriality

Given a (based) continuous map $f:(X,x_0)\to(Y,y_0)$ (that is, $f$ is continuous and $f(x_0) = y_0$), one may define a group homomorphism $f_*:\pi_1(X,x_0)\to\pi_1(Y,y_0)$ by sending each loop $a\in\pi_1(X,x_0)$ to $f\circ a\in \pi_1(Y,y_0)$. It is easy to see that $f_*$ sends homotopic loops to homotopic loops (indeed if $F$ is a homotopy from $a$ to $a'$, then $f\circ F$ is a homotopy from $f\circ a$ to $f\circ a'$), and thus $f_*$ is a well-defined map. Also $f_*$ clearly preserves concatenations, so $f_*$ is indeed a homomorphism.

Furthermore, it is easy to see that if $f$ and $g$ are maps $f:(X,x_0)\to(Y,y_0)$ and $g:(Y,y_0)\to (Z,z_0)$, then: \[(g\circ f)_* = g_*\circ f_*,\] and if $1_X$ is the identity map on $(X,x_0)$ and $1_{\pi(X,x_0)}$ is the identity map on $\pi_1(X,x_0)$, then \[(1_X)_* = 1_{\pi_1(X,x_0)}.\] Thus we may in fact regard $\pi_1$ as a (covariant) functor from the category of based topological spaces to the category of groups.

One can also show that the induced map $f_*$ depends only on the homotopy type of $f$, that is if $f,g:(X,x_0)\to (Y,y_0)$ are (based) homotopic maps that $f_* = g_*$. Indeed, for any loop $a\in\pi_1(X,x_0)$, if $F$ is a based homotopy from $f$ to $g$, then $F\circ a$ is a based homotopy from $f\circ a$ to $g\circ a$, and thus $f\circ a = g\circ a$ in $\pi_1(Y,y_0)$.

Homotopy invariance

In order for the fundamental group to be a useful topological concept, any two spaces that are topologically "the same" must have the same fundamental group. Specifically, if $X$ and $Y$ are homeomorphic then $\pi_1(X)$ and $\pi_1(Y)$ are isomorphic.

We will in fact show that $\pi_1(X)$ and $\pi_1(Y)$ are isomorphic if $X$ and $Y$ satisfy the weaker notion of equivalence: homotopy equivalence.

Say that $(X,x_0)$ and $(Y,y_0)$ are based homotopy equivalent ($(X,x_0)\simeq (Y,y_0)$) with homotopy equivalences $f:(X,x_0)\to (Y,y_0)$ and $g:(Y,y_0)\to (X,x_0)$. (By definition, this means that $g\circ f \simeq 1_X$ and $f\circ g\simeq 1_Y$.) Now consider the induced maps $f_*:\pi_1(X,x_0)\to\pi_1(Y,y_0)$ and $g_*:\pi_1(Y,y_0)\to\pi_1(X,x_0)$. From the previous section we get that: \[f_*\circ g_* = (f\circ g)_* = (1_Y)_* = 1_{\pi_1(Y,y_0)}\] and \[g_*\circ f_* = (g\circ f)_* = (1_X)_* = 1_{\pi_1(X,x_0)}.\] Therefore $g_*$ is the inverse of $f_*$, so in particular $f_*$ must be an isomorphism. Hence $\pi_1(X,x_0)\cong \pi_1(Y,y_0)$.

This gives us a very useful method for distinguishing topological spaces: if $X$ and $Y$ are topological spaces whose fundamental groups are not not isomorphic then $X$ and $Y$ cannot be homeomorphic (and in fact, they cannot be homotopy equivalent). For instance, one can show that $\pi_1(S^1)\cong\mathbb Z$ and $\pi_1(S^2) \cong 0$ (where $S^n$ is the n-sphere), and hence a circle is not homeomorphic to a sphere.

Invalid username
Login to AoPS