Geometric Topology 18

The Fundamental Group of a Graph

Fix the basepoint $x_0 \in K_4$ as before. Recall the loops:

Since pairs like $cc^{-1}$ are path-homotopic to constant loops, we recover path-homotopy:

This gives rise to a relation $[\delta] = [\alpha][\beta][\gamma]$ in $G = \pi_1(K_4,x_0)$. In fact $G$ is generated by any three of these loops, for example $[\alpha],[\beta],[\gamma]$, with no relations between them, and is the free group $F_3$.

The fundamental group of any graph is known to be isomorphic to the free group $F_n$ on $n$ generators, where $n = E - V + 1$ is the number of edges remaining when a spanning tree is removed (each such edge defining a loop). But first we must prove that $\pi_1(S^1) \cong F_1$ ($\mathbb{Z}$ in additive notation), the special case $V = E = n = 1$.

Covering Spaces

Definition: Let $Z,X$ be topological spaces. Then:

is a covering map if each point $x \in X$ lies in some open set $U$ for which $p^{-1}U$ is a disjoint union of open subsets $V_i$ of $Z$ such that $p \vert_{V_i}$ maps $V_i$ homeomorphically onto $U$.

The $V_i$ are 'sheets' of the covering. It follows that a subset $W$ of $X$ is open if and only if $p^{-1}(W)$ is open, so $p$ is continuous and $X$ is the quotient of $Z$ by the equivalence relation $z \equiv z' \iff p(z) = p(z')$.

Examples:

1. $p:\mathbb{R} \rightarrow S^1$ defined by $p(s) = e^{2\pi i s} = (\cos(2\pi s), \sin(2 \pi s))$.
2. $p: S^1 \rightarrow S^1$ with $p(z) = z^k$ where $z \in \mathbb{C}, |z| = 1$, and $k \in \mathbb{Z}$.
3. $p: S^n \rightarrow \hat{S^n}$, where $\hat{S^n}$ is the space of equivalence classes $\{v,-v\}$ so $p$ is $2:1$. The quotient (denoted $\mathbb{RP}^n$) is called real projective space, and parametrizes straight lines through the origin in $\mathbb{R}^{n+1}$. $\mathbb{RP}^2$ is homeomorphic to the surface $P$ with $\chi = 1$.

Lifting Maps Uniquely

Convention: From now on 'map' means 'continuous mapping'.

Definition: Given a covering map $p: Z \rightarrow X$ and a map $f: Y \rightarrow X$, a lift of $f$ is any map $\tilde{f} : Y \rightarrow Z$ such that:

so that:

Proposition: If $Y$ is connected, any two lifts $\tilde{f},\hat{f}$ that agree at one point are equal.

Proof: Suppose $\tilde{f}(y_0) = \hat{f}(y_0)$, and consider the subset:

Let $y \in A$; then $\tilde{f}(y) \in V_i$ for some $i$. Since $p: V_i \rightarrow U$ is a bijection, $\tilde{f},\hat{f}$ must agree on the entire open set:

It follows that $A$ is an open subset of $Y$. A similar argument (with $\tilde{f}(y),\hat{f}(y)$ in disjoint sheets) shows that the complement $Y - A$ is also open. Since $Y$ is connected, $A=Y$. $\blacksquare$

We shall apply this proposition to the case in which $Y = I$ and $f = \alpha$ is a path or loop.