Fundamental Groups of Graphs
due by Monday, Nov 20, 2023
Recall that a graph $\Gamma = \Gamma(V, E)$ consists of sets $V$ (vertices) and $E$ (edges) and functions $\iota: E \to V$ (initial vertex) and a function $E\to E$ denoted $e\mapsto \bar{e}$ (reversing orientation) so that $\bar{\bar{e}} = e$ and $\bar{e} \neq e$ for all $e\in E$. We define $\tau(e)=\iota(\bar{e})$ for an edge $e.$ We associate to $\Gamma$ a topological space $|\Gamma|$ called its geometric realization, which is the quotient
$$\left(V\coprod \left(\coprod_{e\in E} I_e\right)\right)/\sim,$$
where $I_e$ is a copy of the unit interval $[0, 1]$ for each $e\in E$, and $\sim$ the equivalence relation generated by the following relations, where the pair $(s, e)$ for $e\in E$ and $s\in [0, 1]$ denotes the point in $I_e$ corresponding to $s$:
- a. For each
$e\in E$we have$(0, e)\sim \iota(e)$. - b. For each
$e\in E$and$s\in [0, 1]$we have$(s, e)\sim (1-s, \bar{e})$.
A subbgraph $\Gamma'(V', E')$ of a graph consists of sets $V'\subset V$ and $E'\subset E$ so that for $e\in E'$ we have $\iota(e)\in V'$ and $\bar{e}\in E'$. The restrictions of the functions $\iota$ and $e\mapsto \bar{e}$ to $E'$ make $\Gamma'$ a graph.
- Show that there is a natural inclusion of
$|\Gamma'|$into$|\Gamma|$.
We regard $|\Gamma'|$ as a subspace of $\Gamma|$ using this inclusion map. The quotient graph $\Gamma/\Gamma'$ is a graph with vertex set $V_Q=(V\setminus V')\cup \{\Gamma'\}$ (i.e., with a single vertex for all vertices of $\Gamma'$) and edge set $E_Q = E\setminus E'.$ For clarity, we often denote the vertex corresponding to $\Gamma'$ as $v_{\Gamma'}$.
- Show that we have functions
$\iota_Q: E_Q \to V_Q$and$E_Q \to E_Q$induced by$\iota$and$e\mapsto \bar{e}$. - Show that the spaces
$|\Gamma/\Gamma'|$and$|\Gamma|/|\Gamma'|$are homeomorphic.
A tree $T=T(V, E)$ is a non-empty graph so that for vertices $v, v'\in V$ there is a unique reduced edge-path from $v$ to $v'$ (for $v\neq v'$, a reduced edge-path from $v$ to $v'$ is a sequence of edges $e_1, e_2,\dots, e_n$ with $\iota(e_1)=v$, $\tau(e_n)= v'$, and, for $1\leq i< n$, $\tau(e_i)=\iota(e_{i+1})$ and $e_{i+1}\neq \bar{e_i}$). Let $\Gamma= \Gamma(V, E)$ be a graph, $x_0\in V$ a vertex, and $T= T(V_T, E_T)$ a subgraph which is a tree. Suppose further that $V_T= V$ (such a tree exists for any connected graph but we will not prove it). We show below that the quotient map $q: |\Gamma| \to |\Gamma|/|T|$ is a homotopy equivalence.
For $v\in V$, $v\neq x_0$, let $\theta_v$ be the (unique) reduced edge-path from $x_0$ to $v$. For $e \in E\setminus E_T$ let $e_Q$ denote the image of $e$ in $|\Gamma|/|T|$. We use the idenitification $|\Gamma|/|T| = |\Gamma/T|$ repeatedly.
-
Show that there is a continuous map
$p: |\Gamma|/|T| \to |\Gamma|$that maps the vertex$v_T$(i.e., the vertex corresponding to$T$) to$x_0$, and, for each edge$e\in E\setminus E_T$, maps$e_Q$to the path$\theta_{\iota(e)}*e*\overline{\theta_{\tau(e)}}$up to homotopy (homotopy is needed only for reparametrization issues). -
Show that
$q\circ p: |\Gamma|/|T| \to |\Gamma|/|T|$is homotopic fixing the basepoint$v_T$to the identity map. Note that such a homotopy is already determined on$(v_T\times [0, 1])\cup (E_Q\times \{0, 1\})$, and this has to be extended to$e_Q\times [0, 1]$for each$e_Q\in E_Q$in a continuous manner. -
Show that
$p\circ q: |\Gamma| \to |\Gamma|$is homotopic to the identity fixing the basepoint$x_0$. Note that such a homotopy is already determined on$(x_0\times [0, 1])\cup (E\times \{0, 1\})$. This has to be extended to$v\times [0, 1]$for each$v\in V\setminus\{x_0\}$and then to$e\times [0, 1]$for each$e\in E$in a continuous manner. The former can be done using the paths$\theta_v$.