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 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.

  1. 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 $v_{\Gamma'}$.

  1. Show that we have functions $\iota_Q: E_Q \to V_Q$ and $E_Q \to E_Q$ induced by $\iota$ and $e\mapsto \bar{e}$.
  2. 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.

  1. 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).

  2. 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.

  3. 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 $\theta_v$.