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$\theta_v$
.