have: $z(\partial_0c(a))\cdot z(\partial_2c(a))$ $= z(\partial_1c(a))$ $\iff  \ \ z(b(a))\cdot z(b(a)) = z(b(a))$ $\iff$ $z(b(a))=\mathbbm{1}$. (b) follows from (a), Lemma \[Ac:1\] and Lemma \[Ab:10\]b.

We now are in a position to show the connection between the fundamental group of ${\mathcal{P}}$ and ${\mathcal{Z}^1({\mathscr{A}}_{\mathcal{P}})}$. Fix a base 0-simplex $a_0$. Given $z\in{\mathcal{Z}^1({\mathscr{A}}_{\mathcal{P}})}$, define $$\label{Ac:3}
\qquad \pi_z([p]) {\equiv}z(p) \qquad \qquad  [p]\in\pi_1({\mathcal{P}},a_0)$$ This definition is well posed as $z$ is invariant for homotopic paths.

\[Ac:4\] The correspondence ${\mathcal{Z}^1({\mathscr{A}}_{\mathcal{P}})}\ni z\longrightarrow \pi_z$ maps 1-cocycles $z$, equivalent in ${\mathfrak{B}(\mathcal{H}_o)}$, into equivalent unitary representations $\pi_z$ of $\pi_1({\mathcal{P}})$ in $\mathcal{H}_o$. Up to equivalence, this map is injective. If $\pi_1({\mathcal{P}})= 1$, then ${\mathcal{Z}^1({\mathscr{A}}_{\mathcal{P}})}={\mathcal{Z}^1_t({\mathscr{A}}_{\mathcal{P}})}$.

First, recall that the identity $1$ of $\pi_1({\mathcal{P}})$ is the equivalence class $[b(a_0)]$ associated with the 1-simplex degenerate to $a_0$. By Lemma \[Ac:2\] we have that $\pi_z(1)=1$ and that $\pi_z([p]^{-1})= \pi_z([p])^*$. Furthermore, it is obvious from the definition of $\pi_z$, that $\pi_z([p]*[q])=\pi_z([p])\cdot \pi_z([q])$, therefore $\pi_z$ is a unitary representation of $\pi_1({\mathcal{P}})$ in $\mathcal{H}_o$. Note that if $z_1\in{\mathcal{Z}^1({\mathscr{A}}_{\mathcal{P}})}$ and $u\in(z,z_1)$ is unitary, then $u_{a_0}\cdot \pi_z([p]) = \pi_{z_1}([p])\cdot  u_{a_0}$. Now, let $\pi$ be a unitary representation of $\pi_1({\mathcal{P}})$ on $\mathcal{H}_o$. Fix a base 0-simplex $a_0$, and for any 0-simplex $a$, denote by $p_a$ a path with $\partial_1p_a = a$ and $\partial_0p_a 