\[Ab:5\] it is enough to prove the assertions in the case that $p$ is a 1-simplex $b$. (a) Let $c_1$ be the 2-simplex defined as $\partial_2 c_1 = b(\partial_1b)$, $\partial_0c_1 = b$, $\partial_1c_1 = b$ and whose support $|c_1|$ equals $|b|$. Then $c_1\in{\mathsf{C}}(b,b(\partial_1b))$ and $b*b(\partial_1b) \sim b$. The other identity follows in a similar way. (b) Let $c_2$ be the 2-simplex defined as $\partial_0c_2=b$, $\partial_2c_2={\overline{b}}$, $\partial_1c_2= b(\partial_0b)$ and whose support $|c_2|$ equals $|b|$. Then $c_2\in{\mathsf{C}}(b,{\overline{b}})$ and $b*{\overline{b}}\sim b(\partial_0b)$. The other identity follows in a similar way.

We now are in a position to define the first homotopy group of a poset. Fix a base 0-simplex $a_0$ and consider the set of closed paths $\mathrm{P}(a_0)$. Note that the composition and the reverse are internal operations of $\mathrm{P}(a_0)$ and that $b(a_0)\in \mathrm{P}(a_0)$. We define $$\label{Ab:11}
  \pi_1({\mathcal{P}},a_0) {\equiv}\mathrm{P}(a_0) / \sim$$ where $\sim$ is the homotopy equivalence relation. Let $[p]$ denote the homotopy class of an element $p$ of $\mathrm{P}(a_0)$. Equip $\pi_1({\mathcal{P}},a_0)$ with the product $$\qquad [p]*[q] {\equiv}[p*q] \qquad   [p],[q]\in  \pi_1({\mathcal{P}},a_0).$$ $*$ is associative, and it easily follows from previous lemmas that $\pi_1({\mathcal{P}},a_0)$ with $*$ is a group: the identity $1$ of the group is $[b(a_0)]$; the inverse $[p]^{-1}$ of $[p]$ is $[{\overline{p}}]$. Now, 