observing that if $c\in{\Sigma}_2({\mathcal{P}})$, then $c\in{\mathsf{A}}(\partial_1c)$ and $c\in{\mathsf{C}}(\partial_0c, \partial_2c)$. This entails that deformation is a symmetric, reflexive binary relation on the set of paths with the same endpoints. However, if ${\mathcal{P}}$ is not directed, deformation does not need to be an equivalence relation on paths with the same endpoints, because transitivity might fail.\
Given $a_0, a_1\in{\Sigma}_0({\mathcal{P}})$, a *homotopy of paths* in $\mathrm{P}(a_0,a_1)$ is a map $h(i): \{1,2,\ldots,n\} \longrightarrow  \mathrm{P}(a_0,a_1)$ such that $h(i)$ is an elementary deformation of $h(i-1)$ for $1<i\leq n$. We will say that two paths $p,q\in\mathrm{P}(a_0,a_1)$ are *homotopic*, $p\sim q$, if there exists a homotopy of paths $h$ in $\mathrm{P}(a_0,a_1)$ such that $h(1)=q$ and $h(n)=p$. It is clear that a homotopy of paths is an equivalence relation on paths with the same endpoints.\
We now define the composition of paths and the reverse of a path. Given $p=\{b_n,\ldots,b_1\}\in \mathrm{P}(a_0,a_1)$ and $q=\{b'_k,\ldots b'_1\}\in\mathrm{P}(a_1,a_2)$, the *composition* of $p$ and $q$ is the path $p*q\in \mathrm{P}(a_0,a_2)$ defined as $$\label{Ab:4}
q*p{\equiv}\{b'_k,\ldots, b'_1,b_n,\ldots,b_1 \}.$$ Note that $p_1*(p_2*p_3)= (p_1*p_2)*p_3$, if the composition is defined.

\[Ab:5\] Let $p_1,q_1\in\mathrm{P}(a_0,a_1)$, $p_2,q_2\in\mathrm{P}(a_1,a_2)$. If $p_1\sim q_1$ and $p_2\sim q_2$, then $p_2*p_1\sim q_2* q_1$.

Let $h_1:\{1,\ldots,n\}\longrightarrow \mathrm{P}(a_0,a_1)$ and $h_2:\{1,\ldots,k\}\longrightarrow \mathrm{P}(a_1,a_2)$ be homotopies of paths such that $h_1(1)=p_1$, $h_1(n)=q_1$ and $h_2(1)=p_2$, 