Lemma 1

The quiver Γ identifies with a subquiver of $\mathsf{\Gamma}$; the quivers Γ and $\mathsf{\Gamma}$ are connected.

The arcwise connected components of X are the $A_i$, for $i\in I$. The arcwise connected components of $X\times_YX$ are the $(V\times \{i\})\times_Y$ $(V\times \{i'\})$, for $(i, i',V)\in J'$. Consequently, the frame $\mathsf{\Gamma}$ of the pair $(\varpi (p_1), \varpi (p_2))$ is isomorphic to the quiver whose set of vertices is I, whose set of arrows is $J'$, the origin and term mappings being respectively the mappings $j\mapsto i_1(j)$ and $j\mapsto i_2(j)$. The quiver $\mathsf{\Gamma}$ is connected (IV, p. 406). Moreover, this description identifies Γ with a subquiver of $\mathsf{\Gamma}$. Let us also observe that for every arrow $j$ of $\mathsf{\Gamma}$, either $j\in$ Fl($\widetilde{\Gamma}$), or there exists $i\in I$ such that $j= (i, i,A_i)$. It follows that the mapping $\pi_0(\Gamma )\rightarrow \pi_0(\mathsf{\Gamma})$ deduced from the injection of Γ into $\mathsf{\Gamma}$ is bijective, so that Γ is connected.

For every element $i$ of I, choose a point $a(i)$ of $A_i$.

For every element $j= (i, i',V)$ of J, choose a point $b(j)$ in V, a path $B_1(j)$ joining $b(j)$ to $a(i)$ in $A_i$ and a path $B_2(j)$

joining$\overline{j}= (i', i,b(jV)$) tobelongs to$a(i')$ in A$J^{i'}$. Letand let us put$j= (i, ib(',\overline{j}V)$) =an element of$b(j)$, $B_1(j) = B\overline{J}_2$; then$(\overline{j})$ and $B_2(j) = B_1(\overline{j})$. For $j\in J'\cup \overline{J'}$, the paths $\overline{B_1(j)}$ and $B_2(j)$ in Y are composable. Put

It is a path joining $a(i_1(j))$ to $a(i_2(j))$ in Y; one has the relation $B(\overline{j}) = B(j)$.

For every $j= (i, i',V)\in J'$, denote by $p_{j,1}: V\rightarrow A_i$ and $p_{j,2}: V\rightarrow A_{i'}$ the canonical injections; also denote by $\varphi_j:\pi_1(V, b(j))\rightarrow \pi_1(A_i, a(i))$ and $\psi_j:\pi_1(V, b(j))\rightarrow \pi_1(A_{i'}, a(i'))$ the group homomorphisms defined by

and

for $v\in \pi_1(V, b(j))$ (cf. IV, p. 407).

Fix an element $i_0$ of I, as well as a subquiver T in the quiver Γ whose associated graph $\widetilde{T}$ is a maximal tree of the graph $\widetilde{\Gamma}$.

For $i\in I$, let $(i_0, j_1, i_1, . . . , j_n, i)$ be the unique path without backtracking joining $i_0$ to $i$ in the tree $\widetilde{T}$, and put

this is the class of a path joining $a(i_0)$ to $a(i)$ in Y. Let $\alpha_i$ denote the homomorphism of $\pi_1(A_i, a(i))$ into $\pi_1(Y, a(i))$ deduced from the canonical injection, and let $\mu_i:\pi_1(A_i, a(i))\rightarrow \pi_1(Y, a(i_0))$ be the group homomorphism defined by

Finally, let $\mu: F(J)\rightarrow \pi_1(Y, a(i_0))$ be the unique group homomorphism such that one has

for every $j\in J$. There exists a unique group homomorphism

which coincides with $\mu_i$ on $\pi_1(A_i, a(i))$, for every $i\in$ I, and with $\mu$ on F(J).

Let $K'$ be the set of quadruples $(i_1, i_2, i_3,U)$, where $i_1, i_2, i_3$ are elements of I and where U is a pathwise connected component of $A_{i_1}\cap A_{i_2}\cap A_{i_3}$. For every element $k= (i_1, i_2, i_3,U)$ of $K'$ and every pair $(s, t)$ of elements of $\{1,2,3\}$, put $j_{st}(k) = (i_s, i_t,V)$, where V is the pathwise connected component of $A_{i_s}\cap A_{i_t}$ which contains U; this is an element of $J'$.

Let K be the subset of $K'$ formed by the quadruples $(i_1, i_2, i_3,U)$ such that $i_1< i_2< i_3$. For every element $k= (i_1, i_2, i_3,U)$ of K, choose a point $c(k)$ of U, together with paths $C_{12}(k)$, $C_{23}(k)$ and $C_{13}(k)$, such that $C_{st}(k)$ joins $c(k)$ to $b(j_{st}(k))$ in $A_{i_s}\cap A_{i_t}$ for $s, t\in \{1,2,3\}$ with $s < t$.

Put then, for $k\in K$,

(4) L$L^2_3((kk) = B) = B^2_2((jj^{12}_{23}((kk))))**\overline{CC^{1223}((kk))}**CC^{23}_{13}((kk))**BB^1_2((jj^{23}_{13}((kk))))$,.

For $s\in \{1,2,3\}$, let $\lambda_s(k)$ denote the class in $\pi_1(A_{i_s}, a(i_s))$ of the loop $L_s(k)$.