Proposition 1

Let Y be a pathwise connected topological space and let $(A_i)_{i\in I}$ be a covering of Y by non-empty, pathwise connected subsets, indexed by a totally ordered set I. Suppose that the canonical mapping of the sum space of the family $(A_i)_{i\in I}$ into Y satisfies property (VK). Then the homomorphism $\mathsf{M}$ introduced above is surjective, and its kernel is the smallest normal subgroup containing the following elements:

($R_1$)$r_1(j) =j$ for $j$ in Fl(T);

($R_2$)$r_2(j, v) =\varphi_j(v)j\psi_j(v)^{-1}j^{-1}$

for $j= (i, i',V)\in J$ and $v\in \pi_1(V, b(j))$;

($R_3$)$r_3(k) =\lambda_1(k)j_{12}(k)\lambda_2(k)j_{23}(k)\lambda_3(k)j_{13}(k)^{-1}$

for $k\in K$.

Let X be the topological sum of the family $(A_i)_{i\in I}$ and let $f: X\rightarrow Y$ be the mapping induced by the family of canonical injections of each $A_i$ into Y. The mapping $f$ satisfies property (VK) by hypothesis; we shall therefore apply Theorem 1 of No$^o1$ to it. We therefore take up again the notation of this No$^o$ and begin by defining a van Kampen datum for the mapping $f$.

The arcwise connected components of X are the sets $X_i=$ $A_i\times \{i\}$ for $i\in I$. Thus $\pi_0(X)$ is identified with the set I. For each $i\in I$, let $\mathsf{a}(i)$ denote the point $(a(i), i)$ of $X_i$.

Put $Z = X\times_YX$. The arcwise connected components of Z are the sets $Z_j= (V\times \{i\})\times_Y(V\times \{i'\})$, where $j= (i, i',V)$ runs through $J'$. Thus $\pi_0(Z)$ is identified with the set $J'$. Let $J_0$ be the set of elements of J of the form $(i, i,A_i)$, so that the family $(J_0,J,J)$ is a partition of $J'$. For $i\in I$ and $j= (i, i,A_i)\in J_0$, put $\mathsf{b}(j) = (\mathsf{a}(i),\mathsf{a}(i))$ and take for $\beta_1(j)$ and $\beta_2(j)$ the class of the constant path at $\mathsf{a}(i)$. For $j= (i, i',V)\in J\cup \overline{J}$, let $\mathsf{b}(j)$ denote the point $((b(j), i),(b(j), i'))$ of $Z_j$ and let $\beta_1(j)$ and $\beta_2(j)$ denote the classes of the paths $t\mapsto (B_1(j)(t), i)$ and $t\mapsto (B_2(j)(t), i')$ in X.

Put $W = X\times_YX\times_YX$. The arcwise connected components of W are the sets $W_k= (U\times \{i_1\})\times_Y(U\times \{i_2\})\times_Y(U\times \{i_3\})$, where $k= (i_1, i_2, i_3,U)$ runs through $K'$. Thus $\pi_0(W)$ is identified with the set $K'$.

Let $K_0$ denote the set of elements of $K'$ of the form $k= (i, i, i,A_i)$, for $i\in I$. For such an element $k\in K_0$, put $\mathsf{c}(k) = (\mathsf{a}(i),\mathsf{a}(i),\mathsf{a}(i))$ and choose for $\gamma_{st}(k)$ the class of the constant path at $(\mathsf{a}(i),\mathsf{a}(i))$.

Let $K_1$ denote the set of elements of $K'$ of the form $k$ = $(i_1, i_2, i_3,V)$ for which the set $\{i_1, i_2, i_3\}$ has two elements. Let $k$ be an element of $K'$ of the form $(i, i, i',V)$, so that $j= (i, i',V)$ belongs to $J\cup \overline{J}$. We then put

and we take for $\gamma_{13}(k)$ and $\gamma_{23}(k)$ the class of the constant path at $\mathsf{b}(j)$. One defines analogously $c(k),\gamma_{12}(k),\gamma_{13}(k),\gamma_{23}(k)$ for every element $k$ of $K'$.

For every element $k= (i_1, i_2, i_3,U)$ of K, put

for every pair $(s, t)$ of distinct elements of $\{1,2,3\}$, take for $\gamma_{st}(k)$ the image under the mapping $x\mapsto ((x, s),(x, t))$ of the class of the path $C_{st}(k)$ in $A_{i_s(k)}\cap A_{i_t(k)}$.

For every point $x= (x_1, x_2, x_3)$ of $X\times_YX\times_YX$ and every permutation $\sigma \in \mathfrak{S}_3$, put $\sigma (x) = (x_{\sigma^{-1}(1)}, x_{\sigma^{-1}(2)}, x_{\sigma^{-1}(3)})$. Thus an operation of the group $\mathfrak{S}_3$ on W is defined. For every $k= (i_1, i_2, i_3,U)\in K'$ and every permutation $\sigma \in \mathfrak{S}_3$, put analogously $\sigma (k) = (i_{\sigma^{-1}(1)}, i_{\sigma^{-1}(2)}, i_{\sigma^{-1}(3)},U)$; one has

Let $k= (i_1, i_2, i_3,U)$ be an element of $K'$ such that $i_1, i_2, i_3$ are pairwise distinct. There exists a unique permutation $\sigma \in$ $\mathfrak{S}_3$ such that $i_{\sigma^{-1}(1)}< i_{\sigma^{-1}(2)}< i_{\sigma^{-1}(3)}$, so that $\sigma (k) =$ $(i_{\sigma^{-1}(1)}, i_{\sigma^{-1}(2)}, i_{\sigma^{-1}(3)},U)$ belongs to K. For $s\in \{1,2,3\}$, one then puts $c_s(k) =c_{\sigma(s)}(\sigma (k))$ and $c(k) = (c_1(k), c_2(k), c_3(k))$, so that $c(k) =\sigma^{-1}(c(\sigma (k)))$. For $(s, t)\in \{(1,2),(1,3),(2,3)\}$, one defines $C_{st}(k) = C_{\sigma^{-1}(s)\sigma^{-1}(t)}(\sigma (k))$; it is a path joining $c(k)$ to $b(j_{\sigma(s)\sigma(t)}(\sigma (k))) =b(j_{st}(k))$.

Let $g$ denote the morphism of quivers from Γ into $\mathsf{\Gamma}$ which associates with a vertex $i\in I$ of Γ the vertex $X_i= A_i\times \{i\}$ of $\mathsf{\Gamma}$ and with an arrow $j= (i, i',V)\in J'$ of Γ the arrow $Z_j= (V\times \{i\})\times_Y(V\times \{i'\})$ of $\mathsf{\Gamma}$. The mapping Som($g$) is bijective; the mapping Fl($g$) is injective and the graph Γ is connected (IV, p. 410, lemma 1), and the image under $g$ of the subquiver T is a subquiver $\mathsf{T}$ of $\mathsf{\Gamma}$ whose associated graph is a maximal tree of the graph $\widetilde{\mathsf{\Gamma }}$.

The points $\mathsf{a}(i)$, for $i\in I$, the points $\mathsf{b}(j)$, for $j\in J'$, the points $\mathsf{c}(k)$, for $k\in K'$, the classes of paths $\beta_1(j)$ and $\beta_2(j)$, for $j\in J'$, the classes of paths $\gamma_{st}(k)$, for $k\in K'$, the subquiver $g(T)$ of $\mathsf{\Gamma}$ and the element $i_0$ of I define a van Kampen datum for $f$.

Let $\rho$ denote the unique group homomorphism

which induces the isomorphism of $\pi_1(X,\mathsf{a}(i))$ onto $\pi_1(A_i, a(i))$ deduced from the identification of $A_i\times \{i\}$ and $A_i$, for all $i\in I$, and such that one has

$\rho (j) = 1$ for $j\in J_0$

$\rho (j) =j,\rho (\overline{j}) =j^{-1}$ for $j\in J$.

Let $\mathsf{L}$ be the group homomorphism defined in Theorem 1 of IV, p. 408. For $j= (i, i,A_i)\in J_0$, one has $\mathsf{L}(j) = 1 =\mathsf{M}\circ \rho (j)$. Let $j= (i, i',V)$ be an element of J; one has $\mathsf{L}(j) = (\mathsf{M}\circ \rho )(j)$ by definition. Finally, if $j=$ $(i, i',V)$ is an element of $\overline{J},\overline{j}\in J$ and one verifies that

Consequently, one has $\mathsf{M}\circ \rho =\mathsf{L}$.

The homomorphism $\mathsf{L}$ is surjective (loc. cit.), hence the homomorphism $\mathsf{M}$ is also. Since the homomorphism $\rho$ is surjective, the kernel of $\mathsf{M}$ is the smallest normal subgroup of $(_i*_{\in I}\pi_1(A_i, a(i)))*F(J_1)$ which contains the images under $\rho$ of the elements defined by relations ($R_1$), ($R_2$)$, (R_3)$ of theorem 1 (IV, p. 408). The proof will be complete once we have verified that these images are, besides the elements defined by relations ($R_1$)$, (R_2), (R_3)$ of Prop. 1, elements conjugate to them, or conjugate to their inverses, together with the identity element.

Elements $R_1$. — An arrow of the oriented tree $\mathsf{T}$ is of the form $Z_j$, with $j= (i, i',V)\in J$; its image is the element $j$ of F(J).

Elements $R_2$. — Let $j= (i, i',V)\in J'$. If $i=i'$, we have $\rho (\mathsf{r}_2(j, v)) = 1$ for every $v\in \pi_1(A_i,\mathsf{a}(i))$. If $j\in J$, the image of $\mathsf{r}_2(j, v)$ is the element $r_2(j, v) =\varphi_j(v)j\psi_j(v)^{-1}j^{-1}$, for every $v\in \pi_1(Z,\mathsf{b}(j))$. In the remaining case, we have $\overline{j}\in J$ and the equality

implies that $\rho (\mathsf{r}_2(j, v))$ is conjugate to $\rho (\mathsf{r}_2(j, v^{-1}))$.

Elements $R_3$. — Let $k= (i_1, i_2, i_3,U)$ be an element of $K'$.

If $k\in K_0,i_1$ = $i_2$ = $i_3,\lambda_s(k)$ is the class of the trivial path for every $s\in \{1,2,3\},j_{st}(k)\in J_0$ for every pair $(s, t)\in$ $\{(1,2),(1,3),(2,3)\}$. Then $\rho (\mathsf{r}_3(k))$ is the identity element.

Suppose $k\in K_1$. If $i_1=i_2$, then $j= (i_1, i_3,U)\in J\cup \overline{J}$ and we have

whose image under $\rho$ is the identity element. The other cases are dealt with analogously.

Suppose that $i_1, i_2, i_3$ are pairwise distinct. If $i_1< i_2< i_3$, $k\in K$ and the image of $\rho (\mathsf{r}_3(k))$ is the element $r_3(k)$.

Let $\sigma \in \mathfrak{S}_3$ be the permutation carrying 1 to 2 and 2 to 3. We have

One verifies analogously that $\lambda_2(\sigma (k)) =\lambda_1(k)$ and $\lambda_3(\sigma (k)) =\lambda_2(k)$. Hence,

which proves that $\rho (\mathsf{r}_3(\sigma (k)))$ is conjugate to

Let $\tau \in \mathfrak{S}_3$ be the transposition with support $\{1,2\}$. We have

$\lambda_1(\tau (k)) =\lambda_2(k)^{-1},\lambda_2(\tau (k)) =\lambda_1(k)^{-1}$ and $\lambda_3(\tau (k)) =\lambda_3(k)^{-1}$. The equalities

show that $\rho (\mathsf{r}_3(\tau (k)))$ is conjugate to the inverse of

Since the group $\mathfrak{S}_3$ is generated by the permutations $\tau$ and $\sigma$, it follows that, for every $k\in K$ and every $\sigma \in \mathfrak{S}_3,\rho (\mathsf{r}_3(\sigma (k)))$ is conjugate to $\rho (\mathsf{r}_3(k))$ or to its inverse.

Thus Proposition 1 is proved.