Proposition 4

Suppose that the arcwise connected components of Y are open. For each $i\in$ I, let $G_i$ be the quotient of the group $\pi_1(Y_i, b_i)$ by the smallest normal subgroup containing the image of the homomorphisms $\psi_j$, for $j\in \overset{-1}{\varphi}(i)$ ; let $p_i$ denote the canonical surjection of $\pi_1(Y_i, b_i)$ onto $G_i$.

There exists a unique group homomorphism

$\mathsf{P}:_i*_{\in I}G_i*F(J-T)\rightarrow \pi_1($Côn$(f), s)$

such that

$\mathsf{P}(p_i(v)) =$ Int($\gamma_{\sigma(i)}\beta_{\sigma(i)}$)$(v)$ for $i\in I$ and $v\in \pi_1(Y_i, b_i)$,

$\mathsf{P}(j) =\gamma_j\beta_j\beta_{\tau(j)}^{-1}\gamma_{\tau(j)}^{-1}$ for $j\in J-T$.

The homomorphism $\mathsf{P}$ is an isomorphism.

Let $Y'$ denote the union of the arcwise connected components of Y which meet $f(X)$, and let $f': X\rightarrow Y'$ be the mapping given by $x\mapsto f(x)$. The set $Y'$ is an open subset of Y; its arcwise connected components are open. The cone Cone$(f')$ is identified with the arcwise connected component of $s$ in Cone$(f)$. This makes it possible to suppose that $Y = Y'$, in other words that the mapping $\pi_0(f)$ is surjective and $I =\pi_0(Y)$.

For each $j\in J$, let us put $V_j=\alpha '_f(X_j\times ]0,1[)$. On passing to subspaces, the mapping $\alpha '_f$ induces a homeomorphism of $X\times ]0,1[$ onto the complement of $Y\cup \{s\}$ in Cone$(f)$. Hence the sets $V_j$ are the arcwise connected components of Cone$(f)-(Y\cup \{s\})$.

For each $i\in I$, let us put $U_i= (\rho '_f)^{-1}(Y_i)$; this is an open subset of Cone$(f)$, since $Y_i$ is open in Y by hypothesis. For each $j\in \overset{-1}{\varphi}(i)$, one has $f(X_j)\subset Y_i$ and

so that $V_j\cup Y_i$ is an arcwise connected subset of Cone$(f)$ containing $Y_i$. Since $U_i$ is the union of $Y_i$ and of the sets $V_j$, for $j\in \overset{-1}{\varphi}(i)$, it follows that $U_i$ is arcwise connected.

Finally, the set $C'(X) =$ Cone$(f)$ - Y is an open subset of Cone$(f)$; it is contractible to $s$, hence arcwise connected.

The set $C'(X)$ and the sets $U_i$, for $i\in I$, constitute an open arcwise connected covering of Cone$(f)$, to which we shall apply prop. 1 of IV, p. 412. Let $I'$ be the set obtained by adjoining $s$ to I; let it be provided with a total ordering for which $s$ is its smallest element.

For distinct elements $i, i'$ of I, one has $U_i\cap U_{i'}=\emptyset$. For $i\in I$, $C'(X)\cap U_i$ is the union of the sets $V_j$, for $j\in \overset{-1}{\varphi}(i)$; they are connected and pairwise disjoint. The intersection of any three distinct sets of this covering is empty.

The framework Γ of the covering considered has the set $I'$ for its vertices. Its edges are the triplets $(s, i,V_j)$, for $j\in J$ and $i=\varphi (j)$; thus the set of edges of Γ will be identified with the set J.

For $i\in I$, one chooses as base point $\mathsf{a}(i) =a_i\in U_i$; one also puts $\mathsf{a}(s) =s\in C'(X)$.

For $j\in$ J, put $\mathsf{b}(j) =\alpha '_f(b_j,\frac{1}{2})$. Let $B_1(j)$ be the path in $C'(X)$ with origin $\mathsf{b}(j)$ and endpoint $\mathsf{a}(s)$ given by $t\mapsto \alpha '_f(b_j,(1-t)/2)$. Let $B_2(j)$ be the path in $U_{\varphi(j)}$ with origin $\mathsf{b}(j)$ and endpoint $\mathsf{a}(\varphi (j)) =a_{\varphi(j)}$, obtained by juxtaposing the path $t\mapsto \alpha '_f(b_j,(1 +t)/2)$ and the path $B_j$. Then the class of the path $B(j) =\overline{B}_1(j)*B_2(j)$ is equal to $\gamma_j\beta_j$.

We choose $i_0=s$.

We take for maximal oriented tree T the unique oriented tree of Γ whose set of arrows is $\sigma (I)$. We have $\delta (s) =e$, whereas for $i\in I,\delta (i) = [B(\sigma (i))] =\gamma_{\sigma(i)}\beta_{\sigma(i)}$.

Let $j\in J$ and let $i=\varphi (j)$.

The homomorphism $\varphi_j$ of $\pi_1(V_j,\mathsf{b}(j))$ into $\pi_1(C'(X), s)$ is the trivial homomorphism, since $C'(X)$ is contractible to $s$.

The mapping $\alpha '_f$ induces a homeomorphism of $X_j\times ]0,1[$ onto $V_j$; this homeomorphism induces an isomorphism of the group $\pi_1(V_j,\mathsf{b}(j))$ onto the group $\pi_1(X_j, b_j) =\pi_1(X, b_j)$. By passing to subspaces, the mapping $\sigma '_f$ induces a strong contraction of $U_i$ onto $Y_i$, which induces an isomorphism of the group $\pi_1(U_i,\mathsf{a}(i))$ onto the group $\pi_1(Y, a_i)$. By means of these isomorphisms, the homomorphism

is identified with the homomorphism Int($\beta_{\varphi(j)}^{-1}$)$\circ f_*$ of $\pi_1(X, b_j)$ into $\pi_1(Y, a_i)$.

Since $C'(X)$ is contractible to $s$, the homomorphism $\mu_s$ is the trivial homomorphism.

Let $i\in I$. The homomorphism $\mu_i$ of $\pi_1(U_i,\mathsf{a}(i))$ into $\pi_1($Côn$(f), s)$ is identified with the homomorphism of $\pi_1(Y, a_i)$ into $\pi_1($Côn$(f), s)$ obtained by composing the homomorphism Int($\delta (i)$) and the homomorphism of $\pi_1(U_i, s)$ into $\pi_1($Côn$(f), s)$ deduced from the canonical injection of $U_i$ into Côn$(f)$.

Finally, the homomorphism $\mu: F(J)\rightarrow \pi_1($Côn$(f), s)$ is given by

Let $\mathsf{P}'$ be the unique group homomorphism

$\mathsf{P}':_i*_{\in I}\pi_1(Y_i, a_i)*F(J)\rightarrow \pi_1($Cone$(f), s)$

which coincides with the homomorphism Int($\delta (i)^{-1}$) on $\pi_1(Y_i, a_i)$ and with the homomorphism $\mu$ on F(J). By Prop. 1 of IV, p. 412, the homomorphism $\mathsf{P}'$ is surjective and its kernel is the smallest normal subgroup of $*_i\pi_1(Y_i, a_i)*F(J)$ which contains the elements $\mathsf{r}_1(j)$, for $j\in T$, and the elements $\mathsf{r}_2(j, v)$, for $j\in J$ and $v\in \pi_1(V_j,\mathsf{b}(j))$. (There are no elements $\mathsf{r}_3(k)$, since the set K is empty.)

For $j\in T$, one has $r_1(j) =j$. Let $j\in J$ and let $v\in \pi_1(V_j,\mathsf{b}(j))$. In view of the identification of $\pi_1(V_j,\mathsf{b}(j))$ with $\pi_1(X, b_j)$, one has $\mathsf{r}_2(j, v) =j\psi_j(v)j^{-1}$. Let $p$ denote the canonical surjective homomorphism of $*\pi_1(Y_i, a_i)*F(J)$ onto $*G_i*F(J$ - T). For $j\in T$, one has $p(\mathsf{r}_1(j)) =^ie$; for $j\in J$ and $v\in \pi_1(X^i, b_j)$, one has $p(\psi_j(v)) =e$. Hence there exists a unique group homomorphism $\mathsf{P}$ of $*G_i*F(J-T)$ into $\pi_1($Cone$(f), s)$ such that $\mathsf{P}'\circ p=\mathsf{P}$; it is an isomorphism.$^i$