that the normalization conditions $v(1) = 1$ and $r(1) = 1$ where used to obtain [(\[eq:p1\])]{} and [(\[eq:p4\])]{}.

Conversely, the normalization conditions follow from [(\[eq:p1\])]{} - [(\[eq:p4\])]{} in the following manner: first, for $g' = 1$ in [(\[eq:p1\])]{} we obtain $\sigma(h)r(1) = r(1)\bigl(v(1) \rhd
\sigma(h)\bigl) $ for all $h \in H$. Since $\sigma$ is an automorphism we have : $${\label{eq:th2}}
hr(1) = r(1)\bigl(v(1) \rhd h\bigl)$$ for all $h \in H$. Now let $g_{1}'=1$ in [(\[eq:p3\])]{} to obtain: $$\begin{aligned}
r(g_{2}') = r(1)\bigl(v(1) \rhd r(g_{2}')\bigl)
\stackrel{{(\ref{eq:th2})}} {=} r(g_{2}')r(1)\end{aligned}$$ thus $r(1) = 1$. Finally we let $g_{2}' = 1$ in [(\[eq:p4\])]{} to obtain $v(g_{1}') = v(g_{1}')v(1)$, thus $v(1) = 1$.

It remains to be proven that $\psi$ given by [(\[eq:p5\])]{} is an isomorphism if and only if $v: G'\to G$ is a bijective map. Assume first that $\psi$ is an isomorphism. Then $v$ is surjective and for $g_{1}'$, $g_{2}' \in G'$ such that $v(g_{1}') = v(g_{2}')$ we have: $$\psi(1, g_{2}') = \bigl(r(g_{2}'), v(g_{2}')\bigl) =
\bigl(r(g_{2}'), v(g_{1}')\bigl) =
\psi\bigl(\sigma^{-1}(r(g_{2}'))\sigma^{-1}(r(g_{1}')^{-1}),
g_{1}'\bigl)$$ Hence $g_{1}' = g_{2}'$ and $v$ is injective. Conversely, assume that $v$ is bijective. If $\psi(h, g') = (1, 1)$ we obtain that $\sigma(h)r(g') = 1$ and $v(g') = 1 = v(1)$. It follows from here that $g' = 1$ and $\sigma(h) = 1 = \sigma(1)$, i.e. $h = 1$. Hence $\psi$ is injective. Let $(h, g) \in H\, {}_{\alpha}\!\!
\bowtie_{\beta} \, G$ and $g' \in G'$ such that $v(g') = g$. Then $\psi\bigl(\sigma^{-1}(h)\sigma^{-1}(r(g')^{-1}), g'\bigl) = (h,
g)$ i.e. $\psi$ is an isomorphism of groups.

We shall prove now the main result of this section:

**(Deformation of a matched pair)**[\[th:deformation\]]{} Let $(H, G, \alpha, \beta)$ be a matched pair of groups, $(\sigma,
v, r)$ be a triple where $\sigma \in {{\rm Aut}\,}(H)$, $v: G \to G$ is a bijective map, $r: G \rightarrow H$ is a map such that $v (1_G) =
1_G$, $r (1_G) = 1_H$ and 