Also, if $uz\in H$ then $z\in N_G(u)\cap N_G(v)$ and so $vz\in E(G)$. Finally, if $e\neq vy$ for $y\in N_G(u{\overline}{v})$ then $e\in E(G)\cap E(H)$.

Now define $\phi: {\mathcal{M}}(H)\setminus{\mathcal{M}}(G) \to {\mathcal{M}}(G)\setminus{\mathcal{M}}(H)$ by $$\phi(M) = \{r(e): e\in M\}$$

Given $M\in {\mathcal{M}}(H)\setminus{\mathcal{M}}(G)$ note that $\phi(M) \subseteq E(G)$ since $r(e)\in E(G)$ for all $e\in E(H)$. We claim that in fact $\phi(M)\in {\mathcal{M}}(G)\setminus{\mathcal{M}}(H)$. For the first case, suppose that $u\in r(e) \cap r(f)$. Note that if $u\in r(e)$ then $v\in e$ since edges in $H$ containing $u$ are replaced by edges in $G$ containing $v$ instead. So if $r(e) \cap r(f) = u$ then $e\cap f = v$, a contradiction since $e,f\in M$, a matching.

Now suppose that $e, f\in M$ and $r(e) \cap r(f) = v$. There are two possible ways for $v$ to be in $r(e)$. The first is for $u$ to be in $e$ and the second is for $e=va$ for some $a\notin
N_G(u,{\overline}{v})$. However, since $M\in {\mathcal{M}}(H)\setminus{\mathcal{M}}(G)$ it must be the case that $vy\in M$ for some $y \in N_G(u,{\overline}{v})$. Since $M$ is a matching, $vy$ is the only edge incident to $v$. Therefore, $e$ and $f$ must both contain $u$, a contradiction since $M$ is a matching.

Finally suppose that $z\in 