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Put $Q=G/H$ and write $M^H$ for the invariant submodule. The five-term exact sequence in group cohomology associated with the Lyndon–Hochschild–Serre spectral sequence isThe inflation map in group cohomology composes a cocycle on $Q$ with the quotient homomorphism $G\to Q$. The restriction map in group cohomology restricts a cocycle from $G$ to $H$. The quotient action on the middle term is, for $q=gH$ and a one-cocycle $f$,This is independent of the lift and of the representative at the level of cohomology. Finally, the transgression in group cohomology extends a $Q$-invariant class on $H$ to a one-cochain on $G$; its coboundary is $H$-basic and descends to the two-cocycle on $Q$ representing $d_2[f]$. Changing the extension changes that cocycle by a group coboundary.

For the application, choose free generators $x_1,\dots ,x_n$ of $F$ and normal generators $r_1,\dots ,r_n$ of $R$. Since a finite nonabelian simple group $K$ is a perfect group, its abelianization is zero. The five-term sequence for $1\to R\to F\to K\to 1$ with trivial coefficients containsThe first term is zero because $K$ is finite, and the last term is zero because a free group has cohomological dimension one. It remains to prove that restriction is surjective.

The invariant submodule of homomorphisms $R\to \mathbb{Z}$ is exactlyThe images of the relators $r_1,\dots ,r_n$ generate $R/[F,R]$, so such a homomorphism $\lambda$ is determined by the integer vector $b=(\lambda (r_1),\dots ,\lambda (r_n){)}^T$. Let $A$ be the relator exponent-sum matrix. This square integer matrix presents $K^{\mathrm{ab}}$, which is zero, so $A$ is a unimodular matrix. There is therefore an integer vector $v$ satisfying $Av=b$. Define $\psi :F\to \mathbb{Z}$ by assigning to $x_i$ the $i$th entry of $v$. The definition of $A$ gives $\psi (r_j)=\lambda (r_j)$ for every $j$. Since the relator images generate $R/[F,R]$, the restriction of $\psi$ to $R$ equals $\lambda$. Restriction is surjective, exactness now gives