Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group

By [L2], construct the unique homomorphism $f:F(X)\to H$ whose value on each free generator $x$ is $u(x)$.

The free generators generate $F(X)$: if $K=\langle X\rangle\le F(X)$, the map $X\to K$ extends by [L2] to $a:F(X)\to K$, and inclusion $j:K\hookrightarrow F(X)$ makes $j\circ a$ agree with $\operatorname{id}_{F(X)}$ on $X$, so uniqueness gives $j\circ a=\operatorname{id}_{F(X)}$ and $K=F(X)$. By [L3], the canonical quotient map $\pi:F(X)\to\langle X\mid R\rangle$ is surjective; since it sends $X$ to the classes $[x]$, [F2] and [F3] show that these classes generate the presented group.

The hypothesis puts every $r\in R$ in $\ker f$; [L4] makes the kernel normal, so the minimality in [F1] gives $\langle\!\langle R\rangle\!\rangle_{F(X)}\subseteq\ker f$.

By [F4], apply [L1] to factor $f$ uniquely through $F(X)/\langle\!\langle R\rangle\!\rangle=\langle X\mid R\rangle$, obtaining $\overline u$ with $\overline u([x])=u(x)$.

If $h:\langle X\mid R\rangle\to H$ also has $h([x])=u(x)$, then [F3] makes $h\circ\pi:F(X)\to H$ a homomorphism extending $u$, so [L2] gives $h\circ\pi=f=\overline u\circ\pi$; uniqueness of the factorisation in [L1] gives $h=\overline u$.

By [L4], $\operatorname{im}\overline u$ is a subgroup containing every $u(x)$, so [F2] gives $\langle u(X)\rangle\subseteq\operatorname{im}\overline u$. Conversely, put $K=\langle u(X)\rangle$. By [F3], $\overline u^{-1}(K)$ is a subgroup of the domain, and it contains every $[x]$; step 1.2 and [F2] therefore give $\overline u^{-1}(K)=\langle X\mid R\rangle$. Hence $\operatorname{im}\overline u\subseteq K$, so $\operatorname{im}\overline u=\langle u(X)\rangle$. Thus $\overline u$ is surjective exactly when $u(X)$ generates $H$.