Finite nilpotent link quotients are generated by component meridians

Let $H=⟨m_1,\dots ,m_r⟩$. By [F1] and [F2], the images of these meridians generate $G\mathrm{/}{\mathrm{\Gamma }}_{2}G$, so $G=H{\mathrm{\Gamma }}_{2}G$.

For every $k\ge 2$, the quotient ${\mathrm{\Gamma }}_{k}G\mathrm{/}{\mathrm{\Gamma }}_{k+1}G$ is generated by iterated commutators of weight $k$ in the selected meridians. For $k=2$, [F4] makes the pairing $G\mathrm{/}{\mathrm{\Gamma }}_2\times G\mathrm{/}{\mathrm{\Gamma }}_2\to {\mathrm{\Gamma }}_2\mathrm{/}{\mathrm{\Gamma }}_3$ biadditive, so [F1] implies this layer is generated by brackets of the $m_i$. If the claim holds at weight $k$, then [F3] writes the next layer as generated by brackets of a degree-one class with a weight-$k$ class; [F4] makes that pairing biadditive, so [F1] again reduces the degree-one entries to the $m_i$, and the induction hypothesis reduces the other entries to brackets of those meridians. Each such iterated commutator is a word in $H$ lying in ${\mathrm{\Gamma }}_{k+1}G$, so its class is in the image of $H\cap {\mathrm{\Gamma }}_{k+1}G$ in the graded layer.

The conclusion of step 2.1 implies ${\mathrm{\Gamma }}_{k}G\subseteq H{\mathrm{\Gamma }}_{k+1}G$ for every $k\ge 2$, because each generator of the graded layer is represented by a word in the $m_i$ that lies in $H$. Starting from $G=H{\mathrm{\Gamma }}_{2}G$ in step 1.1 and applying these inclusions successively gives $G=H{\mathrm{\Gamma }}_{q}G$ for every finite $q$. Therefore the image of $H$, and hence of the free group map in the statement, is all of $G\mathrm{/}{\mathrm{\Gamma }}_{q}G$. For $q=1$ the quotient is trivial.