 Primitive-element kernel rigidity lemma

If primitive elements have equal restrictions along $\alpha :P\to m$ and $\beta :P\to n$, then these surjections have equal kernels. To rule out $\alpha (a)=\alpha (b)$ with unequal $\beta$ images, identify those two images by $q:n\to n-1$. The set of pairs equal under $\alpha$ and $q\beta$ maps surjectively both to $P$ by projection and to the kernel pair of $q$ by applying $\beta$. Injective restrictions force the kernel-pair matching condition on the primitive element, contradicting descent along $q$.