The Eilenberg–Moore forgetful functor strictly creates coequalizers of $U^T$-split pairs

By [L2], $q$, $Tq$, and $T^{2}q$ are coequalizers of the corresponding images of $f,g$. In particular each is an epimorphism, since every coequalizer is epic by its uniqueness clause.

Because $f$ and $g$ are algebra homomorphisms, $qb:TB\to C$ coequalizes $Tf$ and $Tg$. The universal property of $Tq$ gives a unique $c:TC\to C$ satisfying $cTq=qb$.

After precomposition with the epimorphism $q$, the unit equation $c{\eta }_C=1_C$ is the unit law for $b$. After precomposition with the epimorphism $T^{2}q$, the associativity equation $cT(c)=c{\mu }_C$ is the associativity law for $b$. Hence $(C,c)$ is a $T$-algebra.

The defining equation $cTq=qb$ says exactly that $q:(B,b)\to (C,c)$ is an algebra homomorphism.

If $r:(B,b)\to (D,d)$ is an algebra homomorphism with $rf=rg$, [L2] gives a unique underlying $u:C\to D$ with $uq=r$. Precomposing $uc$ and $dT(u)$ with the epimorphism $Tq$ gives the same map, so $u$ is an algebra homomorphism and is the unique algebraic factorization.

Any algebra structure $c^{\mathrm{\prime }}$ on the supplied apex for which $q$ is an algebra homomorphism satisfies $c^{\mathrm{\prime }}Tq=qb=cTq$, so $c^{\mathrm{\prime }}=c$ because $Tq$ is epic. Together with step 5.1 this is the unique on-the-nose lift required by [L1].