Left adjoints preserve left Kan extensions

Let $\alpha :SF\Rightarrow MK$ be any natural transformation. By [F2], each component ${\alpha }_c:S(Fc)\to M(Kc)$ has a right adjunct ${\alpha }_c^{\mathrm{♭}}:Fc\to R(M(Kc))$, and these components form a natural transformation ${\alpha }^{\mathrm{♭}}:F\Rightarrow (RM)K$. Since $(L,\eta )$ is a left Kan extension, [L1] gives a unique natural transformation $\tau :L\Rightarrow RM$ with ${\alpha }^{\mathrm{♭}}=(\tau K)\circ \eta$.

Let $\sigma :SL\Rightarrow M$ be the left adjunct of $\tau$. Then $(\sigma K)\circ S\eta$ has right adjunct $(\tau K)\circ \eta ={\alpha }^{\mathrm{♭}}$, so by uniqueness of adjuncts it equals $\alpha$. If ${\sigma }^{\mathrm{\prime }}:SL\Rightarrow M$ also satisfied $({\sigma }^{\mathrm{\prime }}K)\circ S\eta =\alpha$, then its right adjunct would satisfy the same factorization equation as $\tau$, and [L1] would force that adjunct to equal $\tau$; applying [F2] again gives ${\sigma }^{\mathrm{\prime }}=\sigma$. Therefore $(SL,S\eta )$ is initial among pairs $(M,\alpha )$, hence a left Kan extension of $SF$ along $K$.