FALSE: a left Kan extension along a fully faithful functor always restricts back to the original functor

The inclusion $K$ is fully faithful because $\mathcal{A}$ is discrete and $\mathcal{D}$ has no non-identity arrows between the image objects $\mathrm{\ell }$ and $r$, so each hom-set map is bijective as required by [F2].

The pair $(G,\lambda )$ is a left Kan extension of $F$ along $K$. Let $M:\mathcal{D}\to \mathcal{E}$ and $\alpha :F\Rightarrow MK$. Because $\mathcal{D}$ has arrows $m\to \mathrm{\ell }$ and $m\to r$, the object $M(m)$ must admit arrows to both $M(\mathrm{\ell })$ and $M(r)$. In $\mathcal{E}$ the only object with arrows to two targets is $M$, via $g:M\to L^{\mathrm{\prime }}$ and $h:M\to R^{\mathrm{\prime }}$. Therefore $M(m)=M$, $M(\mathrm{\ell })=L^{\mathrm{\prime }}$, $M(r)=R^{\mathrm{\prime }}$, and the two span arrows must be sent to $g$ and $h$. So $M=G$. Then ${\alpha }_{\mathrm{\ell }}:L\to L^{\mathrm{\prime }}$ and ${\alpha }_r:R\to R^{\mathrm{\prime }}$ are forced to be $f$ and $k$, whence $\alpha =\lambda$. Since the only endomorphisms of $L^{\mathrm{\prime }}$, $M$, and $R^{\mathrm{\prime }}$ are identities, the only natural transformation $G\Rightarrow M=G$ is the identity. Thus $(G,\lambda )$ satisfies the universal property [F1].

The components ${\lambda }_{\mathrm{\ell }}=f:L\to L^{\mathrm{\prime }}$ and ${\lambda }_r=k:R\to R^{\mathrm{\prime }}$ are not isomorphisms, since $\mathcal{E}$ has no arrows $L^{\mathrm{\prime }}\to L$ or $R^{\mathrm{\prime }}\to R$. Therefore the unrestricted claim is false. By [L1], this also shows that the witness is not pointwise.