FALSE: every Kan extension is pointwise

The inclusion $K$ is fully faithful by [F1], and the pair $(L,\eta )$ with ${\eta }_{\mathrm{\ell }}=1_a$ and ${\eta }_r=1_b$ is a left Kan extension of $F$ along $K$: if $\alpha :F\Rightarrow MK$ exists, then necessarily $M(\mathrm{\ell })=a$ and $M(r)=b$, since $a$ has no non-identity arrow out of it and neither does $b$; and because $M$ must carry the arrows $m\to \mathrm{\ell }$ and $m\to r$ to arrows into $a$ and $b$, necessarily $M(m)=c$, the only object of $\mathcal{E}$ with arrows to both. Thus $M=L$ and $\alpha$ is forced to be the identity on $\mathrm{\ell }$ and $r$, so there is exactly one natural transformation $L\Rightarrow M$.

But $(K\downarrow m)$ is empty: there is no arrow $\mathrm{\ell }\to m$ and no arrow $r\to m$ in $\mathcal{D}$. If $(L,\eta )$ were pointwise, [F2] would make $L(m)=c$ the colimit of the empty diagram, hence an initial object of $\mathcal{E}$. This is impossible by [F4], since there is no morphism $c\to d$. Therefore $(L,\eta )$ is a left Kan extension which is not pointwise.

So the claim that every Kan extension is pointwise is false. The pointwise hypothesis in [F3] is genuinely needed.