Kan Extensions Density and the Free Cocompletion — Examples

The inclusion $i$ is fully faithful by [F1]. The comma category $(i\downarrow 2)$ has two objects, namely the arrows $0\to 2$ and $1\to 2$, and one non-identity morphism from the first to the second, corresponding to the inequality $0\le 1$ in $\mathcal{A}$.

The induced diagram is therefore just $\mathrm{\varnothing }\to \{\ast \}$, whose colimit in $\mathbf{S}\mathbf{e}\mathbf{t}$ is $\{\ast \}$. By [L1], this is the left Kan extension value at $2$.

On the objects $0$ and $1$, the pointwise left Kan extension restricts back to $F$ by [L2]. Hence the left Kan extension along the full subcategory inclusion is the functor sending $0\mapsto \mathrm{\varnothing }$, $1\mapsto \{\ast \}$, and $2\mapsto \{\ast \}$.

By [F1], the functor ${\mathrm{Lan}}_{y}F$ has domain $\mathbf{S}\mathbf{e}\mathbf{t}$, and [F2] verifies the locally small and cocomplete target hypotheses of [L1]. Hence [L1] makes it left adjoint to the functor $G\mapsto \mathbf{G}\mathbf{r}\mathbf{p}(F(\{\ast \}),G)$.

By [L2], the functor $G\mapsto \mathbf{G}\mathbf{r}\mathbf{p}(F(\{\ast \}),G)$ is naturally isomorphic to the underlying-set functor on groups. Therefore ${\mathrm{Lan}}_{y}F$ is a left adjoint to the underlying-set functor.

The published free-group functor is also a left adjoint to the underlying-set functor by [L2], so [L3] identifies ${\mathrm{Lan}}_{y}F$ with that free functor.

In the one-object setting, an object of $(i\downarrow \ast )$ is just an element $g\in G$. A morphism from $g$ to $g^{\mathrm{\prime }}$ is an element $h\in H$ with $g=g^{\mathrm{\prime }}i(h)$ by the comma-category equation, equivalently $g^{\mathrm{\prime }}=gi(h^{-1})$. So the indexing category is the action groupoid for the right $H$-action on $G$, and the induced diagram on $\mathbf{S}\mathbf{e}\mathbf{t}$ sends the arrow $g\to gi(k)$ to the map $x\mapsto k^{-1}x$ on $X$. By [F2], its colimit is the quotient of $G\times X$ by $(g,x)\sim (gi(k),k^{-1}x)$, equivalently by $(gi(h),x)\sim (g,hx)$, namely $G{\times }_{H}X$. Therefore [L1] identifies the induced representation with the left Kan extension.

Dually, an object of $(\ast \downarrow i)$ is again an element of $G$, and a cone to a set $Y$ is exactly a family of maps indexed by $G$ that is equivariant for the $H$-action. By [F2], the limit is therefore the set of $H$-equivariant maps $G\to X$, written ${\mathrm{Map}}_H(G,X)$. Hence [L1] identifies the coinduced representation with the right Kan extension.

For the unique object of $\mathbf{1}$, the comma category $(!\downarrow \ast )$ is just the one-object category $G$ again, by [F1] and [F2]. A cocone from the $G$-action diagram $X:G\to \mathbf{S}\mathbf{e}\mathbf{t}$ to a set $Y$ is exactly a function $q:X\to Y$ constant on $G$-orbits, so its universal example is the quotient map $X\to X\mathrm{/}G$. Therefore [L1] identifies the orbit set with the left Kan extension of $X$ along $!$.

Dually, a cone from a set $Y$ to the action diagram is exactly a function $Y\to X$ landing in the equalizer of all action maps, that is, in the fixed-point set $X^G$. Hence [L1] identifies $X^G$ with the right Kan extension of $X$ along $!$.

The category of elements of $P$ has the three objects $(0,x)$, $(1,y)$, and $(1,z)$, and no non-identity morphisms, by [F1].

Therefore the density diagram of [L1] is the discrete three-object diagram with values $y(0)$, $y(1)$, and $y(1)$, by [F2]. Its colimit is the coproduct $y(0)\bigsqcup y(1)\bigsqcup y(1)$.

Applying [L1] to this presheaf gives exactly that coproduct as $P$.

The inclusion $K$ is fully faithful by [F3].

The pair $(L,\eta )$ with ${\eta }_{\mathrm{\ell }}=1_a$ and ${\eta }_r=1_b$ is a left Kan extension of $F$ along $K$. Indeed, if $\alpha :F\Rightarrow MK$ exists, then $M(\mathrm{\ell })=a$ and $M(r)=b$, because $a$ and $b$ have no non-identity arrows out of them; and the arrows $m\to \mathrm{\ell }$ and $m\to r$ force $M(m)=c$, the only object of $\mathcal{E}$ with arrows to both $a$ and $b$. So $M=L$, and then $\alpha$ is forced to be the identity on $\mathrm{\ell }$ and $r$, which leaves exactly one natural transformation $L\Rightarrow M$. This is the universal property [F1].

But $(K\downarrow m)$ is empty, because there is no arrow from $\mathrm{\ell }$ to $m$ and none from $r$ to $m$ in $\mathcal{D}$. If $(L,\eta )$ were pointwise, [F2] would make $L(m)=c$ the colimit of the empty diagram and hence an initial object of $\mathcal{E}$. That is impossible by [F4], since there is no morphism $c\to d$. So this left Kan extension is not pointwise.

Therefore the claim is false: a fully faithful left Kan extension need not be pointwise.

For a real $x$, the comma category $(i\downarrow x)$ is the preorder of rationals $q\le x$. The induced diagram sends such a $q$ to $(-\mathrm{\infty },q)$ and its colimit in $\mathbf{S}\mathbf{e}\mathbf{t}$ is the union ${\bigcup }_{q\le x,q\in \mathbb{Q}}(-\mathrm{\infty },q)$.

This union is exactly $(-\mathrm{\infty },x)$. If $r<x$, [L2] gives a rational $q$ with $r<q<x$, so $r\in (-\mathrm{\infty },q)$ and hence lies in the union. Conversely every $(-\mathrm{\infty },q)$ with $q\le x$ is contained in $(-\mathrm{\infty },x)$.

Therefore [L1] gives $L(x)=(-\mathrm{\infty },x)$ for the left Kan extension value at every real $x$.