# Geometric morphisms between presheaves

I've read that every functor, $f:C \to D$, generates a geometric morphism from $f^* \vdash f_*:Sets^{C^{Op}} \to Sets^{D^{Op}}$.

But I don't have a feel for how this really works in concrete examples. Take for example the inclusion function between the discrete two element category $|2|$, and the partial order $2 := (\{0,1\}, \leq)$. The induced functor from $Sets^{|2|} \to Sets^2$ associates each pair of sets with a pair of sets and a function between them. But I can't think of any natural functor that would come up with a function between sets out of thin air like this. It would be helpful to me if I could see this functor described concretely.

The general way to calculate the functor $\mathbf{Set}^{C^{op}}\to \mathbf{Set}^{D^{op}}$ is to consider the left Kan extension of a functor $F:C^{op}\to \mathbf{Set}$ along the functor $f^{op}:C^{op}\to D^{op}$. Since $C,D$ are presumably small and $\mathbf{Set}$ is co-complete, such a Kan extension will always exist. The easiest way to calculate it is, for each $d$ in $D$, we take $Lan_{f^{op}} F(d)$ to be the colimit of $F\circ Q:(f^{op}\downarrow d)$ in $\mathbf{Set}$. It will turn out that the existence of $Lan_{f^{op}} F$ for any $F$ is the same as there being a left adjoint to $-\circ f:\mathbf{Set}^{D^{op}}\to \mathbf{Set}^{C^{op}}$.
Understanding what this means in the case of a given poset $P$ is a little harder to visualize. For a functor $G$ on $|P|$ and an element $p$ of the poset, it takes that $p$ to the coproduct $G^*p:=\coprod\{G(q)|q\geq p\}$, and for the inequality $p<p'$, we have the coproduct inclusion of $G^*p'$ into $G^*p$.
• I stumbled across this again and realized I had ignored the part of the question where the point was about understanding the direct image part of a geometric morphism and I wrote out the wrong adjoint! Instead you want to use the dual construction and take $p$ to the limit of $G\circ Q:(d\downarrow f^{op})\to\mathbf{Set}$, where $f:|P|\to P$ is the inclusion of the elements of the poset. It's a pretty straightforward dualization of the one I described above. – Malice Vidrine Sep 23 '17 at 8:32
The induced functor sends a pair of sets $A,B$ to the arrow $i_A:A\to A\sqcup B$. Its right adjoint sends an arrow to its domain and codomain. This actually has a further right adjoint, $A,B\mapsto (p_B: A\times B\to B)$.
• Thank you! Is there simple way to see how that generalizes to an arbitrary poset, $P$, and the corresponding discrete category $|P|$? – Andrew Bacon Mar 10 '17 at 19:19