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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.

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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 $G^*p$.

(Hopefully I got all this the right way around. I sometimes get right and left Kan extensions flipped.)

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  • $\begingroup$ 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. $\endgroup$ – Malice Vidrine Sep 23 '17 at 8:32
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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, $A,B\mapsto (p_B: A\times B\to B)$.

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  • $\begingroup$ Thank you! Is there simple way to see how that generalizes to an arbitrary poset, $P$, and the corresponding discrete category $|P|$? $\endgroup$ – Andrew Bacon Mar 10 '17 at 19:19
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    $\begingroup$ There is a formula for the left Kan extension, which is what this is, and which you can look up: you calculate a colimit over a certain comma category. In small cases like this, you can also work things out for yourself using the adjunction isomorphisms, since you know how to compute the right adjoint. Try the first four the inclusion of either point into the arrow, for instance. $\endgroup$ – Kevin Carlson Mar 10 '17 at 19:58

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