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This might be a silly question, but given a presheaf $\mathscr{F}$ on $Aff$, can we always lift $\mathscr{F}$ (up to isomorphism) to a presheaf $\mathscr{F}^{+}:Sch^{op} \rightarrow Sets$ making the diagram below commute (possibly up to isomorphism)?

$\require{AMScd}$ \begin{CD} Sch^{op} @>\mathscr{F}^{+}>> Sets \\ @AAA @VV\textrm{id}V \\ Aff^{op} @>>\mathscr{F}> Sets \end{CD}

Here's my thought. Because can always write a scheme $X=\textrm{colim} \ \textrm{Spec}(A)$ (a colimit of affines), define $\mathscr{F}^{+}(X):= \textrm{lim} \ \mathscr{F}(\textrm{Spec}(A))$, where the limit is taken over the same index as the colimit. $Sets$ is complete so this limit makes sense. My worry is that two different colimits of affines will not produce the same value in $Sets$, but two isomorphic sets. If this all works out it seems like this should be some instance of a more general categorical phenomenon...

Edit: Diagram updated per suggestion of Derek Elkins.

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    $\begingroup$ The general categorical construction you're looking for is called left Kan extension, although it doesn't always have the second property you asked for. $\endgroup$ – Qiaochu Yuan Dec 14 '17 at 21:34
  • $\begingroup$ Does the presheaf I defined satisfy the properties I’ve specified? $\endgroup$ – LPK Dec 15 '17 at 1:37
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    $\begingroup$ Either you're missing a hypothesis or it's obvious that this can't be done in general. E.g. there is no reason why $\mathscr{F}(\mathrm{Spec}(A) \amalg \mathrm{Spec}(A))$ should be isomorphic to $\mathscr{F}(\mathrm{Spec}(A)) \times \mathscr{F}(\mathrm{Spec}(A))$ $\endgroup$ – Hurkyl Dec 15 '17 at 15:00

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