I am currently reading Sheaves in Geometry and Logic and have hit a little bit of a snag that I can’t figure out. In the book it is claimed that if $ gcolon Blongrightarrow C $ is morphism and $Q$ is a subfunctor of $Hom(_,C)$ then this determines a subfunctor $Q’$ of $Hom(_,B)$. The claim is that $g$ induces $Q’$ from $Q$ "by pullback". I am at a loss as to how this is induced. We want $Q'(D)$ to be a subset of $Hom(D,B)$ for any object $D$ and $Q(f:Drightarrow D’)$ to be the restriction of $Hom(f,B)$ for any arrow $f$, but I have no idea how to use $Q’$ and $g$ to create this new functor.

Mathematics Asked by subobject_classifier on January 2, 2021

1 AnswersThe definition would be $$ Q'(D) = {f: D to B mid gf in Q(D)}. $$ we can easily check that $Q'(h: D to D'): Q'(D') to Q'(D)$ is the restriction of $operatorname{Hom}(h, B)$ . Let $f in Q'(D')$, then $gf in Q(D')$, so $gfh in Q(D)$. This follows because $Q$ is a subfunctor of $operatorname{Hom}(-, C)$. Hence $fh in Q'(D)$, as required.

Correct answer by Mark Kamsma on January 2, 2021

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