Timeline for Right adjoint to pullback functor
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 9, 2020 at 22:12 | comment | added | Todd Trimble | @mattecapu I may have meant to say (but didn't) that since we have $Set/X \simeq (Set/Y)/f$, we just relativize the object of sections construction, replacing $Set$ in the first construction by a new base topos $Set/Y$, and replacing $X$ by $f$. Thus we just internalize the first construction to the new base topos. In effect, $\prod_f: (Set/Y)/f \to Set/Y$ is a bundle of instances of the first object of sections construction parametrized over $y \in Y$, each of the form $\prod_{X_y}: Set/(X_y) \to Set$ where $X_y$ is the fiber $f^{-1}(y)$. This gives $\prod_{x \in f^{-1}(y)} p^{-1}(x)$. | |
| Jan 9, 2020 at 17:44 | comment | added | seldon | Is there an intuition for the second construction? In the first, we get a nice object of sections, in the second this doesn't seem to be that nice. I see it is still an object of sections, but it is $Y$-indexed so on each $Y$-fiber you consider only $p$ sections of that fiber. But this is just a restating of the definition. Is there a wider picture I'm not seeing? | |
| Nov 8, 2012 at 19:37 | vote | accept | CommunityBot | ||
| Nov 8, 2012 at 18:01 | comment | added | Todd Trimble | Look at the next-to-last sentence: we are taking the object of sections of a certain map $p$ in $Set/Y$. Taking advantage of the fact that $Set/Y$ is finitely complete and cartesian closed, we can write down the equalizer formula in $Set/Y$ for $Sect(p)$ exactly as we did in $Set$: it only uses equalizers and exponentials. This $p \mapsto Sect(p)$ is the right adjoint to the pullback functor. | |
| Nov 8, 2012 at 17:24 | comment | added | user2664 | Ok, thanks it's clearer. But I don't see well where the fact that Sets is locally cartesian closed is used (ie the slices are cartesian closed, ie they have all finite products and exponentials if I refer to Awodey's book). Probably when you wrote $(\prod_f p)_y := \prod_{x \in f^{-1}(y)} p^{-1}(x)$ you used that slices have products (finite?) because fibers are the same than morphisms with codomain $X$. But where exponentials are used ? | |
| Nov 8, 2012 at 14:12 | comment | added | Todd Trimble | No, I wanted to analyze a general morphism of the form $X^\ast A \to p$, because by the adjunction you are asking about, these are in bijection with morphisms of the form $A \to \prod_X p$. Here $A$ is an object of $Set$, and $p$ is an object of $Set/X$. | |
| Nov 8, 2012 at 13:37 | comment | added | user2664 | I don't understand when you said "A morphism from $X^{\star}A\rightarrow (p:\,E\rightarrow X)$ ...", I think you want to define $X^{\star}(A\rightarrow B)$ ? | |
| Nov 8, 2012 at 12:37 | history | answered | Todd Trimble | CC BY-SA 3.0 |