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Hello everyone,

Let $\mathcal{C}$ a category which has the pullbacks and $f$:$X\longrightarrow Y$ a morphism in $\mathcal{C}$ then I know that the functor pullback $f^{\star}$:$\mathcal{C}/Y\longrightarrow\mathcal{C}/X$ has a left adjoint which is just the postcomposition by $f$.
If I add that $\mathcal{C}$ is locally cartesian closed it will be possible to show that it has a right adjoint. But I don't see how this right adjoint behaves on object and morphisms in $\mathcal{C}$.

Thanks for your help.

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  • $\begingroup$ The right adjoint you can think of as being a "dependent product". Have you thought about the simple case where $Y = 1$? $\endgroup$ – Zhen Lin Nov 8 '12 at 12:09
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Let's focus on the case where $C = Set$, since this will give the intuition for other cases.

An object $p: E \to X$ in the category $Set/X$ can be thought of as an $X$-indexed set, where over every $x \in X$ there is a fiber $p^{-1}(x)$. Similarly, a morphism in $Set/X$ from $p: E \to X$ to $q: F \to X$ is a global function $h: E \to F$ which takes fibers to fibers, i.e., is an $X$-indexed family of functions $h_x: p^{-1}(x) \to q^{-1}(x)$.

Now, for simplicity, take $Y = 1$ to be a 1-element set, where $Set/1 \simeq Set$. The pullback functor $X^\ast: Set \to Set/X$ takes a set $A$ to the $X$-indexed set where $A_x = A$ for all $x$. A morphism from $X^\ast A \to (p: E \to X)$ is thus a family of functions $h_x: A \to p^{-1}(x)$. Such families are in natural bijection with functions

$$A \to \prod_{x \in X} p^{-1}(x)$$

where on the right we take the product of all fibers together. That basically gives you the right adjoint, and suggests the usual notation for this functor $\prod_X$. More formally, the set $\prod_x p^{-1}(x)$ is constructed as the set of sections $s: X \to E$ of $p: E \to X$; categorically it is the equalizer of a pair of functions

$$Sect(p) \to E^X \stackrel{\to}{\to} X^X$$

which you can work out yourself; basically it's the solution set to the equation $p \circ s = 1_X$. The effect on morphisms is similarly described: $\prod_X h = \prod_{x \in X} h_x$; formally, it can be constructed by taking advantage of the universal property of equalizers.

The situation for the right adjoint to a pullback $f^\ast: Set/Y \to Set/X$ is only slightly more complicated. Intuitively, the right adjoint $\prod_f$ sends an $X$-indexed set $p: E \to X$ to a $Y$-indexed set where for each $y \in Y$, we have

$$(\prod_f p)_y := \prod_{x \in f^{-1}(y)} p^{-1}(x)$$

i.e., don't take the product of all fibers $p^{-1}(x)$, but only over those where $x$ sits over $y$ via the map $f$. Again, this can be constructed more formally by considering $Y$-indexed sets of sections, where we take families of equalizers which implement section equations; here we consider a $Y$-indexed family of diagrams of the form

$$(f \circ p)^{-1}(y)^{f^{-1}(y)} \stackrel{\to}{\to} f^{-1}(y)^{f^{-1}(y)}$$

More compactly, compute the object of sections of $p$ seen as a morphism from $f \circ p$ to $f$ in the category $Set/Y$.

Once the formal categorical details of that have been squared away, it works the same way for any locally cartesian closed category.

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  • $\begingroup$ I don't understand when you said "A morphism from $X^{\star}A\rightarrow (p:\,E\rightarrow X)$ ...", I think you want to define $X^{\star}(A\rightarrow B)$ ? $\endgroup$ – user2664 Nov 8 '12 at 13:37
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    $\begingroup$ 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 $Set/X$. $\endgroup$ – Todd Trimble Nov 8 '12 at 14:12
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    $\begingroup$ 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 ? $\endgroup$ – user2664 Nov 8 '12 at 17:24
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    $\begingroup$ 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. $\endgroup$ – Todd Trimble Nov 8 '12 at 18:01
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    $\begingroup$ @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 $\prod_{x \in f^{-1}(y)} p^{-1}(x)$. $\endgroup$ – Todd Trimble Jan 9 at 22:12

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