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Stream: learning: questions

Topic: Dependent Sum vs Composition

John Onstead (Dec 04 2024 at 19:23):

Real quick question here. As we all know, given a morphism $f: A \to B$ in a category $C$ with pullbacks, there exists a base change functor $f^*: C/B \to C/A$. It has two adjoints: its left adjoint is the dependent sum and its right adjoint is the dependent product construction.

However, recently I came across this article by the nlab which might conflict with this information. It says that the inverse image and direct image (composition) functors between slice categories form an adjoint pair. If the inverse image functor is the base change functor, then this is a direct contradiction since we know from above that the left adjoint to the base change is the dependent sum, not the composition functor. The only possible way this makes sense is if the "inverse image" functor as described on that nlab page is not, in fact, the same as the base change functor. But both are given by pullbacks, so what is the difference?

Mike Shulman (Dec 04 2024 at 19:38):

Dependent sum is the same as composition, from a categorical point of view.

Mike Shulman (Dec 04 2024 at 19:38):

(BTW, dependent product may not exist in a category with pullbacks.)

John Onstead (Dec 04 2024 at 19:46):

Mike Shulman said:

Dependent sum is the same as composition, from a categorical point of view.

I see, thanks!

John Onstead (Dec 04 2024 at 19:46):

I guess it's one of those "concepts with attitudes"!

John Baez (Dec 04 2024 at 21:31):

I think it goes like this - someone will correct me if I'm wrong. Say you have a set over $A$, like

$g \colon X \to A$

This is an object in the slice category $\mathsf{Set}/A$. The set of all elements of $X$ mapping to $a \in A$ is called the fiber over $a$.

Now say we have a function $f: A \to B$.

The composite $f \circ g: X \to B$ is now a set over $B$. The fiber of $f \circ g$ over any point $b \in B$ is a sum, or disjoint union, of fibers of $f$. Which fibers? The fibers over points $a$ with $f(a) = b$.

This construction is called a dependent sum, and it gives the left adjoint @John Onstead is talking about.

John Onstead (Dec 04 2024 at 21:46):

John Baez said:

The composite $f \circ g: X \to B$ is now a set over $B$. The fiber of $f \circ g$ over any point $b \in B$ is a sum, or disjoint union, of fibers of $f$. Which fibers? The fibers over points $a$ with $f(a) = b$.

That's a really cool perspective and justification of the "sum" in "dependent sum". Thanks!

John Baez (Dec 05 2024 at 00:04):

Thanks! I believe the dependent product works similarly, but now the fiber over $b \in B$ is the product of all fibers over points $a \in A$ with $f(a) = b$.