# Intuitive interpretation of “co-bundles” i.e objects of $X\downarrow \mathsf{Set}$?

Objects of a slice category $\mathsf{Set}\downarrow X$ are just set functions into $X$, and they can be identified with the partition they induce on their domain with their fibers. Arrows in the slice category are then just collections of functions between the corresponding fibers at each point. This is the intuitive interpretation I know of bundles - arrows between bundles over a space $X$ are a system of arrows between $X$-indexed fibers. For instance, if we take $f,g:Y\rightarrow X$ then an arrow $h:f\rightarrow g$ transforms one $X$-indexed partition of $Y$ into another.

Now what about $X\downarrow\mathsf{Set}$? Let's make the same identification between functions and the partitions they induce on the domain. I was hoping that maybe arrows here could "reshape" of the partitions, but this time, allowing different "indexing spaces": say an arrow from $f:X\rightarrow Y$ to $g:X\rightarrow Z$ would transform a $Y$-indexed partition of $X$ via the fibers of $f$ into a $Z$-indexed partition via the fibers of $g$. Problem is I don't see how to realize this arrow $h$ as a bunch of components.

So how should one think of $X\downarrow\mathsf{Set}$?

Where do situations in which we want to look at a different "indexing space" arise?

## 1 Answer

I don't know if this is helpful for you, but you can think of $f : X \to Y$ as a labeling of the elements of $X$ by elements of $Y$. For example $f : \{a,b,c,d,e\} \to \{0,1,2\}$, $f(a) = f(c) = 0$, $f(b) = f(e) = 1$, $f(d) = 2$ means $a,c$ are labelled by $0$, $b,e$ are labelled by $1$, and $d$ is labelled by $2$. And then a morphism in $X \downarrow \mathsf{Set}$ is a map between labels: for example if $g : \{0,1,2\} \to \{0,1\}$, $g(0)=g(2) = 0$ and $g(1) = 1$, then all elements that were labelled by $0$ and $2$ are now labelled by $0$ and elements that were labelled by $1$ are now labelled by $1$. (Note that if $(f : X \to Y) \in X \downarrow \mathsf{Set}$ and $g : Y \to Z$ is some map, then there's a unique $(h : X \to Z) \in X \downarrow \mathsf{Set}$ such that $g_*(f) = h$.)

• Not sure if this is what you're looking for. I always find such questions about "intuition" not very well-defined. – Najib Idrissi Oct 18 '15 at 8:38
• hey @Najib , wanna explain this whole subscript * thing for those who do not know category theory; thanks! – Musa Al-hassy Oct 20 '15 at 16:36
• @MusaAl-hassy $g_*$ is the morphism in $(X \downarrow \mathsf{Set})$ from $(f : X \to Y)$ to $(h : X \to Z)$ induced by $g$, i.e. $g_*(f) = g \circ f$. – Najib Idrissi Oct 24 '15 at 16:33
• Thanks for making it clear ^_^ – Musa Al-hassy Oct 24 '15 at 23:43