# A Function as a collection of Arrows

Normally you define a function to be a map on a set. But how about defining a function, in Category Theory, as a collection of arrows?

Take this cateogry

Objects: true, false.

Arrows:

true -> true
false -> false
true -> false
false -> true


Here I had to specify FOUR arrows, but I could summarize this collection of arrows as TWO functions:

id(x) = x    // replaces the first 2
not(x) = !x  //replaces the second 2


If you want to maintain "arrowhood", which is more categoric, you could write a meta-arrow

idArrow(x)  = x -> x
notArrow(x) = x -> not(x)


Either way you compress information.

## Question

Is there a strandard way of expressing these meta-arrows? Is this even part of the theory?

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So what is the question? – JoeyBF Jun 12 '14 at 17:15

You have a set $S$ being acted on by a group $G$, and you are constructing the corresponding action groupoid. This is a standard construction. By the way, you haven't fully specified a category yet, since you haven't specified how your arrows compose.

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They compose as you would expect (b -> c) * (a -> b) = (a -> c), just that I don't want to write out all the combinations. What I am saying is that instead of having to specify each arrow, and each composition rule, this is could be abstracted out by specifying its form without having to use sets and functions. – Cristian Garcia Jun 12 '14 at 18:08
@Cristian: that's fine. Then your category is an action groupoid. – Qiaochu Yuan Jun 12 '14 at 18:10

Given any functor $F : X \to Y$, you can construct the category $Z$ defined by:

Start with the disjoint union of $X$ and $Y$.

For each object $x \in X$ and arrow $F(x) \xrightarrow{f} y$, add an arrow $x \xrightarrow{(x,f)} y$.

Composition is

• $g (x,f) = (X,fg)$ if $g : y \to y'$
• $(x, f) g = (x', f F(g))$ if $g : x'\to x$

This construction has a name, but I forget what it's called. (bridge, maybe?)

Anyways, in the special case that $X$ and $Y$ are sets (i.e. all morphisms are identity morphisms), then $F$ is a function, and this gives a way to view $F$ as a bunch of arrows.

The motivation behind the choice of new arrows and composition law above is that we want to formally add arrows $x \to F(x)$, but have a commutative diagram

$$\begin{matrix} x &\xrightarrow{f}& x' \\ \downarrow & & \downarrow \\ F(x) &\xrightarrow{F(f)}& F(x') \end{matrix}$$

All new arrows are thus an arrow in $X$ followed by one from $X$ to $Y$ followed by an arrow from $Y$. But using the diagram, we can rewrite every such arrow as simply one from $X$ to $Y$ followed by one in $Y$, and I've used that normalization in choosing the representation. We could normalize the other way as well (make the thing in $Y$ the identity arrow)

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If memory serves, this is indeed called a bridge. – goblin Jun 12 '14 at 18:39
Having a hard time picturing it. How would you apply it to my example? – Cristian Garcia Jun 12 '14 at 18:44
It is also called the collage of the profunctor $\hom(F-,-):X^{op}\times Y\to \mathcal Set$. – Berci Jun 12 '14 at 21:03
I would call it the mapping cylinder. – Zhen Lin Jun 13 '14 at 0:57