Universal property of products on actions of sets

Question

This is from Lawvere's Conceptual Mathematics.

Suppose there is a binary operation $\alpha\colon A \times A \to A$ and a point $a_0\colon \mathbf{1} \to A$ and we restrict the notion of an action of $A$ on $X$ to those actions compatible with $\alpha$ and $a_0$ in the sense that the action $\alpha$ corresponds to composition of endomaps of $X$ and $a_0$ acts on $1_X$, i.e.:

1. $\xi(\alpha(a,b), x) = \xi(a, \xi(b, c))$ for all a, b, x.

2. $\xi(a_0, x) = x$ for all x.

The problem is to express these equations as equations between maps:

$A \times A \times X \substack{\rightarrow \\[-1em] \rightarrow} X, X \substack{\rightarrow \\[-1em] \rightarrow} X$

constructed by using $\xi$ and the universal mapping property of products.

I'm having some issues with this problem

1. I have no idea what the stacked arrows mean, this notation hasn't been introduced in the text. I thought it could mean treating $(A \times A) \times X$ and $A \times (A \times X)$ as two different maps, but then how does $X \substack{\rightarrow \\[-1em] \rightarrow} X$ make sense?

2. It is clear to me how $\xi(\alpha(a,b), x) = \xi(a, \xi(b, c))$ correspond to a map from $A \times A \times X \to X$, but I don't know how to incorporate the universal property of products (and then no idea on the second part. It seems like the map should be $\mathbf{1} \substack{a_0 \\[-0.5em] \to} A \to A \times X \to X$. Earlier in the section, they say for each point of $A$, $\alpha$ gives rise to an endomap of $X$... i.e.

$X \substack{\langle \bar{a},1_X\rangle \\[-0.5em] \to} \to A \times X \to X$.

But I still don't see how we can treat that as two maps from $X$ to $X$

My thinking: For the first part, we have the commutative diagram

$$\require{AMScd} \begin{CD} A \times A \times X @>{1_A,\xi}>> A \times X\\ @V{\alpha,1_X}VV @V{\xi}VV \\ A \times X @>{\xi}>> X \end{CD}$$

I thought we would get the equality of the first equation directly from this, so I'm not sure how the universal property of products comes into play.

The only part that looks usable is that we have the maps $(1_A,\xi)$ and ($\alpha,1_X)$... am I supposed to use those as the projection maps then I would get a unique function into $A \times A \times X$? That doesn't feel right to me, they don't feel like the projection functions we normally have.

Any help is appreciated.

Answer 1

From $A\times A\times X$ to $A\times X$ you have two maps, that can be described by their projection $A\times A\times X\to A$ and $A\times A\times X\to X$: this description is what is meant by "universal property of the product", and I think no more is meant.

One of these maps consists in doing $\alpha$ on $A\times A$ and nothing on $X$; this is represented by saying "on the $A$ factor, do $\alpha\circ\pi_{A\times A}$, and on the $X$ factor do $\pi_X$"

The other one consists in doing $\xi$ on $A\times X$ and nothing on the first $A$; this is represented by saying "on the $A$ factor, do $\pi_A$, on the $X$ factor do $\xi\circ \pi_{A\times X}$" (the names of my projection aren't super clear here, but hopefully you get what I mean)

Then by postcomposing these two maps by $\xi$ you get two maps $A\times A\times X\to X$, and if you look at it carefully, 1. tells you exactly that these two maps are equal.

For the second one, you can do a similar analysis on 2. and note that you have a map $X\to 1\times X$ which you can follow up by doing nothing on $X$ and doing $a_0$ on $1$, which you can again express by the universal property of the product to get a map $1\times X\to A\times X$, and then you can follow that up with $\xi$