Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $S$ be the set of maps and $\phi,\psi \in S$. Let $x,y \in \mathbb{Z}$. Suppose that $\phi(x) * \psi(y) = \nu(xy)$ for some $\nu \in S$. Then what would you call such a system of maps?

If that was easy, then what would you call such as system if all of the maps are only partially defined on $\mathbb{Z}$?

Note: $xy$ is the integer product, and $*$ is some binary operator on $\{\phi(x) : x \in \mathbb{Z}, \phi\in S\}$

Grazie.

share|improve this question
    
You have a set of maps, but what is the codomain? –  M Turgeon Mar 23 '12 at 17:45
    
Not known yet. If it helps, let the codomain be the same for each. But it might turn out to be different for each map, so $*$ might be considered a collection of binary operators. It's up in the air right now. –  Enjoys Math Mar 23 '12 at 17:51

1 Answer 1

up vote 1 down vote accepted

The triple $(\phi:\mathbb Z\to X,\psi:\mathbb Z\to X,\nu:\mathbb Z\to Y)$ describes a morphism from $(\cdot):\mathbb Z^2\to \mathbb Z$ to $(*):X^2\to Y$ in the arrow category over $\mathbf{Set}$. It probably doesn't have any short name, if you don't want to call it just a "commuting square".

If the maps are partial, it's the same thing, but over the category of sets and partial functions.

share|improve this answer
    
How can one draw the commuting square for this? –  Enjoys Math Mar 23 '12 at 18:01
    
I used copier paper and a .5 mm mechanical pencil, but here's a feeble attempt with MathJax:$$\begin{array}{ccc}\mathbb Z^2 & \rightarrow^{\phi\times\psi} & X^2 \\ \downarrow \cdot && \downarrow * \\ \mathbb Z & \rightarrow^{\nu} & Y\end{array}$$ –  Henning Makholm Mar 23 '12 at 18:06
    
Cool! What's the syntax to embed that here? –  Enjoys Math Mar 23 '12 at 18:10
    
It's just a \begin{array}...\end{array} with arrows in some of the cells. You can right-click on any rendered formula on the site to see the source markup. (Though for some browsers you have to find a way to dismiss the browser right-click menu that appears on top of the MathJax menu). –  Henning Makholm Mar 23 '12 at 18:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.