Functions on a cover with the property $f(U \cap V) = f(U)\,f(V)$

Question

Let $X$ be a topological space and $\mathscr{U}$ an open covering of $X$. Is there a special name for functions $f : \mathscr{U} \rightarrow \mathbb{k}$, where $\mathbb{k} = \mathbb{R}$ or $\mathbb{C}$ or some such space endowed with multiplication, satisfying \begin{equation} f(\emptyset) = 0 \qquad\text{and}\qquad f(U \cap V) = f(U)\,f(V) \end{equation} for all $U,V \subset \mathscr{U}, U\neq V?$

Answer 1

First of all I remark that since $\operatorname{dom} f=\mathscr{U}$, we need that the cover $\mathscr{U}$ is closed with respect to finite intersections and contain an empty set. Next, it is easy to see that the set $\mathscr{U}$ endowed with a binary operation $\cap$ is a semigroup (even a semilattice).

Let $U,V\in\mathscr{U}$ and $U\ne V$. If $U\subset V$ then $f(U)=f(U\cap V)=f(U)f(V).$

Let $U, V\in\mathscr{U}$ and $f(U)=0$. If $V=U$ then $f(U\cap V)=f(U)=0$. If $V\ne U$ then $f(U\cap V)=f(U)f(V)=0$. Anyway $f(U\cap V)=0$. Thus $\mathscr{U_0}=f^{-1}(0)$ is an ideal of a semigroup $\scr U$.

The map $f: \mathscr{U}\to \mathbb{k}$ is very similar to a homomorphism from the semilattice $\mathscr{U}$ to a set $\mathbb{k}\ni 0$ endowed with a multiplication. Moreover, since $\mathscr{U}$ is a semilattice, from different properties of $\mathbb{k}$ we can derive different properties of $f$, see below.

From now we assume that $\mathbb{k}$ is a semigroup (that is, the multiplication on $\mathbb{k}$ is associative) generated by a set $f(\scr U)$. Then, since $f(U)f(V)=f(V)f(U)$ for any $U,V\in\mathscr{U}$, the semigroup $k$ is commutative. Let $x\in\Bbb k$ be an arbitrary element. Pick a minimal number $n$ such that there exist elements $U_1,\dots, U_n\in\scr U$ such that $x=f(U_1)\cdots f(U_n)$. If $U_i\ne U_j$ for some distinct $i$ and $j$ then we can replace $f(U_i)f(U_j)$ by $f(U_i\cap U_j)$, contradicting the minimality of $n$. Therefore there exists an element $U\in\scr U$ such that $x=f(U)^n$.

This observation allows us to determine the multiplication of $\Bbb k$ as follows. Let $a,b\in\Bbb k$ be arbitrary elements. Let $a=f(U)^n$, $b=f(V)^m$, so some $U,V\in\scr U$ and natural $m$ and $n$. Then $xy=f(U)^nf(V)^m$. The following cases are possible.

1) $U=V$. Then $ab=f(U)^{n+m}$.

2) $U\ne V$, $U\subset V$. Then $f(U)f(V)=f(U\cap V)=f(U)$, so $f(ab)=f(U)^n=a$. By induction we can also show that $f(U)f(V)^k=f(U)$ for each $k\ge 0$.

3) $U\ne V$, $V\subset U$. Then $f(U)f(V)=f(U\cap V)=f(V)$, so $f(ab)=f(V)^m=b$. By induction we can also show that $f(U)^kf(V)=f(V)$ for each $k\ge 0$.

4) $U\not\subset V$, $V\not\subset U$. Then $f(U)f(V)=f(U\cap V)=f(U\cap V)f(U)= f(U\cap V)f(V)$, so $f(ab)=f(U\cap V)$. This implies that if $k,p,q\ge 0$ and $k\ge 1$ or $p,q\ge 1$ then

$$f(U)^pf(U\cap V)^kf(V)^q=f(U)^{p+k}f(V)^{q+k}=f(U\cap V).$$

In particular, if for an element $W\in\scr U$ there exist elements $U,V\in\scr U$ such that $U\not\subset V$, $V\not\subset U$, and $W=U\cap V$ then $f(W)^2=f(W)$.

Also remark that if $U,V\in\mathscr{U}$ and $U\not\subset V$ then $$f(U)f(V)=f(U\cap V)=f(U\cap (U\cap V))=f(U)f(U\cap V).$$

From now we assume that the semigroup $\mathbb{k}$ has no divisors of zero (that is, if $\mathbb{k}^*=\mathbb{k}\setminus\{0\}$ is a subsemigroup of $\Bbb k$, that is for each $a,b\in \mathbb{k}^*$ we have $ab\in\mathbb{k}^*$). Then the set $f^{-1}(\Bbb k^*)=\scr U_1=\scr U\setminus\scr U_0$ is a subsemigroup of the semigroup $\scr U$. So we have that the family $\scr U_1$ is centered and closed with the respect to taking supsets in $\scr U$.

From now we assume that the semigroup $\mathbb{k}^*$ is cancelleative that is if $ac=bc$ for some $a,b,c\in\mathbb{k}^*$ then $a=b$. In particual, this condition holds when $\mathbb{k}$ is $\mathbb{R}$ or $\mathbb{C}$.

Let $U,V\in\scr U_1$ be two distinct elements. Then $U\cap V\ne U$ or $U\cap V\ne V$, so without loss of generality we may suppose that $U\subset V$. Then $f(U)=f(U\cap V)=f(U)f(V)$. Then $xf(U)=xf(U)f(V)$ for each $x\in\Bbb k^*$, so $x=xf(V)$ for each $x\in\Bbb k^*$, that is $f(V)$ is the unit $1$ of the semigroup $\Bbb k$. That is, among any two distinct elements $U,V\in\scr U_1$ either $f(U)=1$ or $f(V)=1$.

If $f(U)=1$ for each $U\in\scr U_1$ then we already completely described the map $f$. Conversely, given an arbitrary partition of $\scr U$ into an ideal $\scr U_0$ and a semilattice $\scr U_1$, it is easy to check that a map $f:\scr U\to\Bbb k$ such that $f(U)=0$, if $U\in\scr U_0$, and $f(U)=1$, if $U\in\scr U_1$, is a homomorphism.

Otherwise let $Z$ be the unique element of $\scr U_1$ such that $f(Z)\ne 1$. Then $f(Z\cap U)=1$ for each $U\in \scr U_1$ different from $Z$, that is $Z\cap U\ne Z$, which means that $Z$ is a maximal element of $\scr U_1$. Conversely, given an arbitrary partition of $\scr U$ into an ideal $\scr U_0$ and a semilattice $\scr U_1$, containing a maximal element $Z$, it is easy to check that a map $f:\scr U_1\to\Bbb k$ such that $f(U)=0$, if $U\in\scr U_0$, and $f(U)=1$, if $U\in{\scr U_1}\setminus\{Z\}$, and $f(Z)=z$, where $z$ is a arbitrary non-zero element of $\Bbb k$, satisfies $f(U\cap V)= f(U)f(V)$ for each distinct $U,V\in\scr U$.