Transitivity axiom for a Grothendieck Topology

Question

I am currently trying to define a Grothendieck Topology on the category Prob which consists of finite probability spaces with measure preserving maps between them.

I declared the covering sieves of an object $U$ to be the sieves generated the projections i.e. the covering sieves $S_V$ are each generated by set $\Pi_V=\{\pi_U:U\times V\to U\}$ where $V$ is another finite probability space. I have already proved that the maximal sieve is covering by considering $V$ to be the terminal object and the stability action follows pretty easily too.

However for the transitivity axiom, I have tried a lot of strategies but none of them work. Note that the inclusion maps $i_U:U\to U\times V$ are not measure preserving so that fact cannot be used.

I also realise that this might not even be a Grothendieck Topology so a counterexample would also be helpful.

Answer 1

I will make some assumptions about the definitions you are using, let me know if you are using a different approach.

A map $f: X \to Y$ is measure-preserving if $\mu(f^{-1}(A)) = \mu(A)$ for each subset $A \subseteq Y$. It is enough to check this for singletons $A$. This means in particular that $\mu(Y)=\mu(X)$ whenever there is a measure-preserving map $X \to Y$.

If you want the projection map $U \times V \to U$ to be measure-preserving, then the natural measure on $U \times V$ seems to be the one with $$\mu(\{(u,v)\})~=~\frac{\mu(\{v\})\mu(\{u\})}{\mu(V)}.$$ If $\mu(V)=\mu(U)$ then this definition is symmetrical, but even then $U\times V$ does not satisfy the universal property of the product in $\mathbf{Prob}$.

This is all very tricky, so I don't think the covering sieves that you mention define a Grothendieck topology (the stability axiom fails).

For example, consider a one-element set $\{\ast\}$ with measure $1$, and a two-element set $U = \{a,b\}$ where $a$ and $b$ have both measure $\tfrac{1}{2}$. Then $U \to \{\ast\}$ generates one of your covering sieves $S$. We can view $S$ as the family of probability spaces of measure $1$ that can be split in two parts of equal measure.

Another measure-preserving map would be $f: V \to \{\ast\}$ with $V = \{x,y\}$, where $x$ has measure $\tfrac{2}{3}$ and $y$ has measure $\tfrac{1}{3}$. Because $V$ can not be split into two equal parts, $f \notin S$.

We compute: $$f^{-1}(S) = \{g: T \to V ~\text{such that}~ fg \in S \}$$

In other words, $f^{-1}(S)$ consists of the measure-preserving maps $g: T \to V$ such that $T$ can be split into two equal parts. For example, we can take $T = \{u,v,w\}$, where $u$ has measure $\tfrac{1}{2}$, $v$ has measure $\tfrac{1}{6}$ and $w$ has measure $\tfrac{1}{3}$. This can be split in half by taking $\{u\}$ and $\{v,w\}$. We can define $g$ via $g(u)=g(v)=x$, $g(w)=y$.

Using that $g$ is contained in $f^{-1}(S)$, you can deduce that $f^{-1}(S)$ is not of the form of one of the covering sieves that you mentioned, so the stability axiom fails.

So constructing an interesting Grothendieck topology on $\mathbf{Prob}$ will be tricky (but interesting), I hope the above helps in finding the right starting point.