I was looking at torsors from nlab, and I got stuck at the trivialization part. I could prove the fact that the set underlying a $G$-torsor $\bar T$ is isomorphic to the set underlying $G$, the so called trivialization of $\bar T$. But then remark 2.13 stumped me. Before I proceed, some context of the notations, definitions and results.
A $G$-torsor $\bar T$ is a set $T$ and a group action $a:G\times T\to T$ such that the map $a\times \pi_2:G\times T\to T\times T$ is an isomorphism (where $\pi_2:G\times T\to T$ is the natural projection on the second coordinate). Then a theorem says that given $T$ is nonempty, then the set underlying $G$ is isomorphic to $T$, and such an isomorphism is called a trivialization. Let $\rho:G\to T$ be a trivialization.
Now onto the problems. They define a 'division' among the elements of $T$ by the map $$d:=\pi_{1} \circ \left( \rho^{-1} \times id \right): T \times T \rightarrow G \times T \rightarrow G$$ which seems kind of fine, yet iffy to me for the following reason. It seems fine because we can assume that the trivialization $\rho$ comes from the choice of some element $t\in T$, we can identity $\rho$ with $t$ itself, and call the map $t$ (sloppy language for sure, but helps me visualize what's going on, maybe). Then, $d(t_1,\ t_2)=\pi_1\circ (t^{-1}\times id)(t_1,\ t_2)=\pi_1(t^{-1}(t_1),t_2)=t^{-1}(t_1)$ which looks like division for sure, but we have forgotten the element $t_2$ altogether. Something fishy is going on, or I am being dumb.
So question 1 : Am I thinking correctly? If yes, where does the element $t_2$ vanish? If not, then what is the correct interpretation of said map?
Moving on, they define $$D:=\rho\circ d:T\times T\to T$$ and call this a 'division structure on $T$'.
Question 2: What is a division structure on a set? Maybe the answer lies in the answer of question 1, but I am really lost.
Finally, the map $D$ looks very odd. Suppose $(t_1,\ t_2)\in T\times T$, then $$D(t_1,\ t_2)=\rho\circ d(t_1,\ t_2)=\rho(\rho^{-1}(t_1))=t_1$$ and thus whatever value I take in the second coordinate, it vanishes into thin air.
Question 3: How does $D$ make sense?
Condensing all the questions into one question, what I really want to ask is : Am I misinterpreting the maps $d$ and $D$, or am I missing something?
Sorry for the long question, but I am really stuck and appreciate any help.
If you've never worked with torsors before it's worth carrying around some examples in your head as you work through all these definitions. Here are two nice ones:
A more straightforward but less categorical definition of a $G$-torsor is that it's a $G$-set $T$ on which $G$ acts freely and transitively; equivalently, such that for each $x, y \in T$ there is a unique $g \in G$ such that $gx = y$. (Take a minute to see why this holds in the two examples I just gave.) This $g$ is the "division" operation.
The definition of division given in the nLab cannot possibly be correct, because it makes no reference to either the action map or the group operation; you are correct that something fishy is going on here. If $\rho : G \to T$ is a trivialization, the condition $gx = y$ becomes $g \rho(h) = \rho(k)$, or equivalently $\rho(gh) = \rho(k)$, so $gh = k$, so $g = h^{-1} k$. So in terms of a trivialization, division is given by
$$d : (x, y) \mapsto \rho^{-1}(x)^{-1} \rho^{-1}(y)$$
which means it can be written as the composite
$$T \times T \xrightarrow{\rho^{-1} \times \rho^{-1}} G \times G \xrightarrow{h^{-1} k} G.$$
A "division structure" is just a map that axiomatizes the properties of the division operation $(h, k) \mapsto h^{-1} k$ on a group $G$.
For context, it's probably worth knowing that the reason the nLab is describing what should be a simple concept so tortuously is that torsors become much more interesting in categories other than $\text{Set}$, and in particular they no longer admit trivializations in general. For example we get nontrivial principal bundles. The set-of-bases example above generalizes to a construction called taking the frame bundle of a vector bundle.