I was reading some introductory theory on G-sets and G-actions and this theorem says that G-sets and G-actions have a very strong equivalence. So far so good. Then it uses a double currying and it's also my first time encountering the currying concept, so I'm confused on the logic behind the double curry. Here's what it has:
$(1)$ The passage from a function of two variables ∶ × ⟶ to a function of one variable ∶ ⟶ $^$, ⟼ (, −) is a standard operation called currying.
Let be a group. The following statements hold:
- Let ( , ) be a ‐set. Then, the function
$(2)$ $\bar{}$ ∶ × ⟶ , (, ) ⟼ ⋅ ∶= $_$ (), for $_\in$ Sym ()
is an action of on .
- Let be a set equipped with an action ∶ × → . Then, the function
$(3)$ $\bar{}$ ∶ ⟶ Sym (), ⟼ ((, −) ∶ ↦ ⋅ )
is a group homomorphism, that is the pair ( , $\bar{}$) is a ‐set.
- Show $\bar{\bar{}}=$ and $\bar{\bar{}}=$.
Here's what is written down as the proof and I add the superscript on the equal to identify the reason why I think the equality holds:
$\bar{\bar{}}_()=^{(1)}\bar{}(,)=^{(2)}_()$
$\bar{\bar{}}(,)=^{(2)}\bar{}_()=^{(1)}(,)$
My confusion is with $(2)$ and the notation. The (2) of is by the definition and the $(2)$ of is due to using the same tactics on the ‐set ( , $\bar{}$). Is the bar in (2) the same as the bar for currying used in (1) and (3)? Due to notation I'd think so, but I can't see why/how.
Thanks in advance!
It's ambiguous, yes. I probably would not have written it that way myself. But it's written clearly enough nonetheless.
There are two different bars. They happen to look exactly alike, so, you have to look carefully at what kind of object that bar is sitting on top of, in order to determine which bar it is.
The first bar is defined in (2), and that's the bar that is put on top of a homomorphism like $\rho$ from $G$ to $\text{Sym}(X)$.
The second bar is defined in (3), and that's the bar that is put on top of an action function like $\mu$ from $G \times X$ to $X$.
For example, in the expression $\bar{\bar \rho}$, $\rho$ is a homomorphism from $G$ to $\text{Sym}(X)$, and so the bar in the sub-expression $\bar\rho$ is the one defined in (2). The effect is that $\bar\rho$ is an action from $G \times X$ to $X$. So, the top level bar in the expression $\bar{\bar\rho}$ is sitting on top of an action function from $G \times X$ to $X$, namely $\bar\rho$, and therefore that top level bar is the one defined in (3). Its effect is that $\bar{\bar\rho}$ is then a homomorphism from $G$ to $\text{Sym}(X)$.
Your goal is therefore to prove that $\rho$ and $\bar{\bar\rho}$ are the exact same homomorphism from $G$ to $\text{Sym}(X)$. I would suggest that the first line of your proof should be "Let $\mu=\bar\rho$. We must show that $\bar\mu=\rho$."