What is the meaning of the parentheses in $\phi^{-1}\left[\{\phi(g)\}\right]=gH=Hg$? I am studying homomorphisms is groups and i saw a theorem saying:
For $g$ in a group $G$, the cosets $gH$ and $Hg$ are the same, and collapsed onto the single element $\phi(g)$ by $\phi$. That is, $\phi^{-1}\left[\{\phi(g)\}\right]=gH=Hg$.
I am confused about the parentheses here. I think $\{\phi(g)\}$ denotes the set, formed by the elements of the mapping, but i am not sure what $[\ ]$ represent. Can anyone tell how should i interpret those parentheses? Or explain the theorem in other words?
Thank you

Some use the notation $f^{-1}[X]$ to denote the preimage of a set $X$ under a map $f:A\to B\supseteq X$: apparently the brackets prevent ambiguity in set theory, as one of my instructors explained to me.
Here is an example justifying the need. Suppose we build the natural numbers $\bf N$ as ordinals:


*

*$0:=\{\}$

*$1:=\{0\}=\{\{\}\}$

*$2:=\{0,1\}=\{\{\},\{\{\}\}\}$

*$3:=\{0,1,2\}=\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}$

*$4:=\{0,1,2,3\}=\cdots$

*$\cdots$

*$n:=\{{\rm ordinals}<n\}$

*$\cdots$


Now let $f:\bf N\to N$ be an injective function. Suppose $f(0)=1$ and $f(1)=0$. Then the inverse function $f^{-1}$ applied to $0$ is $1$, and applied to $1$ is $0$. That is,


*

*$f^{-1}(0)=1$,

*$f^{-1}(1)=0$.


However, the preimages under $f$ of these as sets of ordinals are


*

*$f^{-1}[0]=f^{-1}[\varnothing]=\varnothing=0$,

*$f^{-1}[1]=f^{-1}[\{0\}]=\{f^{-1}(0)\}=\{1\}$.


But $0,1,\{1\}$ are all distinct sets, so $f^{-1}(0)\ne f^{-1}[0]$ and $f^{-1}(1)\ne f^{-1}[1]$. So these sorts of clashes can occur in situations where you have set-valued functions of sets and thus need to differentiate between applying inverse functions and preimages. It's worth mentioning that the ambiguity may arise in the forward direction with plain-old images just as well, so this is just a general feature of set-valued mappings of sets: $f(x)$ and $\{f(u):u\in x\}$ need not be the same set.
However elsewhere in mathematics it is generally unlikely one happens across such issues, so most don't need to bother with brackets as no real ambiguity will arise.
