What is the inverse of this map? Let $G$ be a group. Suppose $f:G\to G/N$ is the canonical map, where $N=\ker f $ is normal in $G$.
I wonder what $f^{-1}(G/N)$ is explicitly in this context. I see some sources say it is a disjoint union of cosets of $N$. Why is it $\{xN:f(x)\in G/N\}$ rather than $\{x:f(x)\in G/N\}$?
Thanks.
 A: Because the inverse of an element of $G/N$ is a subset of $G$, not an element of $G$; specifically, it’s a coset of $N$. In this context $f$ doesn’t actually have an inverse in the usual sense (unless $N$ is trivial); its ‘inverse’ is actually a set-valued mapping.
You see the same thing in a more familiar setting with the function $f:\Bbb R\to[0,\to):x\mapsto x^2$: it has no inverse in the usual sense, but it does have a set-valued inverse that takes $x\in[0,\to)$ to $\{\sqrt{x},-\sqrt{x}\}$.
In most such cases, if $f:X\to Y$, I prefer to look at $f^{-1}$ as a function from $\wp(Y)$ to $\wp(X)$ that takes $A\subseteq Y$ to $f^{-1}[A]=\{x\in X:f(x)\in A\}$; the square brackets are the signal that this is how $f^{-1}$ is being used. (Similarly, for $A\subseteq X$ one can write $f[A]$ for $\{f(x):x\in A\}$, the square brackets indicating that one is thinking of $f$ as a function from $\wp(X)$ to $\wp(Y)$.)
A: The inverse image of the quotient group $G/N$ is the whole group $G$. 
Also, the two sets you mentioned are equal: $f(xn)=f(x)f(n)=f(x).1$, hence $f(xN)=f(x)$.
