Is this proof of the group theory question correct? 
I am working thorough a past question paper on Abstract Algebra (3rd year) and I am uncertain if my proof for ii) is correct or missing anything because it feels quite short.
This is my proof:  
$f(x)=y$
$e_Hf(x)=e_Hy  $
$f(k)f(x)=y$       ...since $f(k)=e $
$f(kx)=y$    ...using homomorphism property
$y=f(kx)$
$Thus f^{-1}(y)=kx$
Is this correct? Thanks a lot.
 A: This is (almost) correct as far as it goes, but it's only half the problem. You have (almost) proved that if $k \in \ker f$ then $kx$ is a member of $f^{-1}(y)$. Your proof isn't quite right at the end when you say "$f^{-1}(y) = kx$" because $f^{-1}(y)$ is a subset of $G$, not an element of $G$. You should say instead
$$
kx \in f^{-1}(y) .
$$
Then you'll be half done. You still must show the reverse inclusion: if $z \in f^{-1}(y)$ then $z = kx$ for some $k$ in the kernel.
A: Your answer is incorrect.
$f^{-1}(y)$ is a set, which is called the preimage of $y$ under $f$.  
Let $a\in f^{-1}(y)$. Then $f(a)=y=f(x)$.
Hence $f(ax^{-1})=1$, which means that $ax^{-1}\in \ker f$.
So $ax^{-1}=k$ for some $k\in \ker f$ and hence $a=kx$.
Thus $f^{-1}(y) \subseteq \{kx \;|\; k\in \ker f\}$.  
Let $k\in \ker f$. Then $f(kx)=f(k)f(x)=1f(x)=f(x)=y$.
Hence $kx\in f^{-1}(y)$.
Thus $\{kx \;|\; k\in \ker f\}\subseteq f^{-1}(y)$.
So we conclude that $f^{-1}(y) = \{kx \;|\; k\in \ker f\}$
A: The symbol $f^{-1}$ causes people a lot of trouble.
It's not a mapping $f(G) \to G$, but rather a mapping:
$f^{-1}: \mathcal{P}(f(G)) \to \mathcal{P}(G)$ (where $\mathcal{P}(S)$ means the power set (or set of all subsets) of $S$).
Technically, we should write: $f^{-1}(\{y\})$ instead of $f^{-1}(y)$, to indicate we are looking for the pre-image of a singleton set, but people are lazy, and this then makes people think that $f^{-1}$ is a function with argument $y$.
To be clear, then:
$f^{-1}(\{y\}) = \{x \in G: f(x) = y\}$.
If $f$ is not one-to-one (which is often the case), $f^{-1}(\{y\})$ can have more than one element, that is, we might have $x_1 \neq x_2$ with $f(x_1) = f(x_2)$, so that the set $\{x_1,x_2\}$ is a subset of $f^{-1}(\{y\})$.
Now if $K = f^{-1}(\{e_H\}$) (note that $K$ here is a set, not an element of $G$), it's pretty clear that the coset $Kx = \{kx: k \in K\}$ gets mapped to simply $\{f(x)\}$ under $f$, via the homomorphism property of $f$:
Take $k \in K$, then for $kx \in Kx$ we have $f(kx) = f(k)f(x) = e_Hf(x) = f(x)$.
Thus if $y = f(x)$, certainly $Kx \subseteq f^{-1}(\{y\})$.
So what $f$ does, is map cosets $Kx$ to singletons $\{x\}$. You can think of this as a kind of "shrinking" where all of $K$ gets shrunk down to the identity of $H$.
The question now is, does $f$ shrink down "more", it is possible that $f^{-1}(\{y\})$ is even "bigger" than just $Kx$?
Put another way, if we have $g \in f^{-1}(\{y\})$, that is $f(g) = y = f(x)$, is it even possible that $g \not\in Kx$?
Again, the homomorphism property of $f$ helps us out:
If $f(g) = f(x)$, then $f(g)[f(x)]^{-1} = e_H$, so that:
$f(g)f(x^{-1}) = e_H$, and thus: $f(gx^{-1}) = e_H$, that is: $gx^{-1} \in K$.
Thus $g = ge_G = g(x^{-1}x) = (gx^{-1)}x \in Kx$, so no, it's not possible.
So we can safely conclude $f^{-1}(\{y\}) = Kx$.
