Let $f: G \to H$ be a group homomorphism with $|\ker(f)|=2$. Show that every element of $H$ has no preimage or exactly $2$ preimages under $f$. 
Let $f: G \to H$ be a group homomorphism such that the order of $\ker(f)$ is $2$. Show that any element of $H$ has either no pre-image or exactly two pre-images under $f$.

I have been trying to prove  this - 
since $|\ker f|= 2$, $\ker f$ contains two elements say $k_1, k_2$. 
Now we need to show each element of $H$ contains exactly $2$ or no pre-images. 
Suppose $y \in H$ such that $y$ has $3$ pre-images, say $x_1,x_2,x_3$.  Then $f(x_1)=f(x_2)=f(x_3)$. 
Since $f$ is a homomorphism, $x₁\cdot x_2^{−1},x_2\cdot x_3^{-1},x_3\cdot x_1^{-1}$ belong to $\ker f$.
Without loss of generality suppose $x_2\cdot x_3^{-1} = x_1\cdot x_2^{−1} = k_1$, and this implies $x_1=x_2$. 
Is this ok? 
 A: That's a good try.
I take it that you're using the fact that proving $A\lor B$ is the same as proving $(\lnot A)\to B$. That's fine.
One problem is that you haven't justified why, if at all, we can assume w.l.o.g. that $x_2x_3^{-1}=x_1 x_2^{-1}$. I'm not sure whether $x_1=x_2$ follows from this either. Even if it did, so would $x_2=x_3$, so you wouldn't have shown there is two pre-images.
Here's how to prove the result:
First of all, consider $e_H$. Since $f$ is a homomorphism, we have $f(e_G)=f(e_Ge_G)=f(e_G)f(e_G)\in H$, which gives $f(e_G)=e_H$. Thus $e_H$ has a pre-image in $G$; indeed, since $\lvert\ker f\rvert=2$ and $K=\ker f=\{ k\in G\mid f(k)=e_H\}$, the existence of a pre-image is guaranteed and we know that there is exactly two of them.
Now let $h\in H$. If $h$ has a pre-image, then $h=f(g)$ for some $g\in G$. Define the set of pre-images of $h$ under $f$ as $f^{-1}(h):=\{g'\in G\mid h=f(g')\}$. Recall that $gK=\{gk\in G\mid k\in K\}$. Now, by definition, we have that 
$$f^{-1}(h)=gK,$$ 
but the map
$$\begin{align}
I(g): K&\to gK \\
k&\mapsto gk
\end{align}$$
is clearly a bijection. Hence $f^{-1}(g)$ has the same number of elements as $K$. Thus $g$ has two pre-images.
