preimages under group homomorphism Let $\varphi : G \rightarrow H$ be a group homomorphism with kernel $K$ and let $a,b \in \varphi(G)$. Let $X = \varphi^{-1}(a)$ and $Y = \varphi^{-1}(b)$. Fix $u \in X$. Let $Z=XY$. Prove that for every $w \in Z$ that there exists $v \in Y$ such that $uv=w$. This is Dummit and Foote exercise 3.1.2.
My attempt:
Suppose we assume that $v = u^{-1}w$
I try to show that $v \in Y$
$\varphi(v) = \varphi(u^{-1})\varphi(w) = a^{-1}\varphi(w)$
If I could somehow show $\varphi(w) = ab$, then $\varphi(v) = b$ so that $v \in Y$ but I think I am going in circles.

Since $w\in Z$, you know that $w=xy$, for some $x\in X$ and $y\in Y$.
By definition, $\varphi(x)=a$ and $\varphi(y)=b$.
Also $\varphi(u)=a$, which implies $u^{-1}x\in\ker\varphi$.
Then
$$
w=xy=u(u^{-1}xy)
$$
Can you finish?
