Show that a cokernel of $(u,v)$ exists(abstract algebra problem). A homomorphism between groups, $f:H\to K$ is said to be an epimorphism if and only if f is surjective
Let G,H be groups and $u,v: G \to H$ be homomorphism of groups. We say a pair $(C,\pi)$ is a corkernel of the pair $(u,v)$ provided $\pi:H\to C $ is an homomorphism of groups satisfying $\pi \circ u=\pi\circ v$ and, for any group L and homomorphism $y:H\to L$ such that $y \circ u=y\circ v$, there is an unique homomorphism $y':C\to L$ such that $y=y'\circ \pi$. show that a cokernel of (u,v) exists and $\pi$ is an epimorphism.
I have spent a long time in this question, but still have no idea how to solve it. Can someone give me some help?
 A: A good strategy for solving a problem like this is to assume you already know the answer and see what that would tell you about what the solution has to look like.  In this case, suppose you have a cokernel $\pi:H\to C$ of $(u,v)$ such that $\pi$ is an epimorphism.  Then $\pi$ has some kernel $K$, and $C$ is actually isomorphic to the quotient $H/K$ (with $\pi$ then being the quotient map).  So instead of trying to construct $H$ and $\pi$ from scratch, you can try to figure out what the kernel $K$ should be.
So, what should $K$ be?  Well, $\pi$ is supposed to satisfy $\pi\circ u=\pi\circ v$, so $\pi(u(g))=\pi(v(g))$ for all $g\in G$.  This is equivalent to saying $\pi(u(g)v(g)^{-1})=1$ for all $g\in G$, or $u(g)v(g)^{-1}\in K$ for all $g\in G$.  Since $K$ has to be a normal subgroup, it must contain the normal subgroup generated by the set of all elements of the form $u(g)v(g)^{-1}$.  On the other hand, knowing that $u(g)v(g)^{-1}\in K$ already guarantees us that $\pi\circ u=\pi\circ v$, so there's not really any reason to expect $K$ to contain any other elements.
So let's try defining $K$ to be the normal subgroup generated by $\{u(g)v(g)^{-1}:g\in G\}$, set $C=H/K$, and $\pi:H\to C$ the quotient map.  Can you prove that this is a cokernel? (You will want to use the universal property of quotients, namely that if $y:H\to C$ is a homomorphism such that $K\subseteq \ker y$, then there is a unique homomorphism $y':H/K\to C$ such that $y=y'\circ \pi$.)
