An offbeat question involving Milnor's $K_2$ has come up recently.  Start with an algebraically closed field $F$ (perhaps required to be of characteristic 0).  Let $G$ be a connected, simply connected simple algebraic group over $F$, for instance $\mathrm{SL}_n(F)$, maybe of rank $\neq 1,2$ to be on the safe side.  Then consider a linear group $H \subset \mathrm{GL}(V)$ with $V$ finite dimensional over $F$, together with an epimorphism $\pi:H \rightarrow G$ of abstract groups whose kernel lies in the center of $H$. (EDIT: Further assume that $H$ is equal to its derived group.  I neglected to include this crucial condition in the question originally posed to me.)

> Are these conditions enough to imply that $\pi$ has trivial kernel?

Note that when $G$ is a special linear group of rank at least 2, its abstract universal central extension is the *Steinberg group* (with generators and relations specified by Matsumoto) and the kernel of the resulting map is $K_2(F)$.   Typically this is an infinite group, uncountable if $F$ is uncountable (Milnor).  However, if we add the assumption that $H$ acts *irreducibly* on $V$, then by Schur's Lemma $\pi$ induces a sort of reverse projective representation of $G$ with image equal to the image of $H$ in $\mathrm{PGL}(V)$.  Older work of Steinberg would allow us to lift this projective representation to an ordinary one, thus mapping $G$ onto $H$.   (See sections 6,7 of Steinberg's Yale lectures <a href="http://www.math.ucla.edu/~rst">here</a>.) 

However, it's apparently undesirable in the original question to make any assumption about complete reducibility of the action of $H$.  So I'm unsure what can be said, given only that $H$ is a linear group.