Pullbacks classified by morphisms of fundamental groups? Let $G$ be a connected compact Lie group, $H$ a connected semisimple compact Lie group with $\tilde{H}$ its universal cover which is compact and simply connected. Let $\phi: G\to H$, and $\pi: \tilde{H}\to H$ and let $G'$ be the corresponding pullback, with $p: G'\to G$ and $\phi': G'\to \tilde{H}$ being the projections, which is the fibered product of $G$ and $\tilde{H}$ over $H$.
In short we consider the commutative diagram
$$\begin{array}{cc}
G'&\stackrel{\phi'}{\longrightarrow}&\tilde{H}\\
\downarrow{p} &&\downarrow{\pi}\\
G&\stackrel{\phi}{\longrightarrow}&H
\end{array}$$
I would like to understand the relationships between the different coverings $p: G'\to G$ and the morphisms $\pi_1(G)\to \pi_1(H)$ of fundamental groups induced by $\phi$. Does the latter classify the former? That is to say is there a bijection between the morphisms of fundamental groups $\pi_1(G)\to\pi_1(H)$ and the coverings $p:G'\to G$? Is the compactness hypothesis important?
Edit1: Here are my thoughts so far
If $\phi_0:G\to H$ and $\phi_1:G\to H$ are homotopic, they are actually conjugate because $G$ is compact. Therefore we get isomorphic pullbacks. But I am not familiar with algebraic topology, so perhaps the use of this result is too strong for the question asked. And I do not know if we actually get a bijection between morphisms of fundamental groups and homotopy class of function from $G$ to $H$ in that case.
 A: Here is the basic facts of the covering theory. In what follows, I will consider only path-connected semilocally-simply-connected topological spaces. (The latter means that for each point $x\in X$ and a neighborhood 
$U$ of $x$, there is a smaller neighborhood $V$ of $x$ such that the natural map $\pi_1(V,x)\to \pi_1(U,x)$ is trivial.)  
Let $p: X\to Y$ be a covering map  topological spaces. It induces a monomorphism of fundamental groups $p_*: \pi_1(X,x)\to G=\pi_1(Y, y)$, $y=p(x)$. The image group $p_*(\pi_1(X,x))$ is the defining subgroup of the covering $p$. Conversely, given any subgroup $H\le G$ there is a unique (up to an isomorphism) covering map $p: X\to Y$ with the defining group $H$. 
Moreover, given a map $f: (Z,z)\to (Y,y)$ and a covering map $X\to Y$ with the defining subgroup $H\le G$, the map $f$ lifts to $X$ if and only if $f_*(\pi_1(Z,z))\le H$. 
All this can be found in Hatcher or Massey or any other half-way decent textbook on algebraic topology. 
One can specialize this setting as follows. Suppose that we have a map $f: (Y,y)\to (Y_1,y_1)$ of (pointed) path connected spaces. Then the kernel $H=ker(\pi_1(h_*))$ is a subgroup of $G= \pi_1(Y,y)$. The trivial subgroup of $\pi_1(Y_1,y_1)$ defines the universal covering $p_1: \tilde Y_1\to Y_1$. The subgroup $H$ defines a certain covering $p: Y'\to Y$. In view of the lifting property mentioned above, the map $f$ lifts to a map $\tilde f: Y'\to \tilde Y_1$ such that $p_1\circ \tilde f= p\circ f$. In other words, this is your fiber product diagram. 
In particular, the map $\phi: \pi_1(Y,y)\to \pi_1(Y_1, y_1)$ uniquely determines the pull-back covering map $Y'\to Y$. Specializing further to the case of topological groups $Y, Y_1$ (say, Lie groups) as in your question, we obtain an injective map from $Hom(\pi_1(Y,y), \pi_1(Y_1, y_1))$ to the set of isomorphism classes of covering maps $Y'\to Y$. This map is, of course, not bijective even though $Y, Y_1$ have abelian fundamental groups (as topological groups do): Given two abelian groups $A, A_1$, not every subgroup $B< A$ is realized as the kernel of a homomorphism $A\to A_1$. 
Even worse, even if there is a homomorphism $\phi: \pi_1(Y,y)\to \pi_1(Y_1, y_1)$ whose kernel is the given subgroup $N< \pi_1(Y,y)$, there is no reason to expect that there is a continuous group homomorphism $Y\to Y_1$ which induces $\phi$. Examples are easy to construct and I leave this to you.  
