If $K(G,1)$ and $K(H,1)$ are Eilenberg-MacLane spaces, show that $K(G\ast H,1)$ is also one.
Definition. A path-connected space whose fundamental group is isomorphic to a given group $G$ and which has a contractible universal covering space is called a $\mathbf{K(G,1)}$ space.
For my attempt I will try to construct a "graph of groups" and put a structure on it such that the resulting space has the properties of a $K(G,1)$ space.
Let $\Gamma$ be a connected, directed graph with three vertices associated, respectively, to $G,H$, and $I$ where $I$ is the trivial group, and connect $I$ to $G$ and $H$ with edges associated with group homomorphisms $\varphi_e$.
Now we can construct a space $K\Gamma$ by associating $K(G,1)$ with vertex $G$, $K(H,1)$ with vertex $H$, and of course the trivial $K(I,1)$ with vertex $I$. Next I propose filling in a mapping cylinder along each edge $e_1,e_2$ which realizes each homomorphism $\varphi_e$. In particular, we would have two mapping cylinders, one glued at either end to $K(I,1)$ and $K(G,1)$ and the other $K(I,1)$ and $K(H,1)$.
Then it is clear to see that our space $K\Gamma$ is a finite CW-complex, and in particular since $K(I,1)$ is trivial, therefore the edge homomorphisms are injections. It follows from Hatcher Theorem 1B.11 that $K\Gamma$ is a $K(G,1)$ space, and as such it also has contractible universal cover by definition.
A standard application of the Van Kampen Theorem then tells us that $\pi_1 (K\Gamma)\cong \pi_1 (K(G,1)) \ast \pi_1 (K(H,1)) \cong G\ast H$.
Hence $K\Gamma$ is really a $K(G\ast H,1)$ space by construction. $\Box$
Is this correct?