# If $K(G,1)$ and $K(H,1)$ are Eilenberg-MacLane spaces, show that $K(G\ast H,1)$ is also one.

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?

• Your definition and question use the term differently. Your definition defines $K(G,1)$ as a class of spaces, but in your question, $K(G,1)$ is a space. The real problem is $K(G*H,1)$ - it is unclear what that means. Are you saying there exists a $K(G*H,1)$ space if there is a $K(G,1)$ space and a $K(H,1)$ space? May 12, 2022 at 17:00
• Yes. By $K(G\ast H,1)$ I mean a path-connected space, with contractible universal covering, which has fundamental group isomorphic to the free product $G\ast H$. The proposition is: Given that two groups $G$ and $H$ are associated to $K(G,1)$ and $K(H,1)$ spaces respectively, prove that there exists a $K(G\ast H,1)$ space. May 12, 2022 at 17:16
• There is an analogous example in Hatcher (pp.88), which shows that the product $K(G,1)\times K(H,1)$ is what he calls a $K(G\times H,1)$ space. The general goal is to extend this to the free product instead of direct product. May 12, 2022 at 17:21

For any group $$G$$, there exists an $$K(G,1)$$ space, so the answer to your question in the title is definitely "yes".
Regarding your explicit construction, I assume you were following page 91/92 of Hatcher, where this is exact construction is given as Example 1B.10? The only mistake I can see in your argument is that the CW-complex $$K(\Gamma,1)$$ certainly need not be finite, since any of the $$K(G,1)$$ or $$K(H,1)$$ may already be infinite.