Is it true that $K(G \ast H, 1) = K(G,1)\vee K(H,1) $?

Question

I know that, unlike the case of the fundamental group (where $\pi_1(X \vee Y) \cong \pi_1(X)\ast \pi_1(Y)$ at least for CW complexes, which are the spaces I care about for the purpose of this question), there is no straightforward formula for the higher homotopy groups of a wedge sum of spaces.

However, I was wondering whether in the case that $\pi_k(X) = \pi_k(Y) = 0$ for all $k \geq 2$ one can deduce that also $\pi_k(X\vee Y) = 0$ for all $k \geq 2$. If this is true, how can one show this? I don't necessarily need a full solution, a hint to get started would be enough.

Thanks in advance!

Answer 1

Yes, this is true (at least, for nice spaces like CW complexes). As a sketch of a proof, observe that you can construct a universal cover for $X\vee Y$ by gluing together copies of a universal cover of $X$ and a universal cover of $Y$ at preimages of the basepoint. (You can in fact give a combinatorial description of how the copies are glued together which parallels the description of $\pi_1(X)*\pi_1(Y)$ in terms of reduced words.) Since the universal covers of $X$ and $Y$ are contractible, it follows easily that the universal cover of $X\vee Y$ has trivial homology in dimensions greater than $1$, and so it is contractible.

(There are other ways you could do the last step besides using homology; for instance, you could contract the copies of the universal covers of $X$ and $Y$ to a point one by one which does not change the homotopy type since they are contractible; here you use the combinatorics of how they are attached to ensure that no copy can get glued together with itself when you contract other copies.)

Answer 2

Yes, see the question and comments at overflow.

The answer there explains the map $$BG\vee BH \to B(G*H)$$ is a weak equivalence more generally for topological groups, or even topological monoids.