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

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.

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.)
• You can finid a nice illustrated example at math3ma.com/blog/a-recipe-for-the-universal-cover-of-x-y. It's also very instructive to just think about the usual tree depiction of the universal cover of $S^1\vee S^1$ as a union of horizontal and vertical segments which are covers of the two circles. Apr 14, 2019 at 16:09
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.