1. Is it true that if $A\_1 ​​\vee A\_2 \vee .. \vee A\_n = B\_1 \vee B\_2 \vee .. \vee B\_m$, where $A\_i, B\_j$ are homotopy types of complexes not decomposable into a bouquet, then the multisets $A\_i$ and $B\_j$ coincide? That is, is it true that a commutative monoid of homotopy types decomposable into a bouquet of a finite number of indecomposable ones is freely generated by indecomposable ones.
2. Is it true that any finite complex decomposes into a finite number of indecomposable ones (and thus all finite complexes are included in the monoid above)? Countable complexes are not necessarily included in it, for example, any countable bouquet. But, for example, all spaces $K(G, n)$ are included, it seems.



Is the decomposition of the homotopy type of a complex into a bouquet unique?
In Hilton&Roitberg paper "On principal $S^3$-bundles over spheres" it's proven that if you have a prime order $p \neq 2,3$ class $\alpha$ in $\pi\_k(S^n)$ that is a suspension, then for a prime $q \neq \pm 1 \, mod \, p$ mapping cones $C(\alpha)$ and $C(q \cdot \alpha)$ satisfy $C(\alpha) \vee S^n \cong C(q \cdot \alpha) \vee S^n$.


Now it's easy to see that those cones are indecomposable, and give counterexample to 1.


For 2, I think Grushko theorem gives that all but finite number of summands are simply-connected, and then you have only finitely many generators in homology so overall any such sum has to be finite. I'll think about that, but my reasoning seems correct to me now.


