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.

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)? Counts are not necessarily included in it, for example, any counted bouquet. But, for example, all spaces $K(G, n)$ are included, it seems - they are simply indecomposable (I think this follows from the Milnor-Hilton theorem, but I don't know the exact argument).

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.

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.