# Is the decomposition of the homotopy type of a complex into a product and into a smash product unique?

1. Is it true that if $$A_1\times A_2\times ... \times A_n = B_1\times B_2\times .. \times B_m$$, where $$A_i, B_j$$ are homotopy types of connected complexes not decomposable into a product, then the multisets $$A_i$$ and $$B_j$$ coincide? What if it is limited only to finite complexes?

2. Is it true that if $$A_1​​\wedge A_2 \wedge .. \wedge A_n = B_1\wedge B_2 \wedge .. \wedge B_m$$, where $$A_i, B_j$$ are homotopy types of connected pointed complexes not decomposable into a smash product, then the multisets $$A_i$$ and $$B_j$$ coincide? What if it is limited only to finite complexes?

It is clear that finite complexes decompose into a finite number of indecomposable (with smash product, the connectedness increases, and with a Cartesian product, the homology groups increase).

Update: in fact, the point is decomposable with respect to the smash of the product, so there are no non-decomposable complexes with respect to the smash of the product. With respect to the direct product from the homotopy groups of the product, the indecomposability of the point is obvious, so my proof remains valid with respect to it.

P.S. Similar question

• There are two nonisomorphic groups $G,H$ such that their products with the integers are isomorphic. If I remember correctly they can be chosen as virtually cyclic, so they are not so big. After applying the classifying space functor $B$ one should get a counterexample to the first question $BG \times S^1 = B(G\times \mathbb{Z})=BH \times S^1$ and $BG$ and $BH$ are indecomposable, and not homotopy equivalent. Apr 25 at 12:32

Let me write $$S^n/p$$ for the cofibre of $$p$$ times the identity map on $$S^n$$, or in other words the mod $$p$$ Moore space with homology in degree $$n$$.
If $$p$$ and $$q$$ are coprime then $$S^n/p\wedge S^m/q$$ is contractible. This generates many counterexamples for your second question. Even if you restrict attention to $$p$$-torsion complexes for a fixed prime $$p$$, we have $$S^n/p\wedge S^m/p^k\simeq S^{n+m}/p\vee S^{n+m+1}/p$$ independent of $$k$$ (except that $$k$$ needs to be at least $$2$$ when $$p=2$$, and at least $$1$$ when $$p>2$$).