# Product of "prime" topological spaces

We call a topological space $(X,\tau)$ product-decomposable if there is an index set $I$ and subsets $X_i\subseteq X$ for $i\in I$ such that $|X_i| > 1$ and $X \cong \prod_{i\in I} X_i$ where each $X_i$ is endowed with the subspace topology.

If a space $(X,\tau)$ is not product-decomposable and $|X|>1$, we call it prime. (Is there established terminology for this?)

If $X_1, X_2$ are prime spaces, is it possible that there are prime topological spaces $Y_1, Y_2$ such that $X_i \not\cong Y_j$ for $i,j\in\{1,2\}$ but $(X_1\times X_2) \cong (Y_1\times Y_2)$?

• If $Y_i$ isn't assumed to be prime (but was assumed to be non-trivial), I guess $TS^2\times \Bbb R^1 \cong S^2 \times \Bbb R^3$ would be an example. Though one would have to show $TS^2$ is prime.
– PVAL
Commented Mar 5, 2015 at 9:13
• Indeed, there are simple contractible 2-dim non-homeomorphic polyhedra P Q, such that P×I and Q×I are homeomorphic. Perhaps Karol Borsuk was the first to consider such and more advance "counter examples", and theorems on unique decomposition as well. An especially deep theorem on unique decomposition was proved by Hanna Patkowska (Borsuk's student). She did it for finite families of 1-dimensional compact ANRs. Commented Mar 5, 2015 at 10:00
• Talking here about subspaces is here unnecessary (it's just an eyesore :-). Commented Mar 5, 2015 at 10:02
• Observe that as a rule, topological spaces are prime--it's a generic situation. Thus restriction to prime spaces (I'd call them indecomposable) is not severe (but yes, one has to talk about non-decomposable/prime spaces in this context). Commented Mar 5, 2015 at 10:07
• The above mentioned polyhedra $\ P\ Q\$ can subspaces of $\mathbb R^2.$ Commented Mar 5, 2015 at 10:16

This is just a cheap extension of PVAL's comment, but I think it works: Taking for granted that $S^2$ and $TS^2$ are prime, which I believe is true, you can write $$TS^2 \times (\mathbb R \amalg S^2) = S^2 \times (\mathbb R^3 \amalg TS^2),$$ and now all four spaces appearing are prime and pairwise not homeomorphic.

• The following at least shows $S^2$ is prime/indecomposable. If $S^2= X\times Y$, then by the Kunneth formula (WLOG) $X$ has the full homology of $S^2$. But if $X\ne S^2$, $X$ embeds inside $\Bbb R^2$ and hence has trivial second homology. The proof for $TS^2$ probably follows a similar pattern (likely the space with non-trivial homology in the product is forced to be homeomorphic to $S^2$ and you would a homeomorphism from $S^2\times \Bbb R^2$ to $TS^2$ with a non-trivial decomposition).
– PVAL
Commented Mar 7, 2015 at 2:17
• One should maybe note that other 4-manifolds (i.e. $\Bbb R^4$) can have very "non-obvious" product structures (i.e. the wishbone space $\times \Bbb R$ or the Whitehead manifold $\times \Bbb R$).
– PVAL
Commented Mar 7, 2015 at 2:20

There is the somewhat trivial example of the following:

Let $X_1 = \{1, \ldots, p\}, X_2 = \{1, \ldots q\}$ with $p, q$ prime and both with the discrete topology. Let $Y_1 = \{1\}, Y_2 = \{1, \ldots, pq\}$ each with the discrete topology as well. Then $X_1 \times X_2 \cong Y_1 \times Y_2$, but the rest of your conditions appear to be satisfied.

• But then isn't $Y_2$ not prime?
– PVAL
Commented Mar 5, 2015 at 8:38
• $Y_2$ is not prime, but it isn't specified that $Y_i$ are prime. Perhaps that is just an omission then... Commented Mar 5, 2015 at 8:43
• Actually, in hindsight, it kind of has to be, since you could always just pick $Y_1 = \{pt\}$ and $Y_2 = X_1 \times X_2$... Commented Mar 5, 2015 at 8:44
• Well, maybe they need not be prime; but at least $|Y_{i}| > 1$.
– jmc
Commented Mar 5, 2015 at 9:11
• C'mon, @SimonRose -- it was silly to consider non-prime option as a serious answer. A comment would be more than enough. Commented Mar 5, 2015 at 10:12