Bing gave a classical example of spaces $X, Y, Z$ such that $X \times Y = Z$, where $X$ and $Z$ are manifolds but $Y$ isn't. The space $Z$ in his example has dimension four. Is it known if this is best possible? In other words, if $X \times Y =Z$ where $X$ is a manifold and $Z$ is a 3-manifold, then is $Y$ a manifold?

In Bing's example, $Z$ is not compact. Is there a compact example in dimension $4$? In the example, $X$ is the real line, so one can also ask if it possible to get a four dimensional example where $X$ is a surface, or where $X$ is compact.


As asked Dusan Repovs (who is an expert in the theory of topological manifolds), and he sent me the following answer:

This is indeed best possible result, since whenever a product of two spaces is a topological manifold, both factors must be generalized manifolds - which in dimensions below 3 are topological manifolds.

Ref.: A.Cavicchioli, F.Hegenbarth and D.Repovš, Higher-Dimensional Generalized Manifolds: Surgery and Constructions, EMS Series of Lectures in Mathematics, European Mathematical Society, Zurich, 2016.

To the second question: there are also compact examples (in dimensions >3): e.g. take the product of the 3-sphere $S^3$ modulo the Fox-Artin wild arc $A$ and $S^1$. This product is homeomorphic to $S^3\times S^1$.

Ref.: R.J.Daverman, Decompositions of Manifolds, Academic Press, Orlando, 1986.

Added in Edit: Answering a comment of John Samples, Dusan Repovs pointed out that the Chapter 29 of Daverman's book contains the following fact: for any $n,m>2$ there are a generalized $n$-manifold $X$ and a generalized $m$-manifold $Y$ which are not topological manifolds, but their product $X\times Y$ is a topological manifold.

  • $\begingroup$ I do not understand how the statement fits together with the example. Is the wild arc a generalized manifold or not? $\endgroup$ – ThiKu Jun 19 '18 at 7:06
  • $\begingroup$ @ThisKu The wild arc is not a generalized manifold (it is a topological arc), but the quotient space of the sphere by the wild acr is a generalized manifold (not being a manifold). $\endgroup$ – Taras Banakh Jun 19 '18 at 7:44
  • $\begingroup$ Now I wonder if there is a pair of (non-manifold) generalized $3$-manifolds whose product gives a manifold, or if one factor will always be a bona fide manifold. $\endgroup$ – John Samples Jun 21 '18 at 22:18
  • $\begingroup$ It should be noted that for some definitions of 'generalized manifold' the properties of separability and metrizability are not assumed. In that setting, there are examples of non-manifold generalized manifolds in dimensions one and two. But factors of separable, metric spaces are separable metric, so since manifolds have these properties everything still works for this factorization problem. I am starting to learn this area, now. $\endgroup$ – John Samples Jun 30 '18 at 2:14

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