TITLE: When factors may be cancelled in homeomorphic products?
QUESTION [37 upvotes]: It is easy to see that if $A\times B$ is homeomorphic to $A\times C$ for topological spaces $A$, $B$, $C$, then one may not conclude that $B$ and $C$ are homeomorphic (for example, take $C=B^2$, $A=B^{\infty}$). The question is: for which $A$ such conclusion is true? I saw long ago a problem that for $A=[0,1]$ it is not true, but could not solve it, and do not know, where to ask. Hence am asking here. The same question in other categories (say, metric spaces instead topological) also seems to have some sense.

REPLY [10 votes]: It is easy to see that if $A\times B$ is homeomorphic to $A\times C$ for topological spaces $A$, $B$, $C$, then one may not conclude that $B$ and $C$ are homeomorphic (for example, take $C=B^2$, $A=B^∞$). The question is: for which $A$ such conclusion is true? 

Witold Rosicki has a lot of results of this sort (usually under some conditions on $B$ and $C$). For instance,
On decomposition of polyhedra into a Cartesian product of 1-dimensional and 2-dimensional factors
On uniqueness of decomposition of 4-polyhedron into Cartesian product of the 2-dimensional factors
On uniqueness of Cartesian products of surfaces with boundary
(with J. Malešič, D. Repovš, A. Zastrow)
There also exist papers of a different flavor on this subject
All lens spaces have diffeomorphic squares (S. Kwasik, R. Schultz)
Non-cancellation and a related phenomenon for the lens spaces (A. J. Sieradski)
As for nice examples, there exist manifolds $M$ such that $M\times I$ is homeomorphic a ball. For instance, Mazur's 4-manifold, as described by Zeeman:

Start with $S^1\times I^3$. In the boundary $S^1\times S^2$, choose a knotted $S^1$ homologous to the first factor. Knotted means that $S^1$ is not isotopic to a 1-sphere $S^1\times y$, $y\in S^2$. Form $M^4$ from $S^1\times I^3$ by attaching a handle to
  $S^1$ (i.e., attach a disk to $S^1$ and then fatten the disk so that its fattened
  boundary is identified with some chosen tubular neighbourhood of $S^1$ in
  $S^1\times S^2$). Form the cube $I^4$ by the same process, only omitting the knotting. The knotting ensures that $M^4\not\cong I^4$. But one extra dimension permits unknotting 
  $M^4\times I\cong I^4\times I$ (by just untwisting the handle).

Zeeman also notes a parallel construction of Whitehead's example with surfaces $\times I$ (mentioned above by Sergei Ivanov): `Start with $S^0\times I^2$. In the boundary $S^0 \times S^1$, choose three linked $S^0$'s, each homologous to the first factor', etc.
A really cool cancellation theorem is about joins of polyhedra, rather than products (H. Morton):

If $A*B\cong A*C$, then  either  $B\cong C$ or else $A\cong pt*A'$, $B\cong pt*X$ and
  $C\cong S^0*X$ for some polyhedra $A'$ and $X$.