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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 $X$.

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 $X$.

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 $M^4\ne I^4$. But one extra dimension permits unknotting $M^4\times I=I^4\times I$ (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.

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 $X$.

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,

http://matwbn.icm.edu.pl/ksiazki/cm/cm72/cm7217.pdf

http://dx.doi.org/10.1016/j.topol.2004.02.013

http://dx.doi.org/10.1016/j.topol.2003.04.002

There also exist papers of a different flavor on this subject

http://dx.doi.org/10.1016/S0040-9383(00)00039-2

http://dx.doi.org/10.1016/0040-9383(78)90014-9

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 $M^4\ne I^4$. But one extra dimension permits unknotting $M^4\times I=I^4\times I$ (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.