I know there are manifolds (with or without boundary) $A$ and $B$ such that $A\times C$ is homeomorphic to $B\times C$ but $A$ is NOT homeomorphic to $B$. 


My question is (in the Diffeomorphism Category)


Is there any infinitely many $A\_i$, with same dimension of course, which are pairwise non-diffeomorphic, but $A\_i\times C$ become all diffeomorphic to each other. 


1) What's the answer to the question under the assumption on $A\_i$, $C$: Smooth Close manifolds.


2) What if we change diffeomorphic to homeomorphic?


3) Is there any example when we require $C$ to be Torus?



Existence of sequence of examples of braking 'Cancellation law in homeomorphic products'
The answer to 2 is yes, there is such an example. In



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> McMillan, D. R., Jr., *Some contractible open $3$-manifolds*. Trans. Amer. Math. Soc. **102** (1962), 373–382.
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there is a construction of uncountably many topologically distinct, contractible (open) $3$-manifolds $M\_\alpha$ such that $M\_\alpha \times \mathbb R$ is homeomorphic to $\mathbb R^4$.


Take a look at [this recent MO question](https://mathoverflow.net/questions/60113/contractible-manifolds) and the [Wikipedia article](http://en.wikipedia.org/wiki/Whitehead_manifold) on the Whitehead manifold for some closely related material.


**Edit.** The answer to 3 is also yes, assuming by "torus" you mean $S^1$. On page 221 of



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> Vogt, E., *Foliations of codimension $2$ with all leaves compact on closed $3$-, $4$-, and $5$-manifolds*. Math. Z. **157** (1977), no. 3, 201–223.
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you can find a construction of infinitely many pairwise non-homeomorphic closed Seifert 3-manifolds whose product with $S^1$ gives the same Seifert 4-manifold.


