If $\mathbf{P}^1$ is replaced by the affine line $\mathbf{A}^1$, this becomes the cancellation problem, and we have a pair of famous Danielewski surfaces ($xy=1-z^2$ and $x^2y=1-z^2$) as a counterexample (though I'm still seeking how to prove that..). I suppose in my case this counterexample might no longer work. 


Also one may replace $\mathbf{P}^1$ by $\mathbf{P}^n$ or other fixed varieties, or ask again after imposing some conditions on $V$ and $W$ (for example dimensions) if there are counterexamples for my question. And more wildly I may ask for what kind of family $X\_n$, we will have the result that $V\times X\_n=W\times X\_n$ implies $V=W$. Any result of these kind of variations of the problem is also welcomed.



$V$, $W$ are varieties. Does $V\times \mathbf{P}^1=W\times \mathbf{P}^1$ imply $V=W$?
This problem was studied by Fujita in his paper ["Cancellation problem of complete varieties", Inventiones Mathematicae 64 (1981).](http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN356556735_0064&DMDID=dmdlog13) 


He showed that the obstruction to cancellation is caused by the Picard schemes, proving the following remarkable result (see Corollary 7 in the cited paper):



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> Let $M$, $V$ and $W$ be compact complex manifolds such that $M \times V \cong M \times W$. Assume that $M$ is projective and that $\textrm{Alb}(M)=0$ or $\textrm{Alb}(V)=0$. Then $V \cong W$.
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In particular, cancellation problem has a positive answer for $M=\mathbf{P}^n$. 


The condition on the Albanese variety is a necessary one; in fact, it is known that cancellation is not always true for abelian varieties.


