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Dec 25, 2019 at 14:35 comment added user43326 OK, there is one more generator in $H^2$ because of the $H^0$ of the base space, but the point is that we don't get tensor product because $x^2\neq 0$.
Dec 25, 2019 at 14:02 vote accept asv
Dec 25, 2019 at 13:54 comment added asv @user43326: The cohomology should be 4-dimensional.
Dec 25, 2019 at 11:58 comment added user43326 @MKO $Z[x]/x^3$ with $deg(x)=2$.
Dec 25, 2019 at 11:56 answer added user43326 timeline score: 5
Dec 25, 2019 at 10:40 comment added asv @GustavoGranja: What is cohomology ring of that space?
Dec 25, 2019 at 10:00 comment added Gustavo Granja A counterexample is the non-trivial S^2 bundle over S^2 (equivalently the blow-up of CP^2 at one point).
Dec 24, 2019 at 10:46 comment added user43326 If you have an isomorphism, it will be compatible with the two maps by naturality. However, in general there can be an algebra extension problem, if I am not mistaken.
Dec 24, 2019 at 10:23 history asked asv CC BY-SA 4.0