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Apr 22, 2014 at 14:51 comment added Oscar Randal-Williams @IgorBelegradek: The simplest example is $B=S^3$ with $x_3$ the nontrivial element of $H^3(S^3;\mathbb{Z}/2)$. As $w_3 = \mathrm{Sq}^1(w_2)$ this cannot be realised.
Apr 22, 2014 at 13:32 comment added Igor Belegradek @nsrt: nice example, and it does not require Wu formula: since $w_1=0$ the bundle is orientable, so $w_2\in H^2(B;\mathbb Z_2)\cong H^1(RP^2;\mathbb Z_2)\cong\mathbb Z_2$ must be a reduction of the Euler class which lives in $H^2(B)=H^1(PR^2)=0$.
Apr 22, 2014 at 13:05 comment added nsrt @IgorBelegradek: There's no two-dimensional bundle over $\Sigma RP^2$ with $w_1=0,w_2\ne 0$.
Apr 22, 2014 at 12:57 comment added Igor Belegradek This gives a potential obstruction. To make it into an answer one needs to produce an example of $B$ and $x_i$'s for which the Wu formula is not satisfied.
Apr 22, 2014 at 8:42 vote accept Oliver Straser
Apr 22, 2014 at 8:42 comment added Oliver Straser What is if they satisfy Wu's formula?
Apr 22, 2014 at 8:38 history answered Oscar Randal-Williams CC BY-SA 3.0