Second Stiefel-Whitney class of $\mathbb{C}\text{P}^2\#\overline{\mathbb{C}\text{P}^2}$

Question

I know that $w_2(\mathbb{C}\text{P}^2\#\overline{\mathbb{C}\text{P}^2})\neq 0$, where $\overline{\mathbb{C}\text{P}^2}$ is $\mathbb{C}\text{P}^2$ with opposite orientation. But how do you prove this? I can't figure it out and I haven't found a single reference justifying it (I suppose it should be quite elementary...).

Answer 1

The second Stiefel Whitney class of $Y=\mathbb{CP}^2$ is non-trivial. Similarly for $Z=\mathbb{CP^2}\setminus \{pt\}$. Namely the inclusion $Z\rightarrow Y$ pulls back the tangent bundle from $Y$ to $Z$, and on cohomology it induces an isomorphism on $H^2(\cdot;\mathbb{Z}_2)$.

Consider now the inclusion $i:Z\rightarrow X$, where $X$ is the space of the OP. The tangent bundle is again pulled back to Z, and $i^*(w_2(X))=w_2(Z)\not=0$. Thus $w_2(X)\not=0$.

This is a more general phenomenon: The first stiefel whitney class of a vector bundle over a $CW$ complex: i.e. if the bundle orientable or not , only depends on the restriction of the bundle over the one skeleton. Something similar is true of the second Stiefel Whitney class over the two-skeleton.