The complex projective plane has nonzero w2 and no Spin structure

By [F1] with $n=2$, $c_1(T{\mathbb{C}\mathbb{P}}^2)=3x$. The coefficient reduction ${\rho }_2$ is a ring homomorphism, so ${\rho }_2(3x)={\rho }_2(x)$ because $3\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}2)$.

The CW structure in [F4] has cellular chains $C_4\cong \mathbb{Z}$, $C_2\cong \mathbb{Z}$, and $C_0\cong \mathbb{Z}$, with all other groups zero. Every cellular boundary is zero because its target is zero, so cellular and singular homology give $H_2({\mathbb{C}\mathbb{P}}^2;\mathbb{Z})\cong \mathbb{Z}$ and $H_1=0$. By UCT with integral coefficients, evaluation identifies $H^2({\mathbb{C}\mathbb{P}}^2;\mathbb{Z})$ with $\mathrm{Hom}(\mathbb{Z},\mathbb{Z})$; since $x$ is an integral cohomology generator by [F2], it evaluates as $+1$ or $-1$ on a homology generator. UCT with ${\mathbb{F}}_2$ coefficients identifies $H^2({\mathbb{C}\mathbb{P}}^2;{\mathbb{F}}_2)$ with $\mathrm{Hom}(\mathbb{Z},{\mathbb{F}}_2)$, and naturality of coefficient reduction makes ${\rho }_2(x)$ evaluate as $1$. Hence ${\rho }_2(x)\mathrm{\ne }0$.

Applying [F3] to the complex tangent bundle and using steps 1.1–1.2 gives $w_2(T{\mathbb{C}\mathbb{P}}^2)={\rho }_2(x)\mathrm{\ne }0$. The manifold is complex-oriented, hence its tangent bundle is oriented; [F5] therefore implies that it admits no Spin lift. The coefficient group has characteristic two and no sign convention changes the nonzero class.