Integral cohomology of real projective space from UCT

By [F1], $H_0=\mathbb{Z}$; for $0<r<m$, $H_r$ is $\mathbb{Z}\mathrm{/}2$ when $r$ is odd and zero when $r$ is even. The top group is $\mathbb{Z}$ for odd $m>0$ and zero for even $m>0$. All groups above $m$ vanish. In particular all homology groups are finitely generated, as required in [F2].

For $j=0$, $H_{-1}=0$, so UCT gives $H^0\cong \mathrm{Hom}(H_0,\mathbb{Z})=\mathbb{Z}$. For even $0<j\le m$, the integer $j-1$ is odd and strictly below $m$, so $H_{j-1}=\mathbb{Z}\mathrm{/}2$ and $H_j=0$. The UCT sequence is $0\to \mathbb{Z}\mathrm{/}2\to H^j\to 0$, giving $H^j=\mathbb{Z}\mathrm{/}2$. For odd $0<j<m$, the adjacent lower group is zero unless $j=1$, when it is $\mathbb{Z}$; either way its Ext is zero. Since $H_j=\mathbb{Z}\mathrm{/}2$ has zero Hom into $\mathbb{Z}$, this gives $H^j=0$.

If $j=m$ is odd, $H_m=\mathbb{Z}$ and $H_{m-1}$ is zero, except at $m=1$ where it is $\mathbb{Z}$; its Ext is zero in both cases. Thus UCT gives $H^m=\mathbb{Z}$. At $j=m+1$, the Hom term is zero and the Ext term is zero because $H_m$ is either zero or free (including $m=0$). For $j>m+1$ both homology inputs vanish. Therefore every cohomology group above $m$ is zero; negative groups vanish by the cochain convention.

For $m=0$ this is precisely the point calculation. For $m=1$ it gives $\mathbb{Z}$ in degrees zero and one; for $m=2$ it gives $H^0=\mathbb{Z}$, $H^1=0$, $H^2=\mathbb{Z}\mathrm{/}2$; for $m=3$ it gives $\mathbb{Z},0,\mathbb{Z}\mathrm{/}2,\mathbb{Z}$ in degrees zero through three. These checks exhibit the shift from odd homological torsion to even cohomological torsion. No orientation-dependent generator is needed for these abstract group identifications. AC is inherited only where [F2] uses the UCT; the integral cell calculation in [F1] is choice-free.