= Solution
Work first over $\mathbb F_2$. The <mod-two cohomology ring of real projective space> is $H^*(\mathbb{RP}^n;\mathbb F_2)=\mathbb F_2[a]/(a^{n+1})$, with $|a|=1$. A map missing a point has zero <mod-two degree of a map between closed manifolds>, so $f^*(a^n)=0$. The group in degree one has only two elements, so $f^*a$ is either zero or $a$. The latter would imply $f^*(a^n)=(f^*a)^n=a^n\ne0$, a contradiction. Therefore $f^*a=0$, and multiplicativity makes $f^*$ zero in every positive cohomological degree. Duality of <homology (mathematics)> and <cohomology> over a field consequently makes $f_*$ zero in every positive degree with mod-two coefficients.
For integral coefficients, <cellular homology of real projective space> gives $H_k\cong\mathbb Z/2$ in odd degrees $0<k<n$, zero in even degrees $0<k<n$, and a top group $\mathbb Z$ if $n$ is odd and zero if $n$ is even. For each torsion group, the natural reduction map to mod-two <homology (mathematics)> is injective by the <universal coefficient theorem for homology>; naturality and the preceding vanishing imply that its integral induced map is zero. When $n$ is odd, the integral top-degree map is also zero because the nonsurjective map has integer <degree of a map between oriented manifolds> zero. All groups above dimension $n$ vanish. Hence \b[$f_*$ is zero on all positive integral <homology (mathematics)> groups] as well.
For $n\ge2$, $\pi_1(\mathbb{RP}^n)=\mathbb Z/2$, and its <abelianization> is already $H_1(\mathbb{RP}^n;\mathbb Z)$. The vanishing just proved therefore implies $f_*\pi_1=0$. By the <lifting criterion for a covering space>, $f$ lifts through the double <covering space> $q:S^n\to\mathbb{RP}^n$ to a map $\widetilde f:\mathbb{RP}^n\to S^n$ with $q\widetilde f=f$.
Choose a point $y$ missed by $f$. Both points of $q^{-1}(y)$ are missed by $\widetilde f$. In particular the lift has image in $S^n\setminus\{z\}\cong\mathbb R^n$, for either such point $z$. This punctured sphere is <contractible>, so contraction there gives a <homotopy> from $\widetilde f$ to a constant map into $S^n$. Composing with $q$ contracts $f$. Thus
$$
\boxed{f\simeq\text{constant}.}
$$
This proves the <nonsurjective self-map of real projective space is null-homotopic> conclusion, and also shows that all positive-degree induced maps vanish with any coefficients.
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