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For $n\ge 2$, a nonsurjective self-map of Real projective space ${\mathbb{RP}}^n$ has zero mod-two degree of a map between closed manifolds. In the mod-two cohomology ring of real projective space, the nonzero degree-one class $a$ has $a^n\ne 0$. If pullback fixed $a$, it would fix $a^n$, contradicting zero degree. Thus the map is trivial on the fundamental group and lifts to $S^n$. A point missed by the original map has both its inverse images missed by the lift. The lift therefore takes values in a punctured sphere, which is contractible, and is null-homotopic. Composing its contraction with the covering projection proves the conclusion.

Work first over ${\mathbb{F}}_2$. The mod-two cohomology ring of real projective space is $H^{\ast }({\mathbb{RP}}^n;{\mathbb{F}}_2)={\mathbb{F}}_2[a]/(a^{n+1})$, with $|a|=1$. A map missing a point has zero mod-two degree of a map between closed manifolds, so $f^{\ast }(a^n)=0$. The group in degree one has only two elements, so $f^{\ast }a$ is either zero or $a$. The latter would imply $f^{\ast }(a^n)=(f^{\ast }a{)}^n=a^n\ne 0$, a contradiction. Therefore $f^{\ast }a=0$, and multiplicativity makes $f^{\ast }$ zero in every positive cohomological degree. Duality of homology and cohomology over a field consequently makes $f_{\ast }$ 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 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 $f_{\ast }$ is zero on all positive integral homology groups as well.

For $n\ge 2$, ${\pi }_1({\mathbb{RP}}^n)=\mathbb{Z}/2$, and its abelianization is already $H_1({\mathbb{RP}}^n;\mathbb{Z})$. The vanishing just proved therefore implies $f_{\ast }{\pi }_1=0$. 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 $\tilde{f}:{\mathbb{RP}}^n\to S^n$ with $q\tilde{f}=f$.

Choose a point $y$ missed by $f$. Both points of $q^{-1}(y)$ are missed by $\tilde{f}$. 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 $\tilde{f}$ to a constant map into $S^n$. Composing with $q$ contracts $f$. ThusThis 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.