= Solution
For connected oriented <closed manifolds> of dimension $n$, the <degree of a map between oriented manifolds> is the integer determined by
$$
\boxed{f_*[M]=(\deg f)[N]\quad\text{in }H_n(N;\mathbb Z),}
$$
where the brackets are the chosen <fundamental classes>. Functoriality gives $\deg(g\circ f)=\deg g\,\deg f$.
Give $S^n$ its boundary orientation as the boundary of the unit ball in $\mathbb R^{n+1}$. The <antipodal map> is the restriction of $-I$. Its ambient determinant is $(-1)^{n+1}$, and it takes the outward normal at $x$ to the outward normal at $-x$. Thus its effect on the boundary orientation has this same sign, giving
$$
\boxed{\deg A=(-1)^{n+1}.}
$$
The original PDF has $A:S^n\to S^n$, correcting the extra prime on the target in the TeX transcription.
For even dimension $n=2k\geq2$, <cellular homology of real projective space> gives $H_n(\mathbb{RP}^n;\mathbb Z)=0$. Thus the composite map on top homology factors through zero, and
$$
\boxed{\deg h=0\quad\text{if }h:S^{2k}\to S^{2k}\text{ factors through }\mathbb{RP}^{2k},\ k\geq1.}
$$
Zero is attained by constant maps.
For odd dimension $n=2k+1\geq3$, <Real projective space> is orientable: the antipodal deck transformation has degree $+1$. Orient it so that the double <covering map> $q:S^n\to\mathbb{RP}^n$ has degree $2$. Since $S^n$ is simply connected, the <lifting criterion for a covering space> gives $f=q\circ\widetilde f$ for a map $\widetilde f:S^n\to S^n$. Hence
$$
\deg(g\circ f)=2\deg g\,\deg\widetilde f,
$$
which is even. Every even integer occurs: collapse the complement of an oriented embedded disk in $\mathbb{RP}^n$ to obtain a degree-one map $g:\mathbb{RP}^n\to S^n$, and choose $\widetilde f$ of any prescribed integer degree $m$. Maps of all integer degrees on spheres are obtained, for example, by suspending the circle maps $z\mapsto z^m$. Setting $f=q\circ\widetilde f$ gives degree $2m$. Thus
$$
\boxed{\{\deg h\}=2\mathbb Z\quad\text{in odd dimensions }2k+1\geq3.}
$$
These are the <degrees of maps factoring through real projective space>. If $k=0$ is permitted in the odd-dimensional clause, there is an exception: $\mathbb{RP}^1\cong S^1$, so every integer degree occurs. The lifting argument requires dimension at least two. The initial connected-manifold degree definition excludes $S^0$; with the usual reduced-homology definition for self-maps of $S^0$, a factorization through the one-point $\mathbb{RP}^0$ has degree zero.
Finally suppose $\deg f=p$ is prime. Fix any prime $q\ne p$ and work over $F=\mathbb F_q$. We claim that $f^*:H^r(N;F)\to H^r(S^n;F)$ is injective. If $a\ne0$, <Poincare duality> supplies $b\in H^{n-r}(N;F)$ with $\langle a\smile b,[N]_F\rangle\ne0$. Naturality of the <cup product> and evaluation gives
$$
\langle f^*a\smile f^*b,[S^n]_F\rangle
=p\,\langle a\smile b,[N]_F\rangle\ne0,
$$
because $p$ is invertible in $F$. Hence $f^*a\ne0$, proving <cohomological injectivity of a map of invertible degree>.
The intermediate cohomology of the sphere is zero, so $H^r(N;F)=0$ for $0<r<n$. Over a field, the <universal coefficient theorem for cohomology> identifies this with the dual of $H_r(N;F)$, so these homology groups vanish as well. The <universal coefficient theorem for homology> then gives
$$
H_r(N;\mathbb Z)\otimes\mathbb F_q=0\quad(0<r<n,\ q\ne p).
$$
Each integral group is a <finitely generated abelian group>, by the permitted finite <CW complex> model. Its decomposition can have no free summand, since that would survive modulo $q$, and no torsion summand divisible by any prime $q\ne p$. Thus it is a finite $p$-primary group. There are only finitely many intermediate degrees, so the exponents of their finite cyclic summands have a common bound $p^a$. Taking $a\geq1$ also covers the case in which all the groups vanish. Therefore
$$
\boxed{\exists a>0:\quad p^a x=0\quad\text{for every }x\in H_r(N;\mathbb Z),\ 0<r<n.}
$$
This proves that a <prime-degree sphere map forces primary torsion>, including one uniform exponent for all the degrees.
Back to article page
