A map \(f: S^n \to S^n\) satisfying \(f(x) = f(-x)\) for all \(x\) is called an even map. Show that an even map \(S^n \to S^n\) must have even degree, and that the degree must in fact be zero when \(n\) is even. When \(n\) is odd, show there exist even maps of any given even degree. [Hints: If \(f\) is even, it factors as a composition \(S^n \to \mathbb{RP}^n \to S^n\). Using the calculation of \(H_n(\mathbb{RP}^n)\) in the text, show that the induced map \(H_n(S^n) \to H_n(\mathbb{RP}^n)\) sends a generator to twice a generator when \(n\) is odd. It may be helpful to show that the quotient map \(\mathbb{RP}^n \to \mathbb{RP}^n/\mathbb{RP}^{n-1}\) induces an isomorphism on \(H_n\) when \(n\) is odd.]
Solution
Let \(f:S^n\to S^n\) be even, so \(f(x)=f(-x)\). Then \(f\) factors through the antipodal quotient: \[ S^n\xrightarrow{q}\mathbb{RP}^n\xrightarrow{g}S^n. \]
1. If \(n\) is even, then \(\deg f=0\).
For even \(n\), \[ H_n(\mathbb{RP}^n;\mathbb Z)=0. \] Hence the induced map \[ H_n(S^n)\xrightarrow{q_*}H_n(\mathbb{RP}^n) \] is zero, and therefore \[ f_*=g_*q_*=0. \] Thus \(\deg f=0\).
2. If \(n\) is odd, every even map has even degree.
For odd \(n\), \(\mathbb{RP}^n\) is orientable and \[ H_n(\mathbb{RP}^n)\cong\mathbb Z. \] The quotient map \(q:S^n\to\mathbb{RP}^n\) is a two-sheeted covering between oriented closed \(n\)-manifolds, so \[ q_*[S^n]=2[\mathbb{RP}^n]. \] If \(g_*\) is multiplication by \(k\), then \[ f_*[S^n]=g_*q_*[S^n]=2k[S^n]. \] Hence \(\deg f=2k\) is even.
3. When \(n\) is odd, every even integer occurs as the degree of an even map.
Collapse the \((n-1)\)-skeleton of \(\mathbb{RP}^n\): \[ c:\mathbb{RP}^n\to\mathbb{RP}^n/\mathbb{RP}^{n-1}\cong S^n. \] For odd \(n\), the top cellular boundary in \(\mathbb{RP}^n\) is zero, so the unique top cell generates \(H_n(\mathbb{RP}^n)\). The quotient map \(c\) carries this top cell homeomorphically to the top cell of \(S^n\), hence \[ c_*:H_n(\mathbb{RP}^n)\xrightarrow{\cong}H_n(S^n). \] Given any \(k\in\mathbb Z\), choose a degree-\(k\) map \[ h_k:S^n\to S^n. \] Then \[ f_k=h_k\circ c\circ q:S^n\to S^n \] is even because it factors through \(q\), and \[ \deg f_k=k\cdot1\cdot2=2k. \] Thus every prescribed even degree occurs.