The FreshBlog

Let us first suppose \( n \) is even. Denoting by \( a \) the antipodal map on \( S^n \), \begin{equation*} \deg f = \deg f \circ a = (-1)^{n+1} \deg f = - \deg f \end{equation*} Thus, \( \deg f = 0 \). Let now \( n \) be odd. Since \( f \) is even, it factors through \( \mathbb{R}P^n \): \begin{equation*} S^n \xrightarrow{\pi} \mathbb{R}P^n \xrightarrow{\widetilde{f}} S^n \end{equation*} By functoriality of \( H_n \), it suffices to show that the induced map \( \pi_* \) takes a generator of \( H_n(S^n) \) to twice a generator of \( H_n(\mathbb{R}P^n) \). To do this, we look at \( S^n \) as a cell complex with \( 2 \) \( i \)-cells for \( i = 0, 1, \dots, n \), namely the two hemipheres glued along the equatorial. Thus, the cellular chain complex of \( S^n \) is \begin{equation*} 0 \longrightarrow \mathbb{Z}^2 \longrightarrow \dots \longrightarrow \mathbb{Z}^2 \longrightarrow 0 \end{equation*} Let \( e_1^i, e_2^i \) be the two \( i \)-cells. Using the cellular boundary formula, we see that \begin{equation*} \partial e_k^i = e_1^{i-1} + (-1)^{i}e_2^{i-1}. \end{equation*} Thus, \( H_n(S^n) \) is generated by the homology class \( \langle e^n_1 - e_2^n \rangle \) and, denoting the only \( n \)-cell of \( \mathbb{R}P^n \) by \( e^n \), we have \begin{equation*} \pi_*(\langle e^n_1 - e_2^n \rangle) = 2 \langle e^n \rangle. \end{equation*} Again by the cellular boundary formula, we see that \( \langle e^n \rangle \) generates \( H_n(\mathbb{R}P^n) \). This completes the proof.