For a map \(f: M \to N\) between connected closed orientable \(n\)-manifolds with fundamental classes \([M]\) and \([N]\), the degree of \(f\) is defined to be the integer \(d\) such that \(f_*([M]) = d[N]\), so the sign of the degree depends on the choice of fundamental classes. Show that for any connected closed orientable \(n\)-manifold \(M\) there is a degree 1 map \(M \to S^n\).
Solution
Choose an embedded closed \(n\)-ball \(B\subset M\) whose interior is orientation-compatible with the chosen fundamental class \([M]\). Collapse the complement of the interior of \(B\) to a point: \[ q:M\longrightarrow M/(M-\operatorname{int}B). \] The quotient is naturally homeomorphic to \[ B/\partial B\cong S^n. \] Choose the final homeomorphism \(B/\partial B\to S^n\) to preserve orientation on the interior of \(B\).
Let \(f:M\to S^n\) be the resulting map. For a point \(y\) in the image of \(\operatorname{int}B\), the preimage \(f^{-1}(y)\) consists of one point and \(f\) is an orientation-preserving local homeomorphism there. By the local-degree formula, \(\deg f=1\). Equivalently, the quotient map sends \([M]\) to the fundamental class of \(B/\partial B\).
Thus every connected closed orientable \(n\)-manifold admits a degree-\(1\) map to \(S^n\).