Show that for a degree 1 map \(f: M \to N\) of connected closed orientable manifolds, the induced map \(f_*: \pi_1 M \to \pi_1 N\) is surjective, hence also \(f_*: H_1(M) \to H_1(N)\).
Solution
Let \[ H=\operatorname{Im}\bigl(f_*:\pi_1(M)\to\pi_1(N)\bigr). \] Let \[ p:\widetilde N\to N \] be the connected covering corresponding to \(H\). By the lifting criterion, \(f\) lifts to \[ \widetilde f:M\to\widetilde N, \qquad p\circ\widetilde f=f. \]
If \(p\) has infinitely many sheets, then \(\widetilde N\) is a connected noncompact \(n\)-manifold. Hence \[ H_n(\widetilde N;\mathbb Z)=0. \] Therefore \[ f_*[M]=p_*\widetilde f_*[M]=0, \] contradicting \(\deg f=1\).
Thus \(p\) is finite-sheeted, say of degree \(d=[\pi_1(N):H]\). Since \(N\) is orientable, so is \(\widetilde N\), and Exercise 9 gives \[ |\deg p|=d. \] Degrees multiply under composition, so \[ 1=|\deg f|=|\deg p|\,|\deg\widetilde f|=d\,|\deg\widetilde f|. \] Hence \(d=1\). Therefore \(H=\pi_1(N)\) and \[ \boxed{f_*:\pi_1(M)\twoheadrightarrow\pi_1(N)}. \] Passing to abelianizations gives the surjection \[ f_*:H_1(M)\twoheadrightarrow H_1(N). \]