Verify that \(f \simeq g\) implies \(f_* = g_*\) for induced homomorphisms of reduced homology groups.
Solution
Suppose \[ f,g:X\to Y \] are homotopic.
1. For every \(n>0\),\[ \widetilde H_n(f)=\widetilde H_n(g). \]
For \(n>0\) one has \[ \widetilde H_n(X)=H_n(X), \qquad \widetilde H_n(Y)=H_n(Y). \] Homotopic maps induce the same homomorphism on ordinary homology, so the induced maps agree.
2. The same holds in degree zero.
For a nonempty space \(Z\), \[ \widetilde H_0(Z)=\ker\bigl(\epsilon:H_0(Z)\to\mathbb Z\bigr), \] where \(\epsilon\) sends each path-component generator to \(1\). Every continuous map commutes with augmentation: \[ \epsilon\,f_* = \epsilon, \qquad \epsilon\,g_* = \epsilon. \] Since homotopic maps satisfy \[ f_*=g_*:H_0(X)\to H_0(Y), \] their restrictions to the augmentation kernel are equal. These restrictions are precisely the induced maps on \(\widetilde H_0\). If \(X\) is empty the assertion is trivial.
3. Hence\[ \boxed{f_*=g_*:\widetilde H_n(X)\to\widetilde H_n(Y)\text{ for all }n.} \]
Combine 1 and 2.