This is an exercise from Hatcher's Algebraic Topology book:
Exercise 2.1.13, pg. 132: Verify that $f≃g$ implies $f_∗=g_∗$ for induced homomorphisms of reduced homology groups.
The formal version of the exercise: If $f,g :(X,A) \to (Y,B)$ are homotopic, then $f_* = g_*:H_n(X,A) \to H_n(Y,B)$.
Is there any gap in the sequence of the proof?
Proof: By Theorem 2.10, we know that if two maps $f,g :X \to Y$ are homotopic, then they induce the same homomorphism $f_* = g_*:H_n(X) \to H_n(Y)$.
So, the prism operator $P:C_n(X) \to C_{n+1}(Y)$ satisfying $\partial P +P \partial =g_\# - f_\#$ clearly sends $C_n(A)$ to $C_{n+1}(B)$.
Thus, $P$ gives a natural prism operator on quotients, i.e., $\tilde{P}: C_n(X,A) \to C_{n+1}(Y,B)$ s.t. $\partial P +P \partial =g_\# - f_\#$.
Therefore $f_*=g_*: H_n(X,A) \to H_n(Y,B).$