Let $X$ be a connected CW-complex, and $A$ be a subcomplex of $X$. The subcomplex $A$ may not be connected. Let $Y$ be a $K(G,1)$ space for some group $G$(i.e., $Y$ is a connected CW-complex having contractible universal cover).
Let $f,g\colon X\to Y$ be two maps such that $f(a)=g(a)$ for all $a\in A$ and $f_*=g_*\colon \pi_1(X,a_0)\to \pi_1(Y,y_0)$ where $a_0\in A$ and $f(a_0)=y_0=g(a_0)$. Now, we know that $f\simeq_{\text{rel } a_0} g$.
Can we say $f$ homotopic $g$ relative to $A$?
This is true without the relativeness restriction, and the procedure is cell-by-cell construction or skeleton-wise construction. So, if I do the same process with the relative-filtration $K^{(n)}:=X^{(n)}\cup A$, can I conclude even then?
I don't think this is true. Let $X=D^1$, $Y=S^1$ and $A=\{0,1\} \subset D^1$. Then maps $[(D^1,0),S^1]_*=0$, so any two maps will induce the zero map on $\pi_1$, but then maps $[(D^1,A),(S^1,s)]$ are classified by $[S^1,S^1]_*$ if I recall correctly, so there are plenty that are not homotopic relative to $A$. To be concrete, maybe take $x \mapsto e^{2 \pi i x}$ and the constant map.