The effacement extension is independent of the effacing morphism

In the homological case, let $p_i:P_i\twoheadrightarrow A$ be two effacements of the relevant target value. Form the epimorphism $p=(p_1,p_2):P_1\oplus P_2\twoheadrightarrow A\mathrm{.}$ By [L2], $P_1\oplus P_2$ is projective. Since $T_n$ is additive, the canonical biproduct identification gives $T_n(P_1\oplus P_2)\cong T_n(P_1)\oplus T_n(P_2),$ and under this identification the map $T_n(p)$ has components $T_n(p_1)$ and $T_n(p_2)$, both zero. Thus $T_n(p)=0$, so $p$ is again an admissible effacement.

The inclusions ${\iota }_i:P_i\to P_1\oplus P_2$ satisfy $p{\iota }_i=p_i$, so they give morphisms from each original effacement to the dominating effacement of step 1.1 over the identity of $A$. By the naturality part of [L1], the component defined from $p_i$ agrees with the one defined from $p$ for each $i$. Hence the components defined from $p_1$ and $p_2$ are equal.

The cohomological case is dual: if $e_i:A\rightarrowtail I_i$ are two effacements, then $e=(e_1,e_2):A\rightarrowtail I_1\times I_2$ is again an admissible effacement by [L2], and the projections $I_1\times I_2\to I_i$ compare it with each original choice. Applying [L1] as in step 2.1 shows that the resulting degree-$(n+1)$ component is independent of the chosen injective effacement.