Effaceable homological delta functors are universal

Start with the given $u_0$ as the degree-zero component.

Suppose by induction that for some $n>0$ we have already constructed natural transformations $u_i:S_i\Rightarrow T_i$ for all $i<n$, and that these form a morphism of homological delta functors through degree $n-1$. For each object $A$, choose an effacement $p_A:P_A\twoheadrightarrow A$ for the target value $T_n(A)$ using [L2]. Then [L3] defines a map $u_n(A):S_n(A)\to T_n(A)$ from that effacement, and [L3] makes it independent of the chosen $p_A$.

To check naturality of the family from step 1.2, compare two chosen effacements over a morphism $f:A\to B$ by a common dominating effacement. The covered naturality from [L3] applies to that dominating choice, and the choice independence from [L3] transports the result back to the original objects. Thus $u_n$ is natural. The same comparison argument, now applied over a short exact sequence, together with the connecting-map compatibility from [L3], shows that adjoining $u_n$ preserves the morphism-of-delta-functors condition in degree $n$.

This constructs a morphism $u:S\to T$ extending $u_0$ in every degree. For uniqueness, let $v_n$ be any other degree-$n$ component compatible with the already fixed lower-degree data. Choose an effacement $p:P\twoheadrightarrow A$ for $T_n(A)$. Because $u_n$ and $v_n$ have the same lower-degree compatibility, their composites with ${\mathrm{\partial }}_n^T$ agree. The map ${\mathrm{\partial }}_n^T$ is monic by [L2], so $u_n(A)=v_n(A)$. Hence the extension is unique in each degree, and [L1] identifies $T$ as universal.