Showing that effaceable delta-functors are universal I am attempting to prove Theorem 1.3A in III.1 of Hartshorne's Algebraic Geometry, which says that $\delta$-functors $T=(T^i)_{i\geq0}$ with each $T^i$ effaceable/erasable for $i\geq 1$ are universal.
My setup is as follows:
Let $T$ be an effaceable $\delta$-functor as above, and considered another $\delta$-functor $\bar{T}$ with a natural transformation $f^0:T^0\implies \bar{T}^0$ between them in the zeroth degree. I want to construct the remaining natural transformations $f^i:T^i\implies \bar{T}^i$. Not having many tools to work with, I plan to construct these by components, that is, for each $A$ in the source category I wish to define morphisms $f^i_A:T^i(A) \to \bar{T}^i(A)$.
I believe I have constructed the first such natural transformation (in an inductively extendable fashion). Letting $A \stackrel{\phi}{\hookrightarrow} M$ be a $T^1$-erasure of $A$ (i.e. an embedding such that $T^1(\phi) = 0$), we can write a short exact sequence
$$ A \stackrel{\phi}{\hookrightarrow} M \stackrel{\rho}{\twoheadrightarrow} N.$$
From the definition of a $\delta$-functor, and the assumed existence of $f^0$, we then have the following "ladder" with exact rows:
$$
  \require{AMScd}
  \begin{CD}
    T^0(A)
    @>>> T^0(M)
    @>>> T^0(N)
    @> \delta^0 >> T^1(A)
    @>0>> T^1(M)
    \\
    @VV f^0_A V
    @VV f^0_M V
    @VV f^0_N V
    @VV f^1_A V
    \\
    \bar{T}^0(A)
    @>>> \bar{T}^0(M)
    @>>> \bar{T}^0(N)
    @> \bar{\delta}^0 >> \bar{T}^1(A)
    @>>> \bar{T}^1(M)
  \end{CD}
$$
where the morphism $f^1_A$ is uniquely induced by the universal property of the cokernel $T^1(A)$ of $T^0(M) \to T^0(N)$.
My issue lies in showing that such an induced morphism is in fact well-defined for a given $A$. The construction appears to heavily depend on the chosen erasure $\phi$, which would be useless for defining the desired natural transformation $f^1$.
Any help is much appreciated.
 A: I refer to this > https://mathoverflow.net/a/260485/105001 for answering your questions.I just give the details and its idea come from others.
if $u:A\longrightarrow M$ and $v:A\longrightarrow N$ are both monomorphisms and such that $T^1(u)=0=T^1(v)$,we can get $w:A\longrightarrow M\bigoplus N$.(I dont know how to draw commutative diagram with diagonal arrows in stackexchange)
let $i_M:M\longrightarrow M\bigoplus N,p_M:M\bigoplus N\longrightarrow M$ be canonical morphisms and $i_N,p_N$ be canonical morphisms for $N$.
Then $w=i_M \circ u + i_N\circ v$. It is obvious $T^1(w)=0$ for $T^1$ is additive functor and $w$ is monomorphism.
then we can easily prove that $f^1:T^1(A)\longrightarrow \bar{T^1}(A)$ from $u:A\longrightarrow M$ is equal to $f^1:T^1(A)\longrightarrow \bar{T^1}(A)$ from $w:A\longrightarrow M\bigoplus N$.
similarly we can easily prove that $f^1:T^1(A)\longrightarrow \bar{T^1}(A)$ from $v:A\longrightarrow N$ is equal to $f^1:T^1(A)\longrightarrow \bar{T^1}(A)$ from $w:A\longrightarrow M\bigoplus N$.
So $f^1$ is well-defined.
