Vtyjkyuk

I am attempting to prove Theorem 1.3A in III.1 of Hartshorne's algebraic geometry, which says that $delta$-functors $T=(T^i)_igeq0$ with each $T^i$ effaceable/erasable for $igeq 1$, are universal.

My setup is as follows:
I have let $T$ be an effaceable $delta$-functor as above, and considered another $delta$-functor $barT$ with a natural transformation $f^0:T^0implies barT^0$ between them in the zeroth degree. I want to construct the remaining natural transformations $f^i:T^iimplies barT^i$. 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 $f^i_A:T^i(A) to barT^i(A)$.

I believe I have constructed the first such natural transformation (in an inductively extendable fashion). Letting $A stackrelphihookrightarrow M$ be a $T^1$-erasure of $A$ (i.e. an embedding such that $T^1(phi) = 0$), we can write a short exact sequence
$$ A stackrelphihookrightarrow M stackrelrhotwoheadrightarrow N.$$
From the definition of a $delta$-functor, and the assumed existence of $f^0$, we then have the following "ladder" with exact rows:

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.