# 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: I have 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: 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.

## 1 Answer

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.

• While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review – Theoretical Economist Sep 17 '18 at 14:18
• @TheoreticalEconomist I have changed my answer but more detailed information need to draw a 3-dimension commutative diagram which I don’t know how to get in stack exchange. – ngegjnjkgn Sep 17 '18 at 15:22