some clearly fact about δ-functor in an abelian category

Question

I am reading Hartshorne's AG and confused about the uniqueness of the universal $\delta$-functor of a functor $F$.

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p206, Remark 1.2.1: If $F$:$\mathfrak{A}\to\mathfrak{B}$ is a covariant functor, then by definition there can exist at most one (up to an isomorphism) universal $\delta$-functor $T$ with $T^0=F$.

I don't really understand “up to an isomorphism”, but I guess it is:

For any short exact sequence $0\to A' \to A \to A'' \to 0 $ in $\mathfrak{A}$, two universal functor $T_1$, $T_2$ of $F$ are isomorphism means there are some natural isomorphisms between $T_1^i$ and $T_2^i$, so I have tried to show that if we give $1_F$ to $F$,the unique sequence $f_1^i:T_1^i \to T_2^i$ and $f_2^i:T_2^i \to T_1^i,i>0$ are isomorphisms in

$$0 \to F(A') \to F(A) \to F(A'') \xrightarrow{\delta _1^0} T_1^1(A') \to \cdots$$ $$0 \to F(A') \to F(A) \to F(A'') \xrightarrow{\delta _2^0} T_2^1(A') \to \cdots$$

Using the commutative diagram at $\delta _1^0$ and $\delta _2^0$, we get $f_1^1 \circ f_2^1 \circ \delta_1^0 =\delta_1^0 $ (and also $f_2^1 \circ f_1^1 \circ \delta_1^0 =\delta_2^0 $).

We need $f_1^1 \circ f_2^1 =1_{T_1^1}$ (and $f_2^1 \circ f_1^1 =1_{T_2^1}$ ), right? But I have no idea about the performance of $f_1^1 \circ f_2^1 $ on $\operatorname{Cok}\delta_1^0$.

Can you help me? Thank you very much for any kind of help!

Answer 1

This is a community wiki answer recording the discussion from the comments, in order that this question might be marked as answered (once this post is upvoted or accepted).

Yeah, the definition of isomorphism here should be that there is a family of natural isomorphisms $\alpha_i:T^i_1→T^i_2$ that respect the $\delta$ morphisms in the sense that $α+{i+1,A′}\circ \delta^i_1=\delta^i_2\circ \alpha_{i,A′′}$ for all short exact sequences $0\to A′\to A \to A′′ \to 0$ and all $i$. – jgon

Then applying universality once to each of $T_1$ and $T_2$, with the identity map from $T^0_1=F=T^0_2$, we get $\alpha_i:T^i_1\to T^i_2$ and $\beta_i:T^i_2→T^i_1$ that commute with the $\delta$ morphisms. Moreover, we have to have $\alpha_i \circ \beta_i=id_{T^i_2}$ and $\beta_i \circ \alpha_i=id_{T^i_1}$ by uniqueness of the morphisms. – jgon

In some sense, the isomorphism follows by very general abstract nonsense about universal objects being unique up to unique isomorphism, and you never need to actually deal with the actual maps or objects in any particular situation. – jgon