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A homological $\delta$-functor is a collection of additive funcors $T_n:\mathcal A\to \mathcal B$ ($n\geq 0$) with $\delta_{n,A,B,C}:T_n(C)\to T_{n-1}(A)$ defined for each short exact sequence $A\to B\to C$ such that $...\to T_{n+1}(C)\to T_n(A)\to T_n(B)\to T_n(C)\to T_{n-1}(A)\to ...$ is exact. And besides, given any morphism of short exact sequences from $A\to B\to C$ to $A'\to B'\to C'$, it commutes with $\delta$.

A morphism $S\to T$ is a system of natural transformations $S_n\to T_n$ that commute with $\delta$.

A homological $\delta$-functor $T$ is said to be universal if for any homological $\delta$-functor $S$ with natural transform $f_0:T_0\to S_0$, there is unique $f:T\to S$ that extends $f_0$.

I want to show that, when $S$ and $T$ are universal and $f_0$ is an isomorphism between $S_0$ and $T_0$, then there are isomorphisms $f_n$ between $S_n$ and $T_n$ so that $S$ and $T$ are isomorphic.

By the uniqueness of $f$ and $g$ , I see that $f_ng_nf_n=f_n$ and $g_nf_ng_n=g_n$. But I can't see why they are isormophisms. Could someone please tell me which property of universal $\delta$-functor I should use?

Yuz
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  • $g \circ f$ and $f \circ g$ are endomorphisms of universal $\delta$-functors, so they are subject to the same uniqueness condition. Use that. – Zhen Lin May 23 '21 at 01:25
  • Do you mean $g\circ f $ is an endomorphism of $T$ that extends $\text{id}$ at $T_0$, therefore it is $\text{id}$? Thank you! – Yuz May 23 '21 at 01:34
  • That’s right. The same tactic works very generally when you have something defined by a universal property like that. – Zhen Lin May 23 '21 at 02:14

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