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Suppose $T$ is a $\delta$-functor and $T_0=F$ for a exact functor $F$ and $T_i=0$ for $i>0$. The aim of the question is to show that $T$ is universal, meaning if there is another $\delta$-functor $S$ and we have a natural transformation $f_0:S_0\to F$, then there exists unique $\{f_i:S_i\to T_i\}_{i>0}$ such that $\{f\}$ defines a morphism of $S\to T$ of $\delta$-functor.

To show it is a morphism, we will have to check the commutativity for each square of the form enter image description here

where $0\to A\to B\to C\to 0$ is an exact sequence. Obviously, $f_i$ has to be natural transformation to zero functor. Verifying the commutativity of the square above is strict forward for $i\geq 2$.

However, there seems to be a problem with $i=1$. The arrows $S_1(C)\to S_0(A)$ and $S_0(A)\to T_0(A)$ are not necessarily zero and as a result the square is not commutative as going through $S_1(C)\to T_1(C)\to T_0(A)$ is a zero arrow.

Did I missed something or misinterpreted the definition of universal $\delta$-functor?

Thanks in advance for answering.

Ivan So
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  • You haven't use the fact that $F$ is exact. Use it to prove that the composition $S_1(C)\to S_0(A)\to T_0(A)$ is zero as well. – Roland May 19 '21 at 06:06
  • Sorry, but explicitly why would that be helpful? – Ivan So May 19 '21 at 11:39
  • If this composition is zero, the square commutes. Isn't it what bothered you ? – Roland May 19 '21 at 11:44
  • Yes, but I mean why exactness of $T_0$ will implies $S_1(C)\to S_0(A)\to T_0(A)$ is an zero arrow. – Ivan So May 19 '21 at 11:45
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    Compose with this morphism with $T_0(A)\to T_0(B)$. You get : $S_1(C)\to S_0(A)\to T_0(A)\to T_0(B)$. Since $S_0(A)\to T_0(A)\to T_0(B)=S_0(A)\to S_0(B)\to T_0(B)$ and that the composite $S_1(C)\to S_0(A)\to S_0(B)$ is zero., we get that $S_1(C)\to S_0(A)\to T_0(A)\to T_0(B)$ is zero. But since $T_0(A)\to T_0(B)$ is a monomorphism by exactness of $T_0$, it follows that $S_1(C)\to S_0(A)\to T_0(A)$ is zero. – Roland May 19 '21 at 11:50
  • Thanks for your patience! – Ivan So May 19 '21 at 11:54
  • @roland Can you make that an answer? – Pedro May 20 '21 at 14:36

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