Are the two morphisms $T^i(C)\to T^{i+1}(A)$ the same for each $i$?

Question

For short exact sequence $0\to A\to B\to C\to 0$,

we have $T^i(C)\to T^{i+1}(A)$ such that the long sequence $$0\to T^0(A)\to T^0(B)\to T^0(C)\to T^1(A)\to \cdots$$ is exact.

For short exact sequence $0\to A\to D\to C\to 0$,

we have $T^i(C)\to T^{i+1}(A)$ such that the long sequence $$0\to T^0(A)\to T^0(D)\to T^0(C)\to T^1(A)\to \cdots$$ is exact.

Are the two morphisms $T^i(C)\to T^{i+1}(A)$ the same for each $i$?

Answer 1

No, in general they will not be the same : when the text says that you have "morphisms $\delta^{i}$ for each short exact sequence", it is understood that the sequence of connecting morphisms $\delta^{i}$ depends on the short exact sequence, and that different short exact sequences could give you different connecting morphisms. Note that by "depending on the short exact sequence", I really mean the two arrows $A\to B\to C$, and not just the central object !

An example where the connecting morphisms would be different is to take two objects $A,C$ such that $Ext^1(C,A)$ is non-trivial, $T_0=Hom_\mathcal{C}(C,\_)$, and $T^{i}$ to be the derived functors of $T_0$. Then two sequences $0\to A\to B\to C\to 0$ that are not isomorphic would give different connecting morphisms $\delta^0:Hom_\mathcal{C}(C,C)\to Ext^1(C,A)$, since the elements $\delta^0(id_C)$ would be different in $Ext^1(C,A)$.

Answer 2

You can check that a cohomological functor defines a functor from the category of short exact sequences in $C$ to the category of long exact sequences in $D$. In this way, different objects (short exact sequences in $C$) goes to (not necessarily) different objects (long exact sequences in $D$).