Suppose we have an exact sequence $0 \to L \to M \to N \to 0$ and a morphism $f \colon A \to B$ of $R$-modules. If $\delta \colon \text{Ext}^{i}_{R}(B,N) \to \text{Ext}^{i+1}_{R}(B,L)$ and $\delta' \colon \text{Ext}^{i}_{R}(A,N) \to \text{Ext}^{i+1}_{R}(A,L)$ are the natural connecting homomorphisms, I'm trying to figure out why
$(*)$ $ \delta' \circ \text{Ext}^{i}(f,N) = \text{Ext}^{i+1}(f,L) \circ \delta? $
My confusion:
If we let $T^{i} = \text{Ext}^{i}(B,-)$ and $U^{i} = \text{Ext}^{i}(A,-)$ we know there are unique natural transformations $\psi^{i} \colon T^{i} \to U^{i}$ such that $\delta' \circ \psi^{i}(N) = \psi^{i+1}(L) \circ \delta$ for each $i \geq 0$ and $\psi^{0} = \text{Hom}(f,-)$. (Northcott, An introduction to homological algebra, pp. 115, theorem 10).
So this would solve my question if I knew that $\psi^{i}(C) = \text{Ext}^{i}(f,C)$ for all $R$-modules $C$. Now this is probably true.. but I don't see why.
The maps $\text{Ext}^{i}(f,C)$ (as I know) are defined by taking injective resolutions of $B$ and $A$ , then taking the chain map induced by $f$ and applying $\text{Hom}(-,C)$ and then cohomology.. now if we define them this way... why would they commute with the connecting homomorphism as in $(*)$?
This follows from the naturality of the connecting homomorphism of a short exact sequence of complexes (homological convention) : if $$ \begin{matrix} 0\longrightarrow &K_{\bullet}&\longrightarrow &L_{\bullet}&\longrightarrow &M_{\bullet}&\longrightarrow 0\\ &~\downarrow\kappa&&~\downarrow\lambda&&~\downarrow\mu&\\ 0\longrightarrow &K'_{\bullet}&\longrightarrow &L'_{\bullet}&\longrightarrow &M'_{\bullet}&\longrightarrow 0 \end{matrix}$$ are two short exact sequences of complexes, and $\kappa,\lambda,\mu$ are chain maps, then the diagram below commutes $$\begin{matrix} \cdots&\longrightarrow&H_{n+1}(M_{\bullet})&\xrightarrow{\delta}&H_n(K_{\bullet})&\longrightarrow&H_n(L_{\bullet})&\longrightarrow&H_n(M_{\bullet})&\xrightarrow{\delta}& H_{n-1}(K_{\bullet})&\longrightarrow&\cdots\\ &&\downarrow\mu&&\downarrow\kappa&&\downarrow\lambda&&\downarrow\mu&&\downarrow\kappa\\ \cdots&\longrightarrow&H_{n+1}(M'_{\bullet})&\xrightarrow{\delta}&H_n(K'_{\bullet})&\longrightarrow&H_n(L'_{\bullet})&\longrightarrow&H_n(M'_{\bullet})&\xrightarrow{\delta}& H_{n-1}(K'_{\bullet})&\longrightarrow&\cdots \end{matrix}$$ The proof is easy, you check that everything commutes using the construction of $\delta$ on the chain level. The same exact statement holds for morphisms between short exact sequences of cochain complexes (cohomological convention).
The $\text{Ext}$ long exact sequences arise in this way: you take injective resolutions $I^{\bullet}_X$ of $X=L,M,N$ that form a short exact sequences of cochain complexes $$ 0\longrightarrow I^{\bullet}_L\longrightarrow I^{\bullet}_M\longrightarrow I^{\bullet}_N\longrightarrow 0 $$ You apply the functors $\mathrm{Hom}(A,-)$ and $\mathrm{Hom}(B,-)$ and get two short exact sequences of complexes that are linked by the natural transformation $f^*:\mathrm{Hom}(B,-)\xrightarrow{-\circ f}\mathrm{Hom}(A,-)$ $$ \begin{matrix} 0\longrightarrow &\mathrm{Hom}(B,I^{\bullet}_L)&\longrightarrow &\mathrm{Hom}(B,I^{\bullet}_M)&\longrightarrow &\mathrm{Hom}(B,I^{\bullet}_N)&\longrightarrow 0\\ &~\downarrow f^*&&~\downarrow f^*&&~\downarrow f^*&\\ 0\longrightarrow &\mathrm{Hom}(A,I^{\bullet}_L)&\longrightarrow &\mathrm{Hom}(A,I^{\bullet}_M)&\longrightarrow &\mathrm{Hom}(A,I^{\bullet}_N)&\longrightarrow 0 \end{matrix} $$ The cohomologies are by definition the $\text{Ext}$ groups you are interested in, and from what precedes we get the commutation of the maps induced by $f$ and the connecting homomorphisms.