Reason for this sequence being exact?

Question

In Altman-Kleinman "Introduction to Grothendieck Duality Theory", page 105, there is the following theorem:

Why the underlined sequence is exact? And then, why from that we can conclude that $coker(\delta)$ is the kahler differential?

Answer 1

By definition, $\text{coker }\delta$ fits into the exact sequence $$I/I^2 \overset{\delta}\longrightarrow \Omega^1_{A/k}\otimes_A B \longrightarrow \text{coker }\delta \longrightarrow 0$$ and so by left-exactness of the contravariant functor $\text{Hom}_B(-,M)$ for a $B$-module $M$, we have an exact sequence $$0\longrightarrow \text{Hom}_B(\text{coker }\delta, M)\longrightarrow \text{Hom}_B(\Omega^1_{A/k}\otimes_A B, M)\longrightarrow \text{Hom}_B(I/I^2,M)$$ Now $M$ also has the structure of an $A$-module, via the quotient map $A\to B$. By tensor-Hom adjunction, we have $$\text{Hom}_B(\Omega^1_{A/k}\otimes_A B, M)\cong \text{Hom}_A(\Omega^1_{A/k},\text{Hom}_B(B,M)) \cong \text{Hom}_A(\Omega^1_{A/k},M)$$ (you could also convince yourself of this isomorphism directly). Since $\text{Hom}_A(\Omega^1_{A/k},M) = \text{Der}_k(A,M)$ (by the universal property defining $\Omega^1_{A/k}$ in terms of derivations), this gives you the desired exact sequence.

There is now an isomorphism $$\text{Hom}_B(\text{coker }\delta,M) \cong \text{Der}_k(B,M) \cong \text{Hom}_B(\Omega^1_{B/k},M)$$ for all $B$-modules $M$, functorial in $M$ (check this). The Yoneda Lemma then states that $\text{coker }\delta$ and $\Omega^1_{B/k}$ must be isomorphic as $B$-modules.