I'm trying to solve exercise 5.14 of these notes.
Let $A$ be an algebra over a commutative ring $k$ and $M$ an $A$-$A$-bimodule. Define on the direct sum module $A\oplus M$ the square zero extension algebra given by the product $$ (a, m) (b, n) = (ab, an + mb + \alpha(a, b)) $$ where $\alpha \colon A \times A \to M$ is a bilinear map and considered the induced map $\alpha \colon A \otimes_k A \to M$. For this product to be associative, it is easily verified that $$ aα(b ⊗ c) − α(ab⊗ c) + α(a⊗ bc) − α(a⊗ b)c = 0 \,, $$ which is precisely the condition $\alpha \in \ker(\delta^2)$, where $\delta^2 \colon \hom(A^{\otimes 2}, M) \to \hom(A^{\otimes 3}, M)$ is the Hochschild coboundary operator. Denote this associative algebra by $T_\alpha(A \oplus M)$. The exercise asks the following.
Let $α, α' ∈ \ker(δ^2)$. Show that there is an isomorphism of algebras $T_α(A ⊕ M) \cong T_{α'}(A ⊕ M)$ which induces the identity on the ideal $M$ and on the quotient $A$ if and only if $α$ and $α'$ determine the same class in $\mathrm{HH}^2(A; M)$ (the Hochschild cohomology of $A$ with coefficients in $M$).
First of all, I’m very confused by the fact that the isomorphism induces the identity on $A$ and $M$, because that means that it is the identity on $A \oplus M$ so we would have $\alpha = \alpha'$.
On the other hand, if I forget about that and consider an algebra homomorphism $F$, after imposing that products map to products what I get is $$ F(ab) = F(a) F(b) $$ for all $a, b ∈ A$, and $$ F(an) + F(mb) + F\alpha(a,b) = F(a) F(n) + F(m) F(b) + \alpha'(F(a), F(b)) $$
Now I don’t know how to use the fact that $F$ is an isomorphism.
Considering the converse implication, the fact that $α$ and $α'$ determine the same class in $\mathrm{HH}^2(A; M)$ means that $\alpha - \alpha' = \delta^1(G)$ for some $G \colon A \to M$, but I don’t really know how to use this fact to define a homomorphism.
EDIT. The equalities above come from a mistake, so they’re not true in general.
Imposing the homomorphism condition we have $$ F((a, m) (a, n)) = F(ab, an + mb + \alpha(a,b)) = F(a, m) F(b, n) = (a, m') (b, n') \,. $$ The last equality comes from the fact that $F$ induces the identity on the quotient $A$. If $a = 0$ or $b = 0$ then, using that $F$ is the identity on $M$ I get $$ an + mb + \alpha(a, b) = an' + m'b + \alpha'(a, b) \,. $$ Thus $$ \alpha(a, b) - \alpha'(a, b) = a(n' - n) + (m' - m)b \,. $$
EDIT. For the converse I’ve thought of defining the map $(a, m) \mapsto (a, m + G(a))$. The fact that this is an algebra homomorphism is equivalent to $\alpha(a, b) - \alpha'(a, b) = a G(b) - G(ab) + G(a) b$. This map is an isomorphism and induces the required identities. Therefore I think I have this implication.
Isomorphisms $\psi$ which are the identity on $M$ are determined by $A\to A\oplus M$, and inducing identity on the quotient means identified with a map $\phi: A\to M$.
so compute that its an algebra isomorphism two ways:
$$\psi(a\oplus 0 \cdot_1 b\oplus 0) = \psi(ab\oplus \delta_1(a,b)) = ab\oplus (\phi(ab) +\delta_1(a,b))$$
$$\psi(a\oplus0)\cdot_2\psi(b\oplus0) = (a\oplus \phi(a))\cdot_2(b\oplus \phi(b)) = ab\oplus (a\phi(b) + \phi(a)b+ \delta_2(a,b))$$
Equating the two, $\phi$ is coboundary between $\delta_1$ and $\delta_2$