Johnson in cohomology of Banach algebra proved the following proposition.
I need to some guidance for the bold part of the following proof. Do you know any papers or book with more details for this part of the proof?


Proposition 8.2. Let $A$ be a commutative amenable Banach algebra and $X$ a commutative Banach $A$-module. Then $H^1(A, X) = 0 = H^2(A, X)$.


Proof. Let $D \in Z^1(A, X)$; then $D \in Z^1(A, X^{∗∗})$ so that $D \in N^1(A, X^{∗∗})$ where this last set
is 0 because $X^{∗∗}$ is a commutative $A$-module. Thus $D = 0 \in N^1(A, X)$ and $H^1(A, X) = 0$.
If $T \in Z^2(A, X)\subseteq Z^2(A, X^{∗∗}) = N^2(A, X^{∗∗})$ then there is $S \in L^1(A,X^{∗∗})$ with $T = δS$.
Let $q$ be the quotient map $X^{∗∗} → X^{∗∗}/X$. 


**??*We then have $δqS = qδS = q(δS − T ) = 0$ so that
$qS \in Z^1(A, X^{∗∗}/X) = 0$ by the result already proved. This shows that $S$ maps into $X$***??


and so
$T = δS \in N^2(A, X)$.



A question about Johnson's theorem on the first and second cohomology of commutative amenable algebras
I will attempt to give an expanded version of Johnson's proof, based on some private unpublished notes I once made (but I claim no originality, I'm sure other people have worked through this themselves).




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The first thing to note is that for any Banach algebra $A$, any quotient map of Banach $A$-bimodules $q: Y \to Z$, and any $n$-cochain $S\in {\mathcal L}^n(A,Y)$, we have the identity
$$
\delta(q\circ S) = q \circ (\delta S)
$$
The proof of this claim is a direct calculation which you should check yourself. In fancier language, $S \to q \circ S$ is a chain map from ${\mathcal L}^\bullet(A,Y)$ to ${\mathcal L}^\bullet(A, Z)$.




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Now, we turn to Proposition 8.2. I follow the notation used in the proof from Johnson's 1972 monograph/paper, with two small modifications: I let $\iota\_X: X \to X^{\*\*}$ denote the canonical embedding, and I let $V$ denote the module $X^{\*\*}/\iota\_X(X)$, so that $q: X^{\*\*} \to V$ is a quotient map of Banach $A$-bimodules.


(1) If $Y$ is any Banach $A$-bimodule and $D\in {\mathcal Z}^1(A,Y)$, then $\iota\_Y\circ D\in {\mathcal Z}^1(A,Y^{\*\*})$. Assuming $A$ is amenable and commutative this means $\iota\_Y\circ D$ is inner; if $Y$ is symmetric then so is $Y^{\*\*}$, hence $\iota\_Y\circ D =0$. Since $\iota\_Y$ is injective, we obtain $D=0$.


(2) Now suppose $A$ is amenable and $X$ is a symmetric Banach $A$-bimodule. Let $T\in {\mathcal Z}^2(A,X)$, then $\iota\_X \circ T \in {\mathcal Z}^2(A,X^{\*\*})$. By amenability, there exists $S\in {\mathcal L}^1(A,X^{\*\*})$ such that $\delta S = \iota\_X \circ T$.


Compose with $q: X^{\*\*} \to V$ on both sides, and note that $q\circ\iota\_X=0$; we obtain (using the "general fact" I mentioned at the start) that
$$ \delta (q\circ S) = q\circ \delta S = q\circ \iota\_X T = 0 $$
and thus $q\circ S \in {\mathcal Z}^1(A, V)$. **By part (1) above** we deduce that $q\circ S =0$. Since $\ker q = {\rm im}\ \iota\_X$, this implies that $S=\iota\_X \circ S\_1$ for some $S\_1\in {\mathcal L}^1(A,X)$. We then have
$$ \iota\_X \circ \delta S\_1 = \delta(\iota\_X \circ S\_1) = \delta S = \iota\_X \circ T $$
and by injectivity of $\iota\_X$ we conclude that $\delta S\_1 = T$. In other words, $T$ is a coboundary, as desired. **Q.E.D.**




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Some further comments:


(a) The assumption that the target module is symmetric is vital. Johnson himself gave examples of commutative amenable $A$ and Banach $A$-bimodules $X$ such that ${\mathcal H}^2(A,X)\neq 0$.


(b) It is a deceptively tricky result of Helemskii that if $A$ is a biprojective Banach algebra (such as $c\_0$ or $\ell\_1$ with pointwise product, or $L^1(G)$ with convolution product where $G$ is a compact group) then ${\mathcal H}^k(A,X)=0$ for all $k\geq 3$ and all Banach $A$-bimodules $X$.


(c) To my knowledge, the following problem remains open:



> 
> Does there exists a commutative amenable Banach algebra $A$ and a symmetric Banach $A$-bimodule $X$ such that ${\mathcal H}^3(A,X)\neq 0$?
> 
> 
> 


