Recently I've been reading this paper, and in the first section on (differential calculus on associative algebras) they reference a theorem from Bourbaki's Algebra I Chapter 3. I have a problem with the proof of this theorem.
We consider a commutative ring $R$, and an associative $R$-algebra $A$.
Definition. If $\Gamma$ is an $A$-bimodule, then we say a homomorphism $g:A\rightarrow \Gamma$ is a derivation of $A$ into $\Gamma$ if $g(ab)=g(a)b+ag(b)$.
Definition. A first order calculus over $A$ is a pair $(B,d)$ where $d$ is a derivation of $A$ into $B$, and $B$ is the minimal left-$A$-module containing $\operatorname{im}d$.
Then, we define an $A$-bimodule homomorphism $$ \begin{aligned} \mu:A\otimes_R A&\rightarrow A\\ \sum_{i\in I}a_i\otimes b_i&\mapsto \sum_{i\in I}a_i b_i. \\ \end{aligned} $$ We also define an $R$-linear map $$ \begin{aligned} d:A&\longrightarrow \operatorname{ker}\mu\subset A\otimes_R A\\ a&\longmapsto 1\otimes a - a\otimes 1 \end{aligned} $$and it can be shown that $d$ is a derivation of $A$ into $\operatorname{ker}\mu$ (we will call this kernel $I$ from now on).
Then the universal first order calculus over $A$ is the pair $(I,d)$.
What I would like to prove is this:
Theorem. For every first order differential calculus $(\Gamma,g)$, there exists a unique epimorphism $p:I\rightarrow \Gamma$ such that $p\circ d=g$.
The paper references Bourbaki for this theorem, but I don't really follow the proof. The first thing I saw in Bourbaki was the claim "for any $A$-bimodule homomorphism $f:I\rightarrow \Gamma$, $f\circ d$ is a derivation of $A$ into $\Gamma$" - which I don't understand at all. We would have that $$ \begin{aligned} f\circ d(ab)&=f(d(ab))\\ &=f(d(a)b+ad(b))\\ &=f(d(a)b)+f(ad(b))\\ &=f\circ d(a)f(b)+f(a)f\circ d(b) \end{aligned} $$but I don't see the logic behind this being a derivation unless we define the $A$ action using $f$.
Any help would be appreciated, thank you!
The last line of your calculation is incorrect: $f$ is not a map of algebras, it is a map of bimodules. This means that $$ f(ax)=af(x)\;\;\;\mbox{and}\;\;\;f(xa)=f(x)a $$ for all $x\in I$ and $a\in A$. Therefore, the last line in your calculation should be $$ (f\circ d)(ab)=(f\circ d)(a)b+a(f\circ d)(b). $$