Proposition 17

The derivation $\delta_A$ has the following universal property: for every $(A, A)$-bimodule $E$ and every $K$-derivation $d : A \to E$, there exists one and only one $(A, A)$-bimodule homomorphism $f : I \to E$ such that $d = f \circ \delta_A$.

Note first that, for every $(A, A)$-bimodule homomorphism $f : I \to E$, $f \circ \delta_A$ is a derivation (no. 7, Proposition 9). Conversely, let $d : A \to E$ be a K-derivation; then we prove first that if there exists an $(A, A)$-bimodule homomorphism $f : I \to E$ such that $d = f \circ \delta_A$, $f$ is uniquely determined by this condition for the definition of $\delta_A$ gives

and our assertion follows from the fact that the image of $\delta_A$ already generates I as a left A-module (Lemma 1): hence of necessity

Conversely, as the mapping $(x, y) \mapsto -x \cdot dy$ of $A \times A$ into $E$ is $K$-bilinear, there exists one and only one $K$-linear mapping $g : A \otimes_K A \to E$ such that $g(x \otimes y) = -x \cdot dy$; it suffices to verify that the restriction $f$ of $g$ to $I$ is A-linear for the left and right A-module structures. The first assertion is obvious since $(xx') \cdot dy = x \cdot (x' \cdot dy)$; to prove the second, note that, if $\sum_i x_i \otimes y_i \in I$ and $x \in A$, then

but since $\sum_i x_i y_i = 0$ by definition of $I$, this completes the proof.

We have thus defined a canonical $K$-module isomorphism $f \mapsto f \circ \delta_A$

the left hand side being the $K$-module of $(A, A)$-bimodule homomorphisms of $A$ into $E$.