Is the Derivation Algebra functorial

Question

Suppose $A$ is a commutative, associative $k$-algebra with unit and $Der(A)\subset End_k(A,A)$ is the algebra of derivations on $A$, that is the subalgebra of endomorphisms, such that

$D(ab)=D(a)b+aD(b)$ for all $a,b\in A$ and $D\in Der(A)$.

Is this functorial on the category of commutative, associative $k$-algebras? If yes, how are the appropriate morphisms $Der(f):Der(A)\to Der(B)$ obtained from $f:A\to B$ and if not .. why? What goes wrong?

Answer 1

$\DeclareMathOperator{\Der}{Der}$The right notion here isn't $\Der(A)$, it's $\Der_k(A,M)$. It is functorial on the category $\mathcal{C}$:

• whose objects are pairs $(A,M)$ where $A$ is a $k$-algebra and $M$ is an $A$-module;
• whose morphisms $(A,M) \to (B, N)$ are pairs $(f, \phi)$ where:
  • $f : A \to B$ is a morphism of algebras;
  • $\phi : f^* N \to M$ is a map of $A$-modules (where $f^* N$ has the structure induced by $f$).

The composition of $(f, \phi) : (A, M) \to (B, N)$ and $(g, \psi) : (B,N) \to (C,P)$ is given by $gf$ and the map $(gf)^* P = f^* (g^* P) \xrightarrow{f^* \psi} f^* N \xrightarrow{\phi} M$.

$\Der_k$ is given by: $$\Der{}_k(A,M) = \{ D : A \to M \mid D(ab) = D(a) b + a D(b) \}$$

Then this is a contravariant functor on $\mathcal{C}$: if $(f,\phi) : (A,M) \to (B,N)$, the induced map $\Der(f,\phi) : \Der(B,N) \to \Der(A,M)$ is given by $$\Der(f,\phi)(D)(a) = \phi(D(f(a)))$$