The tangent mapping in algebra and algebraic geometry - when does it exist? In differential geometry, for a smooth map $f:M\to N$ one gets the pushforward of the tangent sheaf $T_f:TM\to f^* TN$. Question is does the same work in algebra, that is, if I have a map $f:A\to B$ of commutative algebras, does it yield a map $Tf: Der (B)\to B\otimes_A Der(A)$ without additional requirements like the module of Kahler differentials being dualizable or something like that? I guess in the later case it should be a consequence of the usual map $B\otimes_A \Omega_A\to \Omega_B$.
Edit: I just found a very general solution to the problem: Dual Commutes with Base Change
 A: You may find this in Matsumuras book om commutative rings: If $k\rightarrow A \rightarrow^f B$ are maps of commutative unital rings, there is always a cotangent sequence
$$C1.\text{  } B\otimes_A \Omega^1_{A/k}\rightarrow \Omega^1_{B/k} \rightarrow \Omega^1_{B/A}\rightarrow 0,$$
and dualizing you get the following sequence:
$$ 0 \rightarrow Der_A(B) \rightarrow Der_k(B) \rightarrow Hom_B(B\otimes_A \Omega^1_{A/k},B).$$
If $A\rightarrow B$ is flat there is (Mat, 7.11) an isomorphism
$$Hom_B(B\otimes_A \Omega^1_{A/k},B) \cong B\otimes_A Der_k(A).$$
You get the "tangent sequence"
$$ Tf: Der_k(B) \rightarrow B\otimes_A Der_k(A).$$
The sequence C1 exists in complete generality and for this reason people usually use C1.
Example 1. Let $k$ be an algebraically closed field and let $X:=Spec(B), Y:=Spec(A)$ be non-singular varieties of finite type over $k$ in the sense of Hartshorne Chapter III Proposition 10.4. Assume the induced map $f: X\rightarrow Y$ is smooth of relative dimension $n:=dim(X)-dim(Y)$. Then $f$ is flat and you can define the tangent map. The tangent map $Tf$ is surjective at the level of tangent spaces: If $x\in X$ and $y:=f(x)$ it follows
$$Tf_x: T_x(X) \rightarrow f^*T_y(Y)$$
is surjective. Hence the algebraic geometric situation is similar to the differential geometric situation. In many cases you want to study maps that are non-flat and this is why you use C1 in general.
Note: The map $f$ is smooth of relative dimension $n$ iff $\Omega^1_{B/A}$ is locally free of rank $n$, and then you may dualize. There is a canonical isomorphism of $B$-modules
$$ \Omega^1_{B/A}\cong (\Omega^1_{B/A})^{**}$$
between the relative cotangent module and its double dual. Hence the cotangent module and tangent module determine each other.
