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I seem to have problems understanding algebraically why given a map of manifolds $f: M \to N$ we get a bundle map $TM \to f^*TN$.

Now, fiberwise it's all good. But I do not understand how to define on sections, as a map of sheaves of derivations.

More to the point, say we have a map of rings $\varphi: B \leftarrow A$ (which I am thinking of opposite to $f$ above, in other words $A$ is global functions on $N$ and $B$ is global functions on $M$). And say that $\varphi$ is over a ground ring $k$ (everything in site is commutative!).

Now, I do not see how to get a map $Der_k(B,B) \to Der_k(A,A) \otimes_A B$ (the latter being the pullback of $TN$ on $M$).

While there is an obvious map $Der_k(B,B) \to Der_k(A,B)$, given by restriction, (which can be used to prove functoriality of the cotangent bundle) I do not see why we should have that $Der_k(A,A)\otimes_A B = Der_k(A,B)$ (although of course it needn't even hold).

On the other hand one might define first the map on cotangent bundles and then declare this one as the transpose (but I'm trying to get grips with which one should be more "natural", in some very unfair sense).

I guess my last comment is important to me when $M$ and $N$ at the beginning are taken to be singular varieties instead of manifolds.

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I don't have a complete answer, but I'll point out that there's also an obvious map $\varphi_* : \text{Der}_k(A, A) \to \text{Der}_k(A, B)$ by postcomposing with $\varphi : A \to B$, and you can use the universal property of $\text{Der}_k(A, A) \otimes_A B$ to obtain a map $\text{Der}_k(A, A) \otimes_A B \to \text{Der}_k(A, B)$. This is probably the isomorphism you seek. – Zhen Lin May 21 '11 at 23:34
@lin: I tried two approaches: first the one you suggested, but I failed to construct an inverse, and second I tried to plug $Der_k(A,B)$ in the corner of the pushout diagram to see if it satisfied the universal property (but I guess this boils down to constructing an inverse). – JoeFlow May 22 '11 at 10:09
This works if $N$ is smooth. Indeed, since maps of sheaves can be defined locally, and $N$ smooth iff $\Omega^1_{A/k}$ locally free, we may assume WLOG that $\Omega^1_{A/k}$ is free. But then $Der(A, A) \otimes B \to Der(A, B)$ is an isomorphism, and so we are done (b/c we already have a natural map $Der(B,B) \to Der(A,B)$). I don't know if the smoothness hypothesis is necessary. – Tom Bachmann Sep 21 '12 at 23:02
@TomBachmann: We may drop the smoothness assumption but we'll have to assume that $B$ is flat over $A$. – Yang Zhou Jan 18 '13 at 7:00

I don't understand the problem. We have a commutative diagram

$$\begin{array}{cccc} T(M) & \xrightarrow{T(f)} & T(N) \\ \downarrow & & \downarrow \\ M & \xrightarrow{f} & N.\end{array}$$

The universal property of the pullback says that this corresponds to a map $T(M) \to f^* T(N)$ of vector bundles over $M$. This holds in the category of manifolds, as well as in the category of schemes (where in the latter case $T(M) = \mathrm{Spec } ~\mathrm{Sym}~ \Omega^1_M$ is just an affine $M$-scheme, not a vector bundle in general).

The corresponding statement in algebra is: If $A \to B$ is a homomorphism of commutative $k$-algebras, then this extends to a homomorphism of rings $\mathrm{Sym }~ \Omega^1_{A/k} \to \mathrm{Sym } ~\Omega^1_{B/k}$, thus to a homomorphism of $B$-algebras $B \otimes_A \mathrm{Sym }~ \Omega^1_{A/k} \to \mathrm{Sym }~ \Omega^1_{B/k}$. Remember that $\mathrm{Spec}$ reverses the arrows.

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