Differential of a smooth map in terms of sheaves of deriviations

Question

$\newcommand{\C}{\mathscr{C}^\infty}$$\newcommand{\blank}{{-}}$$\newcommand{\from}{\colon}$$\newcommand{\after}{\circ}$$\newcommand{\Der}{\mathrm{Der}}$Recall that for a smooth manifold $M$, there is an equivalence of categories between vector bundles on $M$ and locally free sheaves of $\C_M$-modules, given by mapping a vector bundle $E \to M$ to the sheaf of sections $\Gamma(\blank, E)$. If $p \from E \to M$ and $p' \from E' \to N$ are two vector bundles and $F \from M \to N$ is a smooth map, then a map of vector bundles $E \to E'$ covering $F$ is by definition a smooth map $\Phi \from E \to E'$ such that $p' \after \Phi = F \after p$ which is linear on the fibers. In terms of sheaves, this corresponds to a morphism $\Gamma(\blank, E) \to F^*\Gamma(\blank,E') := \C_M \otimes_{F^{-1}\C_N} F^{-1} \Gamma(\blank, E')$.

In particular, the tangent bundle $T_M$ corresponds to the sheaf of derivations of the sheaf of algebras $\C_M$. If $F \from M \to N$ is a smooth map of manifolds, then it is well-known that the differential of $M$ defines a map of vector bundles $DF \from T_M \to T_N$ covering $F$. Can this morphism be described in terms of the morphism of sheaves $\Der(\C_M) \to F^*\Der(\C_N)$?

Answer 1

a priori no, because $TM,TN$ (as sheaves) are sheaves on different spaces (live in different categories somehow) so there is no morphism between them, only after pulling back via $F$.
However, if $M \hookrightarrow N$ is a closed immersion, you have a map $F^*\Omega_N=\Omega_N|_M \twoheadrightarrow \Omega_M$ of differential forms. That map you can dualise to get a map $TM \rightarrow TN|_M$ which is often an embedding (for instance if everything is smooth). Note this works because you have two sheaves on the same space (namely $M$). This construction comes from algebraic geometry, hope this still helps!