Reduction of tangent space of abelian variety Let $A$ be an abelian variety defined over a number field $K$ with good reduction everywhere. Let $\mathcal P$ be a prime of $K$. 
Does reduction of $A$ modulo $\mathcal P$ induce a well-defined reduction map from the tangent space of $A$ at identity to the tangent space at identity of the reduced variety?
 A: The only natural thing I can think of is the following:
Let $R_{\mathcal{P}}$ be the ring of integers of $K_{\mathcal{P}}$ (the completion at the prime).
Let $S$ be the spectrum of $R_{\mathcal{P}}$.
Since $A$ has good reduction, there is an abelian scheme (the Néron model) $\mathcal{A}$ over $S$, with $A$ as generic fibre, and the reduction as special fibre (which we will denote $A_{0}$.
Now look at maps $S[\varepsilon] \to \mathcal{A}$ that restrict to the identity section $e \colon S \to \mathcal{A}$, where $S[\varepsilon]$ is $S \times \mathrm{Spec}(\mathbb{Z}[\varepsilon]/(\varepsilon^{2}))$. Such maps $S[\varepsilon] \to \mathcal{A}$ restrict to tangent vectors on the generic fibre and the special fibre. So you get maps
$$ T_{e}A \longleftarrow \mathrm{Hom}_{S}(S[\varepsilon],\mathcal{A}) \longrightarrow T_{e}A_{0}. $$

Like René observed below in the comments:

the middle term is a tangent space in its own right (and a free $R_{\mathcal{P}}$-module of rank $\dim A$), which naturally embeds into $T_{e}A$ and has a reduction map to $T_{e}A_{0}$.

A: There is no such natural map. 
Note that both the tangent spaces you are talking about are vector spaces over different fields. 
