Suppose I have a morphism of rings $\varphi: A \to B$, $M$ a $B$-module, $N$ an $A$-module and $f: M \to N$ a homomorphism of $A$-modules (here, $M$ is considered an $A$-module by restriction of scalars). How do I properly "extend" $f$ to a morphism of $B$-modules $F: M \to N \otimes_A B$? You may assume that $A, B$ are fields, and $M, N$ are vector spaces.
I'm asking this because I'm confused about the definition of the map $df_x: T_xX \to T_{f(x)} Y \otimes_{k(f(x))} k(x)$ for a morphism of schemes $f: X \to Y$. I know that we have a $k(f(x))$-linear map $T: \frac{\mathfrak{m}_{f(x)}}{\mathfrak{m}^2_{f(x)}} \to \frac{\mathfrak{m}_x}{\mathfrak{m}^2_x}$ induced from the map of stalks $f^\#_x: \mathcal{O}_{Y, f(x)} \to \mathcal{O}_{X, x}$ - $k(f(x)$ linearity of $T$ follows from the inclusion $k(f(x)) \to k(x)$. Taking the dual of $T$, we have a $k(f(x))$-linear map $T^\lor: T_x X \to T_{f(x)} Y$. How is $df_x: T_xX \to T_{f(x)} Y \otimes_{k(f(x))} k(x)$ defined in terms of $T$ and $T^\lor$? I tried considering the map $T^\lor \otimes_{k(f(x))}Id_{k(x)}$ - but I wasn't sure that $T_xX \otimes_{k(f(x))} k(x) \cong T_xX$.
The dualization here is crucial to the existence of the map in question. The $k(f((x))$-linear map $T:\frac{\mathfrak{m}_{f(x)}}{\mathfrak{m}^2_{f(x)}} \to \frac{\mathfrak{m}_x}{\mathfrak{m}^2_x}$ induces a $k(x)$-linear map $T':\frac{\mathfrak{m}_{f(x)}}{\mathfrak{m}^2_{f(x)}}\otimes_{k(f(x))}k(x) \to \frac{\mathfrak{m}_x}{\mathfrak{m}^2_x}$ by $T'(v\otimes a)=aT(v)$. The map $df_x$ is then defined as the dual of this map $T'$ over $k(x)$, identifying the $k(x)$-dual of $\frac{\mathfrak{m}_{f(x)}}{\mathfrak{m}^2_{f(x)}}\otimes_{k(f(x))}k(x)$ with $\left(\frac{\mathfrak{m}_{f(x)}}{\mathfrak{m}^2_{f(x)}}\right)^\vee\otimes_{k(f(x))}k(x)$ (which only works if either $\frac{\mathfrak{m}_{f(x)}}{\mathfrak{m}^2_{f(x)}}$ or $k(x)$ is finite-dimensional over $k(f(x))$).