I am studying different definitions of the tangent space to a manifold $X$ in a point. When we identify $T_pX \simeq \left(\mathfrak{m}_p/ \mathfrak{m}_p^2\right)^\ast$, how can we express the differential $dF_p$ of a differentiable map $F:(X, p) \to (Y, q)$? The "standard" definition, seeing $T_pX$ as a space of derivations of $\mathcal{C}^\infty_X(p)$, would be $dF_p(\delta)=\delta F^\ast_p$ for every $\delta \in T_pX$.
My idea is that if we call $\alpha_q=\alpha: T_qY \to \left(\mathfrak{m}_q/ \mathfrak{m}_q^2\right)^\ast$ and $\beta_p=\beta: \left(\mathfrak{m}_p/ \mathfrak{m}_p^2\right)^\ast \to T_pX$ the two isomorphism defined by $$\alpha(\delta)(f+ \mathfrak{m}_q^2)=\delta(f), \quad \beta(\omega)(g)=\omega(g-g(p)+\mathfrak{m}_p^2) \quad \forall \delta \in T_qY, f \in \mathfrak {m}_q, g \in \mathcal{C}^\infty_X(p),\omega \in \left(\mathfrak{m}_p/ \mathfrak{m}_p^2\right)^\ast, $$ then the differential $ \left(\mathfrak{m}_p/ \mathfrak{m}_p^2\right)^\ast \to \left(\mathfrak{m}_q/ \mathfrak{m}_q^2\right)^\ast$ should be $\alpha \circ dF_p \circ \beta$, that is, the bottom row of the following diagram:
$\require{AMScd}$ \begin{CD} {T_pX} @>{dF_p}>> T_qY\\ @A{\beta_p}AA @VV{\alpha_q}V\\ \left(\mathfrak{m}_p/ \mathfrak{m}_p^2\right)^\ast @>{??}>> \left(\mathfrak{m}_q/ \mathfrak{m}_q^2\right)^\ast \end{CD}
However I find difficulties finding an explicit expression.
A differentiable map $F : X \to Y$ such that $F(p) = q$ induces a pullback map $\mathfrak{m}_q \to \mathfrak{m}_p$ (the pullback along $F$ of a function vanishing at $q$ vanishes at $p$). This pullback map sends $\mathfrak{m}_q^2$ to $\mathfrak{m}_p^2$ and so induces a map on quotients $\mathfrak{m}_q/\mathfrak{m}_q^2 \to \mathfrak{m}_p/\mathfrak{m}_p^2$, and then you take the linear dual of this map.