# Pullback map on differential forms

Let $X$ and $Y$ be varieties, and let $\phi:X\rightarrow Y$ be a regular map. Let $x\in X$, and $y=\phi(x)$. Write $\Theta_{X,x}$ and $\Theta_{Y,y}$ for the respective tangent spaces of $X$ and $Y$ at $x$ and $y$. Let $k[X]$ and $k[Y]$ denote respectively the ring of regular functions on $X$ and $Y$, and let $\mathfrak{m}_x$ and $\mathfrak{m}_y$ be the maximal ideals of $k[X]$ and $k[Y]$ at the points $x$ and $y$. Then $\phi$ induces a $k$-algebra homomorphism $\phi^*:k[Y]\rightarrow k[X]$ via right composition with $\phi$. This further induces a homomorphism $(d_x\phi)^*:\mathfrak{m}_y/\mathfrak{m}_y^2\rightarrow\mathfrak{m}_x/\mathfrak{m}_x^2$. Let $d_x:k[X]\rightarrow\Theta_{X,x}^*$ be the map that sends a regular function to the linear part of the Taylor expansion of that function at $x$. Then $d_x$ induces a map into $\mathfrak{m}_x/\mathfrak{m}_x^2$. Actually $d_x$ is an isomorphism from $\mathfrak{m}_x/\mathfrak{m}_x^2$ to $\Theta_{X,x}^*$. Let $d_x\phi$ be the dual map of $d_x\phi^*$. Then the above isomorphism implies that $d_x\phi:\Theta_{X,x}\rightarrow\Theta_{Y,y}$. The question here is: given any $f\in k[Y]$, why is $(d_x\phi)^*(d_yf)=d_x(\phi^*f)$?

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A fundamental property of $\phi ^*$is that for any $f\in k[Y]$ we have $\phi ^*(f)(x)=f(\phi (x))$.
In particular this implies that $\phi^*(\mathfrak m_y)\subset \mathfrak m_x$ and thus that $\phi^*(\mathfrak m_y^2)\subset \mathfrak m_x^2$.
There is thus an induced cotangent map $$d_x(\phi^*):\mathfrak m_y/\mathfrak m_y^2 \to \mathfrak m_x/\mathfrak m_x^2: \bar g\mapsto \overline {\phi^*(g)}\quad (g\in \mathfrak m_y)$$
On the other hand, given a function $f\in k[Y]$, we define $d_y(f)$ as the class $d_yf=\overline {f-f(y)}\in \mathfrak m_y/\mathfrak m_y^2$.
$$d_x(\phi^*)(d_yf)=d_x(\phi^*)(\overline {f-f(y)})=\overline {\phi^*(f-f(y))} =\overline {\phi^*(f)-\phi^*(f)(x)}=d_x(\phi^*(f))$$