Corollary 6.10.505791D2DFFAA

Proof. Suppose first that $f$ is a submersion, and let $a\in X$. Choose holomorphic coordinates $y$ on $Y$ centered at $f(a)$. Since the differential of $f$ at $a$ is surjective, the complex Jacobian matrix at $a$ of the components of $y\circ f$, computed in a chart of $X$, has rank $\dim _{f(a)}Y$. By the holomorphic inverse function theorem (proposition 5.2.4(3)), these components are the first coordinates of a holomorphic chart $(z,w)=(y\circ f,w)$ of $X$ near $a$. In these coordinates $f$ is the projection $(z,w)\mapsto z$. Thus $\varphi \circ f=\varphi \circ p_1+0\circ p_2$, and (6.10.3), together with $\mathcal I(0)=\mathcal O$, gives $\mathcal I(\varphi \circ f)=p_1^{-1}\mathcal I(\varphi )\cdot \mathcal O=f^{-1}\mathcal I(\varphi )\cdot \mathcal O_ X$.

In general, the graph $\Gamma =\{ (x,f(x))\} $ is a complex submanifold of $X\times Y$, since in local coordinates $(x,y)$ the functions $(x,y-f(x))$ are holomorphic coordinates in which $\Gamma =\{ y-f(x)=0\} $; and $\iota (x)=(x,f(x))$ is a biholomorphism of $X$ onto $\Gamma $. The projection $q\colon X\times Y\to Y$ is a submersion, so $\mathcal I(\varphi \circ q)=q^{-1}\mathcal I(\varphi )\cdot \mathcal O_{X\times Y}$ by the first case. Since $\varphi \circ q\circ \iota =\varphi \circ f$, (6.10.2) applied to $\Gamma $ gives $\mathcal I(\varphi \circ f)\subset (q^{-1}\mathcal I(\varphi )\cdot \mathcal O_{X\times Y})|_\Gamma ^{\mathrm{im}}$ after transport by $\iota $. The right side is generated by the functions $g\circ q\circ \iota =g\circ f$ with $g$ a local section of $\mathcal I(\varphi )$, so it equals $f^{-1}\mathcal I(\varphi )\cdot \mathcal O_ X$. $\square$