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Let $\varphi: V \to W$ be a map between projective varieties $V \subset P^n$ and $W \subset \mathbb{P}^m$ given by $\varphi([x_0 : \ldots : x_n]) = [\varphi_0([x_0 : \ldots : x_n]): \ldots : \varphi_m([x_0 : \ldots : x_n])]$, where the $\varphi_i$ are homogeneous polynomials of the same degree that don't vanish simultaneously at any point of $V$.

I would like to show that this is a morphism in the sense of Hartshorne's definition in I.3: a continuous map such that for every open $U \subset W$ and for every regular function $f: U \to k$, the function $f\varphi$ is regular on $\varphi^{-1}(U)$.

How do I do this? Also, can this be generalized? Thanks in advance.

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Alternatively to Jared's solution, if you dehomogenize in any neighborhood you are just getting polynomial maps between affine varieties. Since being a morphism is local, this should be enough. – Alex Youcis Aug 6 '13 at 20:47
up vote 2 down vote accepted

We want to show that for any point $p\in\varphi^{-1}(W)$, there is an open neighborhood $U$ of $p$ such that the composition $f\circ\varphi=h$ on $U$, where $h$ is a well-defined (non-vanishing denominator) rational function of degree $0$. Now, since $f$ is regular, there is an open neighborhood $U'$ around $\varphi(p)\in W$ and a well-defined rational function $g$ of degree $0$ such that $f=g$ on $U'$. Now, let $U=\varphi^{-1}(U')$ and let $h=g\circ\varphi$, noticing that $g\circ\varphi$ is indeed a rational function of degree $0$ because the component functions of $\varphi$ are homogeneous of the same degree.

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Thanks! Very clear. If I'm not mistaken, this also works without modification for quasi-projective $V$ and $W$. – Ricardo Buring Aug 6 '13 at 21:18

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