# Definition of regular map

I am following Shaferevich's Basic Algebraic Geometry 1.Here quasiprojective variety means open subset of a projective closed set. He defines regular maps between quasiprojecive varities (in section 4.2) as locally $f:U\rightarrow \mathbb A_i^m$ is regular. My question is how to define regular map $f:X\rightarrow Y$, where $X$ may be open subset of $\mathbb A^n$ or closed subset of $\mathbb A^n$ or quasiprojective and $Y$ may be open subset of $\mathbb A^n$ or closed subset of $\mathbb A^n$ or quasiprojective. If both $X$ and $Y$ are closed subset of some affine space then I know $f$ is given by polynomials. But what about the other cases.

I know closed subset or open subset of of $\mathbb A^n$ can be regarded as quasiprojective in $\mathbb P^n$ (after identifying $\mathbb A^n$ with one of $\mathbb A_i^n$ in $\mathbb P^n$ and by identifying I mean set theoretically, at least Shaferevich says so[Discussion after Lemma $1.1$ in section $4.1$] ). So to check the map is regular do I need to identify both $X$ and $Y$ to the subsets of $\mathbb P^n$ and $\mathbb P^m$ respectively and then check the corresponding map is regular?

Let me discuss one example:

I need to show $\mathbb A^1-\{0\}$ is isomorphic to $Z(T_1T_2-1)$ in $\mathbb A^2$. Lets identify $\mathbb A^1-\{0\}$ with $X=\{(1:x) :x\in k^* \}\subset\mathbb P^1$ and $Z(T_1T_2-1)$ with $Y=\{(1:x:y):x,y\in k,xy=1\}$ so that both $X$ and $Y$ are quasiprojective varieties. Then define $$$f:X\rightarrow Y\\ (1:x)\mapsto (1:x:\frac{1}{x})$$$

$$$g:Y\rightarrow X\\ (1:x:y)\mapsto (1:x)$$$

If coordinates of $\mathbb P^1$ are $S_0, S_1$ and that of $\mathbb P^2$ are $T_0, T_1, T_2$ then $f$ is given by $1, \frac{S_1}{S_0}, \frac{S_0}{S_1}$ and $g$ is given by $1, \frac{T_1}{T_0}$. Hence $f$ and $g$ are regular and also they are inverse of each other.

Is this method correct or I am making some mistake or doing this in a complicated way?

There isn't really any reason to regard these varieties, one of which is affine and the other which is (a priori) quasi-affine as quasi-projective, you define regular functions in the exact same way. So yes your method is correct but there is no need to use projective spaces or coordinates here. You can simply regard these as subsets of $\mathbb{A}^1$ and $\mathbb{A}^2$, respectively.
Also, your map is well-defined only because you it maps part of a particular affine coordinate chart in $\mathbb{P}^1$ to a particular affine chart in $\mathbb{P}^2$, in general for projective varities you would have to check some gluing conditions.
• Consider the map from $\mathbb{P}^1$ to itself, given by $x\mapsto 1/x$ in the chart $x\neq 0$, and $y\mapsto 1/y$ in the chart $y\neq 0$. This is regular on each chart, but we need to show that it is a well-defined map on $\mathbb{P}^1$, that is, we need to show that the maps "glue together" properly. What this means is that we need to show for a point $[x,y]$ for $x,y\neq 0$, regarding this in either chart and applying the map will give the same result. Feb 10 '17 at 23:00