3.2 Regular functions and morphisms

Regular functions in topological spaces

We want to consider functions defined on topological spaces. In theory we could define any functions. However in a topological space we want them to be consistent with the topology defined. In particular, we want to be able to tell what functions are defined on each open subset and we want to have a mechanics of projecting a function from open set to it's subset. This brings us to the definition of pre-sheaf

def: Pre-sheaf, pre-ringed space

$$\begin{align*} &\sphericalangle \\ &(X, T) \in \mathcal T(S) \\ &\mathcal O_X:T^o \to \mathcal R, U \mapsto \mathcal O_X(U) \\ &\mathcal O_X(\empty) = \{0\} \\ &\forall U,V \in T^o, U \subseteq V: \\ &\exists \rho_{V, U}: \mathcal O_X(V) \rightsquigarrow_R \mathcal O_X(U), f \mapsto f_{|U} \\ &\rho_{U,U} = \text{id} \\ &W \in T^o, U \subseteq V \subseteq W \implies \rho_{W,U}= \rho_{V,U} \circ \rho_{W,V}\\ \hline \\ &(S, T, \mathcal O_X) \in \mathcal R^{sp}_{pre} \\ &\mathcal O_X - \text{pre-sheaf} \\ &\rho - \text{restriction} \end{align*}$$

The pre-sheaf describes the set of functions that are having some global property. For example a set of constant functions. But we want to be more flexible: define functions locally and then make sure that the function that is defined by gluing local definitions is still a viable function. Here's the definition that allows to do just that:

def: Sheaf, ringed space

$$\begin{align*} &\sphericalangle \\ &(X, T, \mathcal O_X) \in \mathcal R^{sp}_{pre} \\ &\forall U, U_i \in T^o, f_i \in \mathcal O_X(U_i): U = \bigcup_i U_i, f_{i|U_i \cap U_j}=f_{j|U_i \cap U_j} \implies \\ &\exists! f \in \mathcal O_X(U): f_{|U_i}=f_i \\ \hline \\ &(X, T, \mathcal O_X) \in \mathcal R^{sp}- \text{ ringed space} \\ &\mathcal O_X - \text{ sheaf } \\ &f \in \mathcal O_X - \text{ section } \end{align*}$$

Example: Pre-sheaf and Sheaf

Consider a ring of constant functions on $\mathbb R$. Then $\rho$ in this case is trivial so it's a pre-sheaf. However it's not a sheaf. Consider functions $f_1 = 1$, $f_2 = 2$ and open sets $U_1:=(0, 1), U_2:= (1, 2), U = U_1 \cup U_2$ (open in Euclidean topology). Then trivially $f_{1|U_1 \cap U_2}=f_{2|U_1 \cap U_2}$ since $U_1 \cap U_2=\empty$ but there's no function in the pre-sheaf that equals $1$ on $U_1$ and $2$ on $U_2$ since functions are globally constant.

To make this pre-sheaf a sheaf in a ringed space we add a piecewise constant functions on open sets.

Finally we'd like to define local rings on topology, that is rings of functions defined locally in the neighborhood of some point.

def: Stalk at point P

$$\begin{align*} &\sphericalangle \\ &(X, T, \mathcal O_X) \in \mathcal R^{sp} \\ &P \in X \\ &\forall U, V \in T^o, P \in U \cap V, f \in \mathcal O_X(U), g \in \mathcal O_X(V): \\ &(U, f) \sim_s (V, g) \iff \exists W \in T^o, P \in W, W \subseteq U \cap V: f_{|W}=g_{|W} \\ \hline \\ &\mathcal O_{X, P}:=\{[U, f]_{\sim s}, U \in T^o, P \in U, f \in \mathcal O_X(U)\} - \text{stalk} \\ &f \in \mathcal O_{X, P} - \text{germ} \end{align*}$$

Morphisms in topological spaces

Addinitionally we'd like to define an analogue of homomorphisms (continuous functions) and isomorphisms (homeomorphisms) for topological spaces.

def: Continous function

$$\begin{align*} &\sphericalangle \\ &(X_1, T_1) \in \mathcal T \\ &(X_2, T_2) \in \mathcal T \\ &f: X_1 \to X_2 \\ &X \in T_2^c \implies f^{-1}(X) \in T_1^c \\ \hline \\ &f: X_1 \rightsquigarrow_T X_2 - \text{continous function} \end{align*}$$

def: Homeomorphism

$$\begin{align*} &\sphericalangle \\ &(X_1, T_1) \in \mathcal T \\ &(X_2, T_2) \in \mathcal T \\ &f: X_1 \leftrightarrow X_2 \\ &f: X_1 \rightsquigarrow_T X_2 \\ &f^{-1}: X_2 \rightsquigarrow_T X_1 \\ \hline \\ &f: X_1 \cong_T X_2 \end{align*}$$

So now we defined homomorphisms of topological spaces. What about ringed spaces?

def: Pullback operator

$$\begin{align*} &\sphericalangle \\ &(X_1, T_1) \in \mathcal T \\ &(X_2, T_2) \in \mathcal T \\ &\phi: X_1 \rightsquigarrow_T X_2 \\ &U \in T_2^o \\ &f: U \to \overline F \\ \hline \\ &\phi^*f: \phi^{-1}(U) \to \overline F, f \to f \circ \phi \\ &\phi^* - \text{pullback operator} \end{align*}$$

def: Ringed space morphism

$$\begin{align*} &\sphericalangle \\ &(X_1, T_1, \mathcal O_{X_1}) \in \mathcal R^{sp} \\ &(X_2, T_2, \mathcal O_{X_2}) \in \mathcal R^{sp} \\ &\phi: X_1 \rightsquigarrow_T X_2 \\ &U \in T_2^o \implies \phi^*(\mathcal O_{X_2}(U)) \subseteq \mathcal O_{X_1}(\phi^{-1}(U)) \\ \hline \\ &\phi: X_1 \rightsquigarrow_{Rsp} X_2 \\ \end{align*}$$

def: Ringed space isomorphism

$$\begin{align*} &\sphericalangle \\ &(X_1, T_1, \mathcal O_{X_1}) \in \mathcal R^{sp} \\ &(X_2, T_2, \mathcal O_{X_2}) \in \mathcal R^{sp} \\ &\phi: X_1 \rightsquigarrow_{Rsp} X_2 \\ &\exists \phi^{-1}: X_2 \rightsquigarrow_{Rsp} X_1 \\ &\phi \circ \phi^{-1} = \text{id}_{X_1} \\ &\phi^{-1} \circ \phi = \text{id}_{X_1} \\ \hline \\ &\phi: X_1 \cong_{Rsp} X_2 \\ \end{align*}$$

Affine regular functions

First we want to prove that any polynomial $\mathbb A^n \to \mathbb A^1$ is continous in Zariski topology.

Proposition 3.2.1: Polynomial is continuous

$$\begin{align*} &\sphericalangle \\ &\mathbb A^n \\ &f \in \overline F[x_1, \ldots, x_n] \\ \hline \\ &f: \mathbb A^n \rightsquigarrow_T \mathbb A^1 \end{align*}$$

Proof

Consider closed set $X \subseteq \mathbb A^1$ with ideal $I(X)$. Then $f^{-1}(X)=\{x \in \mathbb A^n: f(x) \in X\}$. But $f(x) \in X \iff \forall g \in I(X): g(f(x)) = 0$. In other words if $I(X)=\lang g_1, \ldots, g_n \rang_I$ then $I(f^{-1}(X))=\lang g_1 \circ f, \ldots, g_n \circ f \rang_I$. Since $g_i \circ f$ is a polynomial the set $f^{-1}(X)$ is closed.

$\square$

Note that polynomial $f/g$ will be also continous on open sets where $g(x) \ne 0$. Since we're dealing with polynomials in algebraic geometry using it's very natural to define sheaf in the following way:

def: Affine sheaf

$$\begin{align*} &\sphericalangle \\ &\mathbb A^n \\ &(X, T) \in \mathcal T^{c, -}(X) \\ &U \in T^o \\ \hline \\ &\mathcal O_X(U):=\{\phi: U \to \overline F: \exists f, g \in \overline F[x_1, \ldots, x_n]: \forall Q \in U: \phi(Q)= \frac{f(Q)}{g(Q)}, g(Q) \ne 0\} \end{align*}$$

Using this definition we obviously have stalk defined as well. While these definitions work well they're sometimes sowewhat hard for analyzing functions in affine space. We'd like to have an easiler definition. Further we'll describe regular functions on a set $U$ that will correspond to affine sheaf on $U$ and regular functions at point $P$ that will macth affine stalk at point $P$.

We start by considering polynomials $\overline F[x_1, \ldots, x_n]$ on $\mathbb A^n$ and their behavior on subspaces.

def: Affine coordinate ring

$$\begin{align*} &\sphericalangle \\ &\mathbb A^n \\ &(X, T) \in \mathcal T^{c, -}(X) \\ \hline \\ &A(X):=\overline F[x_1, \ldots, x_n] / I(X) \\ &A(X) - \text{coordinate ring} \end{align*}$$

note

Since irreducible closed sets correspond to prime vanishing ideals, $A(X) \in \mathcal R^{\mathcal {ID}}$.

Consider $p, q \in A(X)$ then $p_{|X}=q_{|X} \iff (p-q)_{|X}=0 \iff p-q \in I(X)$. Thus coordinate ring is a class of equivalence of polynomial functions.

Example: Coordinate ring for elliptic curve

Consider $(\mathbb A^n, T^c \in \mathcal Z^c(\mathbb A^n)), \overline F = \mathbb C$. Define $X:=Z(\lang y^2-x^3+x \rang_I)$. Then $A(X)=\mathbb C[x, y]/\lang y^2-x^3+x \rang_I$. Note that if we consider a restriction of functions $f, g \in \mathbb C[x, y]$ to $X \to \mathbb C$ then there's no difference between $f = y^2-x^3+x+2$ and $g=2$ in these restrictions (they have the exact same values on all of $X$). Accordingly $f$ and $g$ lie in the same coset in $A(X)$.

def: Function field

$$\begin{align*} &\sphericalangle \\ &(X, T) \in \mathcal T^{c, -}(X) \\ &A(X):=\overline F[x_1, \ldots, x_n] / I(X) \implies A(X) \in \mathcal R^{\mathcal {ID}} \\ \hline \\ &F(X):=\mathfrak F(A(X)) \end{align*}$$

Now when we have the definition of polynomial functions we can go further and define locally regular functions:

def: Regular functions at point P

$$\begin{align*} &\sphericalangle \\ &\mathbb A^n \\ &(X, T) \in \mathcal T^{c,-}(X) \\ &\mathcal O^r_{X, P}:=\{\frac{f}{g} \in F(X), g(P) \ne 0\} \\ \hline \\ &f \in \mathcal O^r_{X, P} - \text{regular function on } X \text{ at } P \\ \end{align*}$$

note

Consider ideal $\mathfrak m_{X, P}:=\{f \in A(X), f(P)=0\}$. It's easy to see that it's the kernel evaluation homomorphism $\varepsilon_P$ so $A(X)/\mathfrak m_{X, P}\cong_R \overline F$ and so $\mathfrak m_{X, P}$ is a maximal ideal. So we can make another definition of regular function at point using localization:

$$\mathcal O^r_{X, P}:=A(X)_{\mathfrak m_{X, P}}$$

Having this definition we can trivially define regular function on any open set as:

def: Regular function on open set

$$\begin{align*} &\sphericalangle \\ &\mathbb A^n \\ &(X, T) \in \mathcal T^{c,-}(X) \\ &\mathcal O^r_X(U):=\bigcap_{P \in U}\mathcal O^r_{X,P} \\ \hline \\ &f \in \mathcal O^r_{X}(U) - \text{regular function on } X \text{ on } U \\ \end{align*}$$

note

Another way of defining regular functions is $\mathcal O^r_{X}(U):=\{\frac{f}{g} \in \mathfrak F(A(X)), \forall P \in U: g(P) \ne 0\}$

note

Since $X \in T^o$ we have $\mathcal O^r_{X}(X)$ also defined.

Example: Glued regular function

As in the example with piecewise constant functions not any regular function can be expressed as a quotient of polynomials globally. Consider $X=Z(x_1x_4 - x_2x_3)$ and open set $U=U_1 \cup U_2, U_1 = X \cap \{x_2 \ne 0\}, U_2 = X \cap \{x_4 \ne 0\}$. Consider $f=\frac{x_1}{x_2}$ on $U_1$ and $f = \frac{x_3}{x_4}$ on $U_2$. These two functions ($\frac{x_1}{x_2},\frac{x_3}{x_4}$) coincide on $X$ as long as they're defined so $f$ is regular. On the other hand it cannot be represented simply as a quotient of polynomials.

Proposition 3.2.2: Regular function at distinguished subsets is a coordinate ring localization

$$\begin{align*} &\sphericalangle \\ &\mathbb A^n \\ &(X, T) \in \mathcal T^{c, -}(X) \\ &f \in A(X) \\ &X_f:=X \setminus Z(f) = \{P \in X: f(P) \ne 0\} \\ &A(X)_f:=\{\frac{g}{f^r}, g\in A(x), r \ge 0\} \\ \hline \\ &\begin{align*} & \mathcal O_X(X_f) \cong_R A(X)_f \hspace{0.5cm} \tag{a}\\ & \mathcal O_X(X) \cong_R A(X) \hspace{0.5cm} \tag{b}\\ \end{align*} \end{align*}$$

Proof

a.

$\impliedby$

If $\phi \in A(X)_f$ then $\phi=\frac{g}{f^r}, g \in A(X)$. Since $f^r \in A(X)$ as well and $f^r \ne 0$ on $X_f$ so $\phi \in \mathcal O_X(X_f)$.

$\implies$

Consider $\phi \in \mathcal O_X(X_f)$ and denote

$$J:=\{g \in A(X): g\phi \in A(X)\}$$

Note that this is quotient ideal $(A(X): \phi)$, in particular it's and ideal, i.e. $J \lhd_R A(X)$. If we prove that $f^r \in J$ for some $r$ then we're done.

Consider arbitrary $P \in X_f$, we know that $\phi \in \mathcal O^r_{X,P}$ so $\phi = \frac{h}{g}$ with $g \ne 0$ in a neighborhood of $P$. By definition of $J$ it means that $g \in J$ so $J$ contains some function $g: g(P) \ne 0$. That means that $P \notin Z(J) \cap X = Z(I(X)+J)$ or $P \in X \setminus Z(I(X)+J)$. Since we took arbitrary $P$ it means that $X_f \subseteq X \setminus Z(I(X)+J)$ or equivalently $Z(I(X)+J) \subseteq Z(f)$. This implies $\sqrt{\lang f \rang_I}=I(Z(f))\subseteq I(Z(I(X)+J))=\sqrt{I(X)+J}$. Note that $I(X) = 0$ in $A(X)$ so we have $\exists r: f^r \in J$

b.

Follows from $(a)$ by taking $f=1$

$\square$

note

It might seem surprising that a set of regular functions which is $f/g$ where $f, g$ are polynomials is isomorphic to just polynomials (coordinate ring). However if you consider a case of $\mathbb A^n$ for example. It's clear that there could be no $f/g$ with $\forall Q \in \mathbb A^n: g(Q) \ne 0$ since $g$ has always roots in algebraically closed $\overline F$. So we really only having plain polynomials as regular functions.

So we defined the regular functions in $\mathbb A^n$, next we want to relate this definition to sheaf and stalks that we defined in a topological space earlier.

Using this definition the stalk for affine space is already defined. Now let's prove that germ is the same as regular function at point $P$:

Proposition 3.2.3: Stalk at P is the set of regular functions at P

$$\begin{align*} &\sphericalangle \\ &\mathbb A^n \\ &(X, T) \in \mathcal T \\ &P \in X \\ &U \in T^o \\ \hline \\ &\begin{align*} &\mathcal O^r_{X,P} \cong_R \mathcal O_{X,P} \hspace{0.5cm} \tag{a}\\ &\mathcal O^r_{X}(U) \cong_R \mathcal O_{X}(U) \hspace{0.5cm} \tag{b}\\ \end{align*} & \end{align*}$$

Proof

a.

We'll estabish a bijective mapping between two sets, proving isomorphism is left as an exercise.

$\implies$

Consider $\phi \in \mathcal O^r_{X,P}$ it means that $\phi = \frac{f}{g}, g(P) \ne 0$. A set $X_g=\{Q \in X: g(Q) \ne 0\}$. We'll map this function to the germ $[X_g, \phi']_{\sim s}$ where $\phi'(Q)=f(Q)/g(Q)$. Note that $X_g \in T_o, P \in X_g,\phi' \in \mathcal O_X(X_g)$ as required. Since $\frac{f}{g}$ is the same on any open set and $\frac{af}{ag}=\frac{f}{g}$ as a polynomials and pointwise, this mapping is well-defined.

$\impliedby$

Take some $\phi \in \mathcal O_{X,P}$ and consider $(U, \frac{f}{g}) \sim_s (U', \frac{f'}{g'})$. We map this germ to $\frac{f}{g}$ and we want to prove that in this case $\frac{f}{g}=\frac{f'}{g'}$.

$(U, \frac{f}{g}) \sim_s (U', \frac{f'}{g'})$ means that $\exists W \subseteq U \cap U': \forall Q \in W:\frac{f}{g}(Q)=\frac{f'}{g'}(Q)$ or $\forall Q \in W: f'g(Q)=fg'(Q)$. Consider $Z(f'g-fg')$. We know that this is a closed set and $W \subseteq Z(f'g-fg')$. But by $(3.1.1)$ we have $\overline W = X$ so $Z(f'g-fg')=X$ and thus $f'g=fg'$ on $X$ or equivalently $f'g-fg' \in I(X)$ which means $f'g=fg'$ in $A(X)$.

b.

Trivially follows from $(a)$

$\square$

Projective regular functions

As in affine case let's start by defining sheaf of functions:

def: Projective sheaf

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &(X, T) \in \mathcal T^{c, -}(X) \\ &U \in T^o \\ \hline \\ &\mathcal O_X(U):=\{\phi: U \to \overline F: \exists d \ge 0, f, g \in \overline F[x_0, \ldots, x_n]_d: \forall Q \in U: \\ &\phi(Q)= \frac{f(Q)}{g(Q)}, g(Q) \ne 0\} \end{align*}$$

note

We require $f$ and $g$ to have the same homogeneous degree, otherwise their ratio is not well-defined (depends on the representative of projective point)

Next, we'll define regular functions

def: Projective coordinate ring

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &(X, T) \in \mathcal T^{c, -}(X) \\ \hline \\ &S(X):=\overline F[x_0, \ldots, x_n] / I(X) \\ &S(X) - \text{projective coordinate ring} \end{align*}$$

note

Unlike in affine case, projective coordinate rings doesn't specify valid functions on the sets of $X$. This is because these functions are not well-defined

note

Just like in the previous section we consider $S(X)$ as a graded ring $S(X)=\bigoplus_{d \ge 0}S(X)_d$.

def: Projective regular functions at point P

$$\begin{align*} &\sphericalangle \\ &\mathbb A^n \\ &(X, T) \in \mathcal T^{c,-}(X) \\ &\mathcal O^r_{X, P}:=\{\frac{f}{g}, f, g \in S(X)_d, g(P) \ne 0\} \\ \hline \\ &f \in \mathcal O^r_{X, P} - \text{regular function on } X \text{ at } P \\ \end{align*}$$

Regular functions on a set is defined in the same way as in affine case.

Note that in case of projective regular functions $\mathcal O^r_X(X)=\overline F$. Why? We can use the same reasoning as in affine case. The denominator must be constant, otherwise it will have a root in $X$ and thus will be undefined. But since the numerator degree is the same as denominator degree it's also constant.

Morphisms

We already know what is a ringed space morphism. Let's formulate criteria that are true in affine space.

Proposition 3.2.4: Affine ringed spaces morphisms criteria

$$\begin{align*} &\sphericalangle \\ &(X_1, T_1, \mathcal O_{X_1}) \in \mathcal R^{sp} \\ &(X_2, T_2, \mathcal O_{X_2}) \in \mathcal R^{sp} \\ &\phi: X_1 \rightsquigarrow_T X_2 \\ \hline \\ &\text{The following are equivalent:}\\ &\begin{align*} & U \in T_2^o \implies \phi^*(\mathcal O_{X_2}(U)) \subseteq \mathcal O_{X_1}(\phi^{-1}(U)) \hspace{0.5cm} \tag{a}\\ & \phi^*(\mathcal O_{X_2}(X_2)) \subseteq \mathcal O_{X_1}(X_1) \hspace{0.5cm} \tag{b}\\ & \forall P \in X_1: \phi^*(\mathcal O_{X_2, \phi(P)}) \subseteq \mathcal O_{X_1, P}\hspace{0.5cm} \tag{c}\\ \end{align*} \end{align*}$$

Proof

$(a) \implies (b)$

Trivial, take $U=X_2$

$(b) \implies (c)$

Take $f \in \mathcal O_{X_2, \phi(P)}$. Since by $(3.2.2): \mathcal O_{X_2, \phi(P)} \cong \mathcal O^c_{X_2, \phi(P)}$ we have $f = g/h, g, h \in A(X_2)=\mathcal O^c_{X_2}(X_2)=\mathcal O_{X_2}(X_2), h(f(P)) \ne 0$. Since $g, h \in \mathcal O_{X_2}(X_2)$ we know that $\phi^*f, \phi^*g$ are in $O_{X_1}(X_1)=O^c_{X_1}(X_1)=A(X_1)$ so because $\phi^*h(P)\ne 0$, we have $\phi^*f=\frac{\phi^*g}{\phi^*h} \in \mathcal O^c_{X_1,P}=\mathcal O_{X_1,P}$

$(c) \implies (a)$ Follows immediately from $\mathcal O_X(U)=\bigcap_{P \in U}\mathcal O_{X,P}$

$\square$

Consider a morphism $\phi: X_1 \to X_2$ where $X_1 \subseteq \mathbb A^n$, $X_2 \subseteq \mathbb A^k$. Then we can describe $\phi: (x_1, \ldots, x_n) \mapsto (\phi_1(x_1, \ldots, x_n), \ldots, \phi_k(x_1, \ldots, x_n))$. If we take $f \in \overline F[x_1, \ldots, x_k]/I(X_2)$ then $\phi^*f=f(\phi_1(x_1, \ldots, x_n), \ldots, \phi_k(x_1, \ldots, x_n))$. Note that $\phi^*f$ must be a polynomial so $\phi_i$ is a polynomial and it must be well-defined for each $f \pmod {I(X_2)}$ which happens when $\forall P \in X_1: (\phi_1(P), \ldots, \phi_2(P)) \in X_2$. So for affine and projective spaces we'll have the following form of a regular morphism:

$$\phi: (x_1, \ldots, x_n) \mapsto (\phi_1(x_1, \ldots, x_n), \ldots, \phi_k(x_1, \ldots, x_n)) \\ \phi_i \in A(X_1) = \mathcal O_{X_1}(X_1) \\ \forall P \in X_1: \phi(P) \in X_2$$

In fact, there's a one to one correspondence between irreducible closed sets morpishsms $Y_1 \rightsquigarrow_{Rsp} Y_2$ and coordinate rings $\overline F$-algebra homomorphisms $A(Y_1) \rightsquigarrow_A A(Y_2)$. (a statement we will not prove here)

We can extend the definition of morphism to open subsets. Since open subset is dense this will still be a valid relation but with more flexibility:

def: Rational map

$$\begin{align*} &\sphericalangle \\ &(X_1, T_1) \in \mathcal T^{c,-} \\ &(X_2, T_2) \in \mathcal T^{c,-} \\ &\exists U \in T_1^o, U \ne \empty, f: U \rightsquigarrow_{Rsp} X_2 \\ \hline \\ &f: X_1 \rightsquigarrow_{Rat} X_2 \end{align*}$$

Next we want define an analogue of surjective map:

def: Dominant rational map

$$\begin{align*} &\sphericalangle \\ &(X_1, T_1) \in \mathcal T^{c,-} \\ &(X_2, T_2) \in \mathcal T^{c,-} \\ &f: X_1 \rightsquigarrow_{Rat} X_2 \\ &\overline {f(X_1)}= X_2 \\ \hline \\ &f - \text{dominant map} \end{align*}$$

Finally as usual there's a notion of isomorphisms:

def: Birational map

$$\begin{align*} &\sphericalangle \\ &(X_1, T_1) \in \mathcal T^{c,-} \\ &(X_2, T_2) \in \mathcal T^{c,-} \\ &f: X_1 \rightsquigarrow_{Rat} X_2 \\ &\exists g: X_2 \rightsquigarrow_{Rat} X_1\\ &f \circ g = \text{id} \\ &g \circ f = \text{id} \\ \hline \\ &f: X_1 \rightsquigarrow_{Rat} X_2 \end{align*}$$

Note that now that we have a rational map that is defined on open set, we're free to consider ratios of polynomials and thus rational map has a form:

$$\phi: (x_1, \ldots, x_n) \mapsto (\phi_1(x_1, \ldots, x_n), \ldots, \phi_k(x_1, \ldots, x_n)) \\ \phi_i \in \mathcal O(U), U \in T^o \\ \forall P \in U: \phi(P) \in X_2$$

Or we can say that rational map is a map from $X_1$ to $X_2$ represented by ratios of polynomials that are defined everywhere but in some closed subset of $X_1$.

Actually these both forms for morphisms and regular functions make sense since regular functions in essence are morphisms to $\mathbb A^1$ and $\mathbb P^1$ respectively so we have a product of morphisms in this case.

Finally note that in case of projective rational map:

$$\phi: [x_0, \ldots, x_n] \mapsto [\phi_0(x_0, \ldots, x_n), \ldots, \phi_k(x_0, \ldots, x_n)] \\ \phi_i \in \mathcal O_{X_1}(U) \\ \forall P \in U: \phi(P) \in X_2$$

We can surely multiply each coordinate of a projective point by the same constant and get the same point. So for a point $P \in X_1 \setminus U$ we can find some $g_P \in \mathcal O^r_{X_1, P}$ such that $[g_P(P)\phi_0(P), \ldots, g_P(P)\phi_k(P)] \in X_2$ we say that the rational map is regular at point $P$. If it's regular for any point in $X_1 \setminus U$ then it's a morphism.