Question with the definition of rational maps (Algebraic geometry) I am getting a little confused with the definition of rational maps. From this Wikipedia page and on books, a rational map between two varieties is a morphism from a nonempty open set $U \subseteq X$ to $Y.$ Two rational maps are considered the same if they agree on the intersection of their domains. A morphism is the restriction of some polynomial to the open set.
Now, I am having some difficulties matching these definition with the example of the affine variety $X$ given by the curve $x^2+y^2 = 1$ as a subset  of $\mathbb A_\mathbb C^2.$ The function $f(x,y) = \frac{1-y}{x} = \frac{x}{1+y}$ is claimed by the Wikipedia page to be rational. However, none of these two expression of $f$ looks like the restriction of a polynomial.
Did I understand something incorrectly?
 A: question: "However, none of these two expression of f looks like the restriction of a polynomial. Did I understand something incorrectly?"
Answer: You find a definition of regular functions and rational maps in Hartshornes book Chapter I.3. If $k$ is an algebraically closed field and $Y \subseteq \mathbb{P}^n_k$ is a quasi projective variety, a function
$$f: Y \rightarrow k$$
is "regular" iff for any $y \in Y$ there is an open set $y\in U \subseteq Y$ and homogeneous polynomials $u,v\in k[x_0,..,x_n]$ of the same degree with $v$ non-zero on $U$ and $f=\frac{u}{v}$ as function on $U$.
Note that the quotient $\frac{u}{v}$ is a well defined function on $U$.
Let $V \subseteq Y$ be an open set and let $\mathcal{O}_Y(V)$ be the set of regular functions on $V$. It follows $\mathcal{O}_Y(V)$ is a $k$-algebra.
A "morphism" of varieties $X,Y$ is a continuous map $\phi: X \rightarrow Y$ such that for any open set $V \subseteq Y$ and any function $f\in \mathcal{O}_Y(V)$ it follows
$$ f \circ \phi: \phi^{-1}(V) \rightarrow k$$
is a regular function.
"The function $f(x,y) = \frac{1-y}{x} = \frac{x}{1+y}$ is claimed by the Wikipedia page to be rational."
The function $f(x,y)$ is regular on the set $V$ where $x \neq 0$ (and $1+y \neq 0$).
In your case the polynomial $x^2+y^2-1$ is irreducible and the ring $A.=k[x,y]/(f)$ is an integral domain. A rational function is an element of the quotient field $K(A)$ and your function $f$ lives in $K(A)$.
