Why is this function regular? In this English translation of Serres FAC, on page 62 he is giving the structure of $\mathbb{P}_{r}(K)$. 
He sets $t_i$ to be the $i$-th coordinate function on $K^{r+1}$, and defines $$V_i = \{ x \in K^{r+1} | t_i(x) \neq 0\}.$$ Then $U_i$ is the projection of $V_i$ from $K^{r+1}$ to $\mathbb{P}_{r}(K)$, taking the latter to be defined in the usual way by forming the quotient with the equivalence relation.
I do not understand the assertion "the function $t_j/t_i$ is regular on $V_i$ and invariant for $K^{*}$, thus defines a function on $U_i$.”
In order to check this function is regular we use the definition Serre gives on page page 39 that a function $\phi \colon U \to V$ is regular on $U$ if 


*

*$U, V$ are both locally closed subspaces of $K^{r}$.

*$\phi$ is continuous.

*If $x \in U$ and $f \in \mathscr{O}_{\phi(x),V}$ then $f \circ \phi \in \mathscr{O}_{x,U}$
Can someone please help me verify the claims made in the quoted sentence above? I am not even sure what we are taking the codomain of $t_j/t_i$ to be, I am assuming $K$? I am sure these are trivial matters, but I do not understand.
 A: In our case we have a map $p:V_i\to K$, with coordinates $t=(t_1,...,t_{r+1})$ on $V_i$ and $z$ on $K$. Now take $x=(x_1,...,x_{r+1})\in V_i$ and an element $f\in \mathcal{O}_{p(x),K}$, which is nothing else but an element of $k(z)$ of the form $\frac{g(z)}{h(z)}=\frac{\Sigma \lambda_m z^m}{\Sigma \mu_n z^n}$ with $h(p(x))\neq 0$. 
Now consider $(f\circ p)(t)=\frac{g\circ p}{h\circ p}(t)=\frac{\Sigma \lambda_m (\frac{t_j}{t_i})^m}{\Sigma \mu_n (\frac{t_j}{t_i})^n}$. Note that $g\circ p$ and $h\circ p$ are not polynomials in $t_1,..., t_{r+1}$, but we can multiply by a sufficently big $t_i^k$ giving us a representation of $f\circ p(t)=\frac{t_i^k\cdot g\circ p(t)}{t_i^k\cdot  h\circ p(t)}$ as quotient of polynomials. Now we only have to show that $(t_i^k\cdot h\circ p)(x)=x_i^k\cdot h(p(x))\neq 0$. But we know that $h(p(x))\neq 0$ by the assumption that $f$ is regular on $p(x)$ and $x_i^k\neq 0$ by definition of $V_i$.
For continuity you might want to apply a similar idea. If  $f(z)\in k[z]$ and Z=V(f), you want $p^{-1}(Z)=V(g)$ for some polynomial $g\in k[t_1,...,t_{r+1}$], so you can consider $g=f\circ p$ and then...
