For me, an affine variety is an irreducible closed subset of some $A_k^n$. A quasi-affine variety is a non-empty open subset of an affine variety. A projective variety is an irreducible closed subset of some $P_K^n$. A quasi-projective variety is a non-empty open subset of a projective variety. A variety is everyone of the above: an affine variety, or a quasi-affine variety, a projective variety, or a quasi projective variety.
If $X\subseteq A_K^n$ is an affine variety, then the quotient $A(X)=K[x_1,\dots,x_n]/I(X)$ is the affine coordinate ring of $X$, where $I(X)$ is the ideal of the polynomial that vanish on $X$. Let $U$ be a non empty open subset of $X$, let $f\colon U\to K$ be a function, where $K$ is an algebraically closed field. I say that $f$ is regular in $a\in U$ if there exists an open subset $V$ of $U$ such that $a\in V$ and exist $h,k \in A(X)$ such that $f_{|V}=\frac{h}{k}_{|V}$ and $k(y)\neq 0$ for all $y\in V$. I say that $f$ is regular in $U$ if $f$ is regular in all point of $U$. Then, $O_X(U)$ is the ring ($K$-algebra) of the regular functions in $U$.
If $X\subseteq P_K^n$ is a projective variety, then the quotient $R(X)=K[x_0,\dots,x_n]/I(X)$ is the projective coordinate ring of $X$, where $I(X)$ is the ideal generated by the homogeneous polynomial that vanish on $X$. Let $U$ be a non empty open subset of $X$, let $f\colon U\to K$ be a function, where $K$ is an algebraically closed field. I say that $f$ is regular in $a\in U$ if there exists an open subset $V$ of $U$ such that $a\in V$ and exist $h,k \in R(X)$, homogeneous of the same degree, such that $f_{|V}=\frac{h}{k}_{|V}$ and $k(y)\neq 0$ for all $y\in V$. I say that $f$ is regular in $U$ if $f$ is regular in all point of $U$. Then, $O_X(U)$ is the ring ($K$-algebra) of the regular functions in $U$. [Note that $R(X)$ is also a graduated ring because $K[x_0,\dots,x_n]$ is a graduated ring and $I(X)$ is an homogeneous ideal]
If $Y$ is a quasi affine variety and $X$ is the affine variety of which $Y$ is an open subset, and $U$ is an open subset of $Y$, then $U$ is open in $X$ and I define $O_Y(U):=O_X(U)$. (I think this is the case, but my book doesn't say it explicitly, so I am not sure).
If $Y$ is a quasi projective variety and $X$ is the projective variety of which $Y$ is an open subset, and $U$ is an open subset of $Y$, then $U$ is open in $X$ and I define $O_Y(U):=O_X(U)$. (I think this is the case, but my book doesn't say it explicitly, so I am not sure).
Finally, if $X$ and $Y$ are varieties, $f\colon X\to Y$ is a morphism of varieties if $f$ is continuous and for every open subset $V$ of $Y$, $$f^*\colon O_Y(V)\to O_X(f^{-1}(V))$$ is well- defined, i.e. $g\circ f$ is in $O_X(f^{-1}(V))$ for every $g$ in $O_Y(V)$.
My question is:
Let $f\colon X\to Y$ be a morphism between the varieties $X$ and $Y$. Let $Z:= \overline{f(X)}$ be the closure of $f(X)$ in $Y$. Then, why $f\colon X\to Z$ is morphism according to my definitions above?
I know $f\colon X\to Z$ is continuous. Then, i have to show that $Z$ is also a variety. I have distinguished two cases (i know that the image of an irreducible space by a continuous function is also irreducible, and that the closure of an irreducible subset is also irreducible):
- If $Y$ is affine variety, then $Z$ is an affine variety. If $Z$ is projective variety, than $Z$ is also projective vaeriety
- If $Y$ is quasi affine, then let $W$ be the affine variety of wich $Y$ is an open subset. Then the closure of $Z$ in $W$ is affine variety, and $Z$ is open in the closure of $Z$ in $W$, so $Z$ is quasi affine. The same if $Y$ is quasi projective.
My problem is to show the condition
for every open subset $V$ of $Z$, $$f^*\colon O_Z(V)\to O_X(f^{-1}(V))$$ is well- defined, i.e. $g\circ f$ is in $O_X(f^{-1}(V))$ for every $g$ in $O_Z(V)$, knowing that this is true for the open subset of $Y$, i.e. for every open subset $V$ of $Y$, $$f^*\colon O_Y(V)\to O_X(f^{-1}(V))$$ is well- defined, i.e. $g\circ f$ is in $O_X(f^{-1}(V))$ for every $g$ in $O_Y(V)$.
So, what am I missing? I'm struggling on this.
EDIT 26/11/2017
My strategy was: take an open subset $V$ of $Z$ and a regular function $g\colon V\to K \in O_Z(V)$. Then, $V=Z\cap V'$ where $V'$ is an open subset of $Y$. I want to find a regular function $g':V' \to K\in O_Y(V')$ such that $g'_{|f(f^{-1}(V))}=g$. If I do that, then $g'\circ f$ is in $O_X(f^{-1}(V'))$, so $(g'\circ f)_{|f^{-1}(V)}$ is in $O_X(f^{-1}(V))$, so $g\circ f=(g'\circ f)_{|f^{-1}(V)}$ is in $O_X(f^{-1}(V))$ Q.E.D.
So the question is: how can I find a regular function $g':V' \to K\in O_Y(V')$ such that $g'_{|f(f^{-1}(V))}=g$?
EDIT 26/11/2017, n°2
Another idea was to try with a "local approach". Take an open subset $V$ of $Z$ and a regular function $g\colon V\to K \in O_Z(V)$. I want to show that $(g\circ f)\colon f^{-1}(V)\to K$ is in $O_X(f^{-1}(V))$. Let $y$ be in $f^{-1}(V)$.
I want to show that $g\circ f$ is a quotient of polynomials (or a quotient of homogeneous polynomials of the same degree if $X$ is a (quasi) projective variety) in an open neighborhood of $y$.
So $f(y)\in V$. But $g$ is regular in $f(y)$, so there exists an open subset $W$ of $V$ with $f(y)\in W$, and exist $h,k$ "polynomials" such that $k\ne 0 $ in $W$ and $$g_{|W}=\frac{h}{k}_{|W}$$ But then $f^{-1}(W)$ is open in $X$ and $y \in f^{-1}(W)$ and we have that $$(g\circ f)_{|f^{-1}(W)}=\frac{h\circ f}{k \circ f}_{|f^{-1}(W)}$$ The problem is that now I don't know if $h,k$ are regular functions on $Y$, i.e. elements of $O_Y(Y)$ (for example if $Y$ is a projective variety, than $h$ and $k$ are homogeneous polynomials of the same degree, but they don't define regular functions, not even functions). In other words, I want to say that $h\circ f$ and $k \circ f$ are regular functions on $X$ (or, on an open subset $U$ of $X$ containing $y$), such that they are both quotient of polynomials and so $g\circ f$ is quotient of polynomials in an open neighborhood of $y$, but I don't know how to say that.
