Is $f$ a morphism if its graph is a variety Let $X\subseteq\mathbb{P}^n,Y\subseteq\mathbb{P}^m$ be two algebraic varieties in the sense that they are the zero sets of some homogeneous polynomials. Moreover assume that $f:X\rightarrow Y$ is a bijective function (just a function, not necessarily polynomial). Consider the graph of $f$ : 
$$\Gamma=\{(x,y)\in X\times Y\mid f(x)=y\}.$$ If $\Gamma$ is an algebraic variety (as in the above sense) then can I deduce that $f$ is a polynomial map?
 A: No. Here is an example: let $g: Y \to X$ be any bijective morphism which is not an isomorphism and let $f$ be the function $g^{-1}$. For instance, let $Y = \mathbb{P}^1$ and $X = V(X_0 X_2^2 - X_1^3) \subseteq \mathbb{P}^2$ (with coordinates $X_0,X_1,X_2$) and $g: Y \to X$ by $[s:t] \mapsto [s^3: st^2 : t^3]$. The graph of the function $g^{-1}$ is just the graph of $g$, which is a variety as $g$ is a morphism. It is elementary to check that $g$ is bijective on $\mathbb{C}$-points. However, $g$ is not an isomorphism, since $Y$ is smooth but $X$ is singular at $[1:0:0]$. Hence, $f$ is not a morphism of varieties.
In the affine chart $\{X_0 \neq 0\}$, this is the map $\mathbb{A}^1 \to V(y^2 = x^3)$ by $t \mapsto (t^2,t^3)$. Its ''inverse'' would be $(x,y) \mapsto y/x$, which is not a morphism at $(0,0)$.
In general, one has a diagram 
$$X \leftarrow \Gamma \rightarrow Y$$
where the arrows are given by projections, which are always morphisms. Even if these morphisms are bijective, the left morphism may not have an inverse.
