Let $X,Y$ be two irreducible, projective $k$-schemes. $k$ is assumed to be algeraically closed. Consider a dominant morphism $f: X \to Y$ between them which is not an isomorphism! 


Let $\mathcal{L}$ be an invertible sheaf on $Y$ and $\mathcal{G}:=f^\*\mathcal{L}$ it's pullback. Assume that $\mathcal{G}$ is very ample; that is $\mathcal{G}$ induces a closed embedding $g: X \to \mathbb{P}^n= \operatorname{Proj} \ H^0(X, \mathcal{G})$.


$f^\*$ induces a $k$-map $f^\*: H^0(Y,\mathcal{L}) \to H^0(X,f^\*\mathcal{L})=H^0(X,\mathcal{G})$.


Question: Under assumption that $\mathcal{G}$ is very ample: why the $k$-map $f^\*: H^0(X,\mathcal{L}) \to H^0(Y,\mathcal{G})$ cannot be surjective?



Pullback map on global sections surjective
If the map $f^\* \colon H^0(Y,\mathcal{L}) \to H^0(X,f^\*\mathcal{L})$ is surjective then there is a commutative diagram
$$
\require{AMScd}
\begin{CD}
X @>>> \mathbb{P}(H^0(X,f^\*\mathcal{L})^\vee)
\\
@VfVV @VVV
\\
Y @>>> \mathbb{P}(H^0(Y,\mathcal{L})^\vee),
\end{CD}
$$
where the right vertical arrow is an embedding and the bottom arrow is a priori rational. But since $f$ is proper and dominant, it is surjective, hence the bottom arrow is regular. Moreover, it follows that $f$ is a closed embedding. But then $f$ is an isomorphism. 


EDIT.
If you want to check the surjectivity of $H^0(Y,\mathcal{L}) \otimes \mathcal{O}\_Y \to \mathcal{L}$ at point $y \in Y$, choose $x \in X$ such that $y = f(x)$, and consider the commutative diagram
$$
\require{AMScd}
\begin{CD}
H^0(Y,\mathcal{L}) @>>> \mathcal{L}\_y 
\\
@V f^\* VV @|
\\
H^0(X,f^\*\mathcal{L}) @>>> (f^\*\mathcal{L})\_x.
\end{CD}
$$
The left vertical arrow is surjective by assumption, and the bottom arrow is surjective, because $f^\*\mathcal{L}$ is very ample. Therefore, the top arrow is surjective as well.


