Let $X$ and $Y$ be **affine** varieties over $\mathbb C$, and consider a 
morphism $f:X\to Y$ and the induced homomorhism 


$$ \varphi=f^\*:B=\mathbb C[Y]\to A=\mathbb C[X]. $$


It is very easy to see that if $\varphi$ is surjective then $f$ is injective. 
The converse is not true, as the inclusion $\mathbb C\setminus\{0\}\to\mathbb C$ shows. 


My question is:



> 
> Assume $f$ is a **finite** injective morphism. Is it true that $\varphi$ must be surjective?
> 
> 
> 


Here *finite* means that $A$ is integral over $B$ or, equivalently, that $A$ is a fintely generated $B$-module. Further, we are allowed to assume that $X$ and $Y$ are irreducible, and that $A$ and $B$ are free polynomial rings, if that helps. 


One idea was to localize at an arbitrary maximal ideal $\mathfrak m$ of $B$,
but I do not know if this works. 



Is the induced ring homomorphism surjective for a finite injective morphism between affine varieties?
No - consider the normalization of a cuspidal rational curve. 


I don't understand your condition that $A$ and $B$ are free polynomial rings: do you mean that $X$ and $Y$ are isomorphic to $\mathbb{C}^n$ for some $n$? In that case the answer is positive. More precisely, the answer is positive if $B$ is integrally closed. Indeed, the injectivity and finiteness of $f$ implies that the quotient fields of $A$ and $B$ are the same, so that $A$ is contained in the integral closure of $B$. 


