$k$-algebra homomorphism of the polynomial ring $k[x_1,\dots,x_n]$ Let $\phi:k[x_1,\dots,x_n]\mapsto k[x_1,\dots,x_n]$ be a $k$-algebra homomorphism with $\phi(x_i)=f_i$, where $k$ is algebraically closed and has characteristic zero. I have the following questions:
(1) If $\phi$ is surjective, i.e. $k[f_1,\dots,f_n]=k[x_1,\dots,x_n]$, is $(f_1,\dots,f_n)$ a maximal ideal in $k[x_1,\dots,x_n]$? Conversely, if $(f_1,\dots,f_n)$ is a maximal ideal in $k[x_1,\dots,x_n]$, is $\phi$ surjective?
(2) If $\phi$ is surjective, is $\phi$ injective? If this is not true, is there any counterexample?
(3) Now, let $X,Y$ be two affine variety, $\mu :X\mapsto Y$ is a morphism (polynomial map), $\mu$ induce a $k$-algebra homomorphism between the coordinate ring of $X,Y$
 $$\mu^*:A(Y)\mapsto A(X):f\mapsto f\circ \mu.$$
If $\mu$ is injective, is $\mu^*$ always surjective? at least in the case where $X=Y=\mathbb{A}^n$?
 A: (2) Yes, if $\phi$ is surjective, it is injective.
Indeed, the dual  morphism $f=\phi^*:\mathbb{A}^n_k\to \mathbb{A}^n_k$ is a closed embedding and since $f(\mathbb{A}^n_k)\subset \mathbb{A}^n_k$ are irreducible of dimension $n$ they are equal and $f$ is an isomorphism (since  $\mathbb{A}^n_k$ is reduced ).
Thus $\phi$ is an isomorphism too and, in particular, $\phi$ is injective.    
(1) If $\phi$ is surjective, it is an isomorphism  by answer (2) so that, yes, $(f_1,\dots,f_n)$ is a maximal ideal in $k[x_1,\dots,x_n]$ .
Edit
The converse is false: the ideal $(x_1,(x_1+1)x_2)\subset k[x_1,x_2]$ is maximal but  the corresponding morphism  $\phi: k[x_1,x_2]\to k[x_1,x_2]$ determined by $\phi(x_1)=f_1=x_1,\phi(x_2)=f_2=(x_1+1)x_2$ is not surjective, since $x_2$ is not in the image of  $\phi$ .
(3) is false.  A counterexample is obtained  by considering the cusp $C\subset \mathbb A^2_k$ of equation $y^2=x^3$  and the bijective morphism given by $$\mu:\mathbb A^1\to C:t\mapsto (t^2,t^3)$$ This bijective morphism is not an isomorphism and the dual algebra morphism $\mu^*:k[x, y]/(y^2-x^3)\to k[t]$ is injective but not surjective: its image is $k[t^2,t^3]$.   
Remark
Notice that all of  the above is valid for an arbitary field $k$ of any characteristic, algebraically closed or not.
