Our main objective is to interpret $F:V \to W$ a morphism as a map $F:maxSpec \Bbb C[V] \to maxSpec \Bbb C[W]$, $V,W$ are algebraic varieties.
Now from $F:V \to W$ using Hilbert Nullstelensatz we have the pullback $F^*: \Bbb C[W] \to \Bbb C[V]$. Now indeed given a point $p \in V$ we think it as a maximal ideal $m_v \in maxSpec \Bbb C[V]$. Then the image of $p$ under $F$ corresponds to the maximal ideal $(F^*)^{-1}(m_v) \in maxSpec \Bbb C[W]$ This is where I have the problem in showing that $(F^*)^{-1}(m_v)$ is a maximal ideal in $\Bbb C[W]$ which is not true in general.
My attempt: Suppose $(F^*)^{-1}(m_v)$ is not maximal then $\exists m_w \subset \Bbb C[W]$ maximal s.t $(F^*)^{-1}(m_v) \subsetneq m_w$ then $m_w$ corresponds to a point $q \in W$.
Let us comeback in the pullback part as we should have all the ingredients ready in our hand: for any $g \in \Bbb C[W]$ $(F^*)(g)=g \circ F$ so for any $g \in (F^*)^{-1}(m_v)$ we have $(F^*)(g)=g \circ F \in m_v$.
Now can we proceed from here, I think if we can prove $F^*$ is surjective then also we are done.
Rings and algebras are assumed to be commutative.
Let $f\colon A\to B$ be a $k$-algebra homomorphism. If $J$ is a maximal ideal of $B$ and $I=f^{-1}(J)$, then there is an injective $k$-algebra homomorphism $$ A/I\to B/J $$ where the codomain is a field. If $B$ is finite-dimensional over $k$, then so are $B/J$ and $A/I$. Hence $A/I$ is an artinian ring which is a domain, hence a field. So $I$ is maximal.