Math Genius: When can we have the same number of (general) solution as the projection?

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$newcommandC{mathbb{C}}$Let $a_1,dots,a_n$ be parameters in $C$ and let
$f_1,dots,f_n$ be non-constant algebraically independent polynomials in $n$-variables over $C$.

For a general choice of $a_i$ (i.e. $(a_1,dots,a_n)$ belonging to a Zariski open set in the affine space $C^n$) we know that the system

$$f_i = a_i qquad i=1,dots,n$$

is a complete intersection, i.e. there are finite , say $k$, solutions in $C^n$ (I’m not sure what I should use to justfy this, I guess some argument in ellimination theory would justify this).

My question is the following:

Suppose $pi: C^n to C^2$
is a projection on say the first two coordinates.

Can I also claim that for a general choice of $a_i$ the projection of the solutions of

$$f_i = a_i qquad i=1,dots,n$$

with respect to $pi$ yields $k$ points in $C^2$? The projection should be the ellimination ideal of the ideal generated by the polynomials $langle f_i-a_i : i=1,dots,n rangle$ to polynomials in the first two variables. But can I also have a control on the ellimination
ideal (I want it to have the same number of vanishing points as the original ideal) for general $a_i$?

If so, why?

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