Let $Y$ be an affine variety of dimension $d$ inside the affine space $\mathbb{A}^n$. Then $A(Y) = k[x_1,\dots,x_n]/I_Y=:k[\bar{x}_1,\dots,\bar{x}_n]$. By the Noether normalization theorem, there exist $d$ elements of $k[\bar{x}_1,\dots,\bar{x}_n]$, say $f_1,\dots,f_d$ such that $A(Y)$ is integral over $k[f_1,\dots,f_d]$ and the $f_i$ are algebraically independent over $k$. More specifically, the $f_i$ elements can be taken to be linear combinations of the $\bar{x}_i$. If we denote by $f$ the $d$-tuple $(f_1\dots,f_d)$ and by $\bar{x}$ the $n$-tuple $(\bar{x}_1,\dots,\bar{x}_n)$ then there exists a fat full rank matrix $B$ such that $f = B \bar{x}$, where $f,\bar{x}$ are thought of as column vectors. Then this matrix $B$ gives a morphism of varieties $\psi: Y \rightarrow \mathbb{A}^d$, corresponding to the $k$-algebra monomorphism $k[y_1,\dots,y_d] \rightarrow A(Y)$ that takes $y_i$ to $\sum_j B_{ij} \bar{x}_j$.
Question: Why is $\psi$ surjective in the category of sets? I.e. why given any $a=(a_1,\dots,a_d) \in \mathbb{A}^d$ we can find a $\xi \in Y$ such that $a = B \xi$?
PS: Let $\xi=(\xi_1,\dots,\xi_n)$ be the point of $\mathbb{A}^n$ that i wish it to be the preimage of $a$ under $B$ and moreover i wish it to be inside $Y$. Then i need to show that $\xi$ must satisfy the equations $a = B \xi$ and $p(\xi)=0$ for every generator $p$ of $I_Y$. However, it is not clear to me why this system of equations has a solution in general.
One needs $k$ to be algebraically closed, for surjectivity on the level of sets. Let $(a_1, \ldots, a_d) \in \mathbb{A}^d$. This corresponds to the maximal ideal $(y_1 - a_1, \ldots, y_d - a_d) \in \operatorname{mSpec}(k[y_1, \ldots, y_d])$. Since $k[y_1, \ldots, y_d] \subseteq A(Y)$ is integral, there is a maximal ideal $m \in \operatorname{mSpec}(A(Y))$ such that $m^c = (y_i - a_i)$. Since $k$ is algebraically closed, $m = (x_1 - b_1, \ldots, x_n - b_n)$ for some point $(b_1, \ldots, b_n) \in Y$, so $\psi : Y \to \mathbb{A}^d$ sends $\psi(b_1, \ldots, b_n) = (a_1, \ldots, a_d)$.
On the other hand, taking $k = \mathbb{R}$, and $Y = \mathbb{R}[x]/(x^2+1)$, we have $\dim Y = 0$, but $Y \to \mathbb{A}^0_{\mathbb{R}} = \{pt\}$ is not surjective on sets, since $Y$ is in fact the empty set inside $\mathbb{A}^1_{\mathbb{R}}$ (viewed as a classical variety, not a scheme).