Let $K$ be an algebraic closed field. Let $A$ be an affine variety of $K^{n}$, and $f:A\rightarrow K^{m}$ be a polynomial map. Prove that closure of $f(A)$ equal to $V(\langle I(A),Y_{1}-f_{1},...,Y_{m}-f_{m}\rangle_{K[X_{1},...X_{n},Y_{1},...Y_{m}]}\cap K[Y_{1}...Y_{m}]):=V(B\cap C)$. Where $f_{i}$ are the components of $f$,$V(I)$ denote zero set of ideal $I$, $I(X)$ denote vanishing ideal of set $X$.
I am stuck at both directions. I want to to show that $I(f(A))=\langle B\cap C\rangle$, but what I can show is only $(A,\overline{f(A)})\subset V(B\cap C)$, this is clearly not enough. Can anyone help or give a hint?
First, note that $V(I(A),Y_1-f_1,\cdots,Y_m-f_m)\subset K^{n+m}$ is the graph $(a,f(a))$ of $f$, which is isomorphic to $A$ via the projection to $K^n$. So you're really trying to understand why if $X\subset K^{n+m}$ is a variety then the closure of $\pi(X)$, the projection of $X$ to $K^m$, is the variety cut out by $I(X)\cap K[Y_1,\cdots,Y_m]$.
For one direction, any polynomial in the $Y_i$ which vanishes on $\pi(X)$ will vanish on $X$ since it doesn't depend on the $X$-coordinates, so $I(\pi(X))\subset I(X)\cap K[Y_1,\cdots,Y_m]$. On the other hand, if there is a polynomial $f$ in the $Y_i$ which vanishes on $X$, then it must vanish on $\pi(X)$, since $f$ does not depend on the $X$ coordinates. So $I(X)\cap K[Y_1,\cdots,Y_m]\subset I(\pi(X))$. So $V(I(X)\cap K[Y_1,\cdots,Y_m])=V(I(\pi(X)))=\overline{\pi(X)}$ and we're done.