Suppose that $h: X\to Y$ is a morphism between varieties and $h^*:A(Y)\to A(X)$ is induced morphism, if $p\in k[x_1,...,x_m]$ then $p\circ h\in I(X)$ if and only if $p\in I(\overline{h(X)})$.
I know that $I(X)$ are all the polynomials in $k[x_1,...,x_n]$ that cancel out in $X$, so what I think is that if $p\circ h\in I(X)$ then $p\in I(h(X))$, but why $p\in I(\overline{h(X)})$?
The vanishing set of a polynomial is a closed set (by definition of the Zariski topology), so if $p$ vanishes on $h(X)$ it must vanish on the closure $\overline{h(X)}$ (since the vanishing set of $p$ is a closed set that contains $h(X)$.