Problem I.4.7 in Hartshorne

Question 1: Let us represent a point of $\mathbb{A}^n$ as the action of an $n$-tuple of regular functions on the point itself. These regular functions are just the coordinate functions on $\mathbb{A}^n$. So we have that $a=(a_1,\dots,a_n) = (y_1,\dots,y_n)(a_1,\dots,a_n)$. Alternatively, the point $a$ is the image of itself under the identity morphism $(y_1,\dots,y_n)$ of $\mathbb{A}^n$. Now we can define a morphism $f^*: U \rightarrow Y$ by having $f$ act on the $n$-tuple of regular functions $(y_1,\dots,y_n)$ as $(a_1,\dots,a_n) \mapsto (f(y_1),\dots,f(y_n))(a_1,\dots,a_n)$, where $U=\cap_i U_i$ and $U_i$ is the open set where $f(y_i)$ is defined. For $g=f^{-1}$ let $V_i$ be the open set where $g(y_i)$ is defined and let $V=\cap_i V_i$. This gives a morphism $g^*: V \rightarrow X$. To see what is the effect of $g^*$ on the point $(f(y_1)(a),\dots,f(y_n)(a))$, observe that $(f(y_1),\dots,f(y_n))(a)$ and so by definition $f^*$ is induced by the action of $g$ on the $n$-tuple of regular functions $(f(y_1),\dots,f(y_n))$ giving $g^*((f(y_1),\dots,f(y_n))(a)) = (g(f(y_1)),\dots,g(f(y_n)))(a) = (y_1,\dots,y_n)(a)=(a)$.

Question 2: In general $f^*(U) \cap V$ will not be an open set. To remedy this, replace $U$ by $U^* := (f^*)^{-1}(V) \cap U$ and $V$ by $V^* :=(g^*)^{-1}(U^*) \cap V$. Then $f^*, g^*$ induce an isomorphism of $U^*,V^*$.