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Let $Y\subseteq \mathbb{A}^n$ be an affine variety, $\bar{Y}$ be its projective closure in $\mathbb{P}^n$. In GTM 52 of Robin Hartshone, there is a problem of finding generator of $I(\bar{Y})$(problem 2.9)

Suppose that $\varphi : \mathbb{A}^n \rightarrow U_0 $ be the homeomorphism between two spaces, then is $\varphi(Y)$ is the projective closure of $Y$ ?

How can I imagine about the projective closure of an affine variety, topologically ?

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Projective $n$-space has a natural open covering by $n+1$ copies $\Bbb A^n_i$ of affine space; we can consider $\Bbb A^n_0\subseteq\Bbb P^n$ to be defined by the map $(a_1^{(0)},\ldots,a_n^{(0)})\mapsto[1:a_1^{(0)}:\ldots:a_n^{(0)}],$ $\Bbb A^n_1\subseteq\Bbb P^n$ defined by $(a_1^{(1)},\ldots,a_n^{(1)})\mapsto[a_1^{(1)}:1:a_2^{(1)},\ldots,a_n^{(1)}],$ and so on.

Given a subvariety $Y\subseteq \Bbb A^n$, to find its projective closure, we first want to choose some identification of $\Bbb A^n$ with an isomorphic subspace of $\Bbb P^n.$ A natural way to do this is via $\Bbb A^n\cong\Bbb A^n_i$ for any choice of $i.$ Then $\overline Y$ will be the Zariski closure of $Y\subseteq\Bbb A^n_i\subseteq\Bbb P^n,$ defined by the homogenization of the defining equations for $Y$ in the coordinates of $\Bbb A^n_i$ with respect to $x^i.$

In your notation, I believe you have chosen $U_0=\Bbb A^n_0.$ But, to answer your first question, it is not necessarily true that $\varphi(Y)=\overline Y;$ instead we identify $Y$ with $\varphi(Y)$ and $\varphi(Y)^{\mathrm{cl}}=\overline Y$, where $\mathrm{cl}$ denotes Zariski closure.

I usually like to think of projective closure as "compactification" of an affine thing, i.e. the affine $Y$ will be Zariski dense in the closed projective $\overline Y.$

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