Fix an algebraically closed † ground field $k$ of any characteristic. I want to use the classical definition of projective $n$-space $\mathbb{P}^n$ as set quotient of $\mathbb{A}^{n+1}\setminus 0$ by the action of $k^*$, and for its Zariski topology the definition that closed sets are the "zero sets" of homogeneous polynomials in $n+1$ variables considered as functions from $\mathbb{P}^n$ to {$0,1$}.

I'm not satisfied with my understanding of why this is the same as the quotient topology from the map
$\mathbb{A}^{n+1}\to \mathbb{P}^{n}$. This means proving the map is surjective, continuous, and that a set with closed pre-image is closed, the first two properties being clear. The last part is the least trivial, meaning that:

If a set $S\subseteq \mathbb{P}^n$ has closed preimage $\widehat{S}$ in $\mathbb{A}^{n+1}\setminus 0$, i.e. $\widehat{S}$ is cut out by some polynomials $f_1,\ldots,f_r$, then $S$ is closed in $\mathbb{P}^n$, i.e. $\widehat{S}$ is cut out by some homogeneous polynomials $g_1,\ldots,g_s$.

I'm bad at algebra, so I can't seem to start doing anything here that doesn't look ugly/hard... I really want to "see" what's going on and avoid just quoting results about invariants without understanding how they work in this "toy" example.

Thanks for the help, anyone!

† Follow up: the algebraic closed hypothesis is not necessary, since David/Kevin's proof works for any infinite field, and for a finite field our spaces (as defined here) are discrete so the result is trivial.

Fix an algebraically closed † ground field $k$ of any characteristic. I want to use the classical definition of projective $n$-space $\mathbb{P}^n$ as set quotient of $\mathbb{A}^{n+1}\setminus 0$ by the action of $k^*$, and for its Zariski topology the definition that closed sets are the "zero sets" of homogeneous polynomials in $n+1$ variables considered as functions from $\mathbb{P}^n$ to {$0,1$}.

I'm not satisfied with my understanding of why this is the same as the quotient topology from the map
$\mathbb{A}^{n+1}\setminus 0\to \mathbb{P}^{n}$. This means proving the map is surjective, continuous, and that a set with closed pre-image is closed, the first two properties being clear. The last part is the least trivial, meaning that:

If a set $S\subseteq \mathbb{P}^n$ has closed preimage $\widehat{S}$ in $\mathbb{A}^{n+1}\setminus 0$, i.e. $\widehat{S}$ is cut out by some polynomials $f_1,\ldots,f_r$, then $S$ is closed in $\mathbb{P}^n$, i.e. $\widehat{S}$ is cut out by some homogeneous polynomials $g_1,\ldots,g_s$.

I'm bad at algebra, so I can't seem to start doing anything here that doesn't look ugly/hard... I really want to "see" what's going on and avoid just quoting results about invariants without understanding how they work in this "toy" example.

Thanks for the help, anyone!

† Follow up: the algebraic closed hypothesis is not necessary, since David/Kevin's proof works for any infinite field, and for a finite field our spaces (as defined here) are discrete so the result is trivial.

Fix an algebraically closed ground field $k$ of any characteristic. I want to use the classical definition of projective $n$-space $\mathbb{P}^n$ as set quotient of $\mathbb{A}^{n+1}\setminus 0$ by the action of $k^*$, and for its Zariski topology the definition that closed sets are the "zero sets" of homogeneous polynomials in $n+1$ variables considered as functions from $\mathbb{P}^n$ to {$0,1$}.

I'm not satisfied with my understanding of why this is the same as the quotient topology from the map
$\mathbb{A}^{n+1}\to \mathbb{P}^{n}$. This means proving the map is surjective, continuous, and that a set with closed pre-image is closed, the first two properties being clear. The last part is the least trivial, meaning that:

If a set $S\subseteq \mathbb{P}^n$ has closed preimage $\widehat{S}$ in $\mathbb{A}^{n+1}\setminus 0$, i.e. $\widehat{S}$ is cut out by some polynomials $f_1,\ldots,f_r$, then $S$ is closed in $\mathbb{P}^n$, i.e. $\widehat{S}$ is cut out by some homogeneous polynomials $g_1,\ldots,g_s$.

I'm bad at algebra, so I can't seem to start doing anything here that doesn't look ugly/hard... I really want to "see" what's going on and avoid just quoting results about invariants without understanding how they work in this "toy" example.

Thanks for the help, anyone!

Fix an algebraically closed † ground field $k$ of any characteristic. I want to use the classical definition of projective $n$-space $\mathbb{P}^n$ as set quotient of $\mathbb{A}^{n+1}\setminus 0$ by the action of $k^*$, and for its Zariski topology the definition that closed sets are the "zero sets" of homogeneous polynomials in $n+1$ variables considered as functions from $\mathbb{P}^n$ to {$0,1$}.

I'm not satisfied with my understanding of why this is the same as the quotient topology from the map
$\mathbb{A}^{n+1}\to \mathbb{P}^{n}$. This means proving the map is surjective, continuous, and that a set with closed pre-image is closed, the first two properties being clear. The last part is the least trivial, meaning that:

If a set $S\subseteq \mathbb{P}^n$ has closed preimage $\widehat{S}$ in $\mathbb{A}^{n+1}\setminus 0$, i.e. $\widehat{S}$ is cut out by some polynomials $f_1,\ldots,f_r$, then $S$ is closed in $\mathbb{P}^n$, i.e. $\widehat{S}$ is cut out by some homogeneous polynomials $g_1,\ldots,g_s$.

I'm bad at algebra, so I can't seem to start doing anything here that doesn't look ugly/hard... I really want to "see" what's going on and avoid just quoting results about invariants without understanding how they work in this "toy" example.

Thanks for the help, anyone!

† Follow up: the algebraic closed hypothesis is not necessary, since David/Kevin's proof works for any infinite field, and for a finite field our spaces (as defined here) are discrete so the result is trivial.