[proofplan] The chart $\varphi_0$ identifies a point of $U_0 \subset \mathbb{P}^n_k$ with its affine coordinates, normalising the homogeneous coordinates so that $x_0 = 1$. Once this normalisation is in place, the homogeneous polynomial $f(X_0, \ldots, X_n)$ acts on the chart by setting $X_0 = 1$, which is exactly the dehomogenisation $f(1, Y_1, \ldots, Y_n) \in k[Y_1, \ldots, Y_n]$. The vanishing condition for $V \cap U_0$ then translates verbatim into the vanishing condition for the dehomogenised set in $\mathbb{A}^n_k$. Both inclusions of the equality are immediate from the dehomogenisation/rehomogenisation correspondence; the only subtlety is to verify well-definedness of evaluating a homogeneous polynomial on a normalised point of $U_0$. [/proofplan]

[step:Set up the chart and the dehomogenisation map] The chart $U_0 = \{[x_0 : \cdots : x_n] \in \mathbb{P}^n_k : x_0 \neq 0\}$ is an open subset of $\mathbb{P}^n_k$, and the map \begin{align*} \varphi_0 : U_0 &\to \mathbb{A}^n_k \\ [x_0 : x_1 : \cdots : x_n] &\mapsto \left(\frac{x_1}{x_0}, \ldots, \frac{x_n}{x_0}\right) \end{align*} is well-defined: scaling the homogeneous coordinates by $\lambda \neq 0$ scales each ratio $x_i/x_0$ by $\lambda/\lambda = 1$, so the image is independent of the representative. Equivalently, every point of $U_0$ has a unique representative of the form $[1 : y_1 : \cdots : y_n]$ obtained by dividing through by $x_0$, and $\varphi_0$ reads off the last $n$ coordinates of this representative. For a homogeneous polynomial $f \in k[X_0, \ldots, X_n]$, the \textbf{dehomogenisation with respect to $X_0$} is the polynomial \begin{align*} f^{\mathrm{deh}}(Y_1, \ldots, Y_n) := f(1, Y_1, \ldots, Y_n) \in k[Y_1, \ldots, Y_n], \end{align*} obtained by substituting $X_0 = 1$ and renaming the remaining variables. [/step]

[step:Translate the vanishing condition on $U_0$ into a polynomial condition on dehomogenisations]Let $[1 : a_1 : \cdots : a_n]$ be the normalised representative of a point $[p] \in U_0$, so $\varphi_0([p]) = (a_1, \ldots, a_n) \in \mathbb{A}^n_k$. We claim that for any homogeneous polynomial $f \in k[X_0, \ldots, X_n]$, \begin{align*} f \text{ vanishes at } [p] \iff f^{\mathrm{deh}}(a_1, \ldots, a_n) = 0. \end{align*} Indeed, the value of a homogeneous polynomial at a projective point is well-defined as a vanishing/non-vanishing question (not as an absolute scalar): for $f$ of degree $d$ and any $\lambda \neq 0$, \begin{align*} f(\lambda x_0, \lambda x_1, \ldots, \lambda x_n) = \lambda^d f(x_0, x_1, \ldots, x_n), \end{align*} so $f(p) = 0$ for one representative iff for any. We are free to use the normalised representative $(1, a_1, \ldots, a_n)$, giving \begin{align*} f \text{ vanishes at } [1 : a_1 : \cdots : a_n] \iff f(1, a_1, \ldots, a_n) = 0 \iff f^{\mathrm{deh}}(a_1, \ldots, a_n) = 0. \end{align*} This is the desired equivalence.[/step]

[guided]We need to make precise sense of "evaluating a homogeneous polynomial at a projective point" before we can say it equals dehomogenisation evaluated at the affine coordinates. \textbf{Why can't we evaluate $f$ at a projective point in the obvious way?} A point of $\mathbb{P}^n_k$ is an equivalence class $[x_0 : \cdots : x_n]$, with the same point represented by $(\lambda x_0, \ldots, \lambda x_n)$ for any $\lambda \neq 0$. A general polynomial $f$ does not have a well-defined value on equivalence classes: $f(\lambda x_0, \ldots) \neq f(x_0, \ldots)$ in general. \textbf{What is preserved is the vanishing condition.} For a homogeneous polynomial of degree $d$, the rescaling identity \begin{align*} f(\lambda x_0, \ldots, \lambda x_n) = \lambda^d f(x_0, \ldots, x_n) \end{align*} shows that the value of $f$ at different representatives of the same projective point differs by a nonzero scalar $\lambda^d$. Hence $f(p) = 0$ depends only on the projective point $[p]$, not on the choice of representative. This is the only sense in which homogeneous polynomials are "functions on $\mathbb{P}^n_k$" — they have well-defined zero sets. \textbf{Choosing the normalised representative.} On the chart $U_0$, every point has a canonical representative $(1, a_1, \ldots, a_n)$ obtained by dividing through by the nonzero $x_0$. Concretely, given $[x_0 : x_1 : \cdots : x_n]$ with $x_0 \neq 0$, take $\lambda = 1/x_0$: \begin{align*} [x_0 : x_1 : \cdots : x_n] = [1 : x_1/x_0 : \cdots : x_n/x_0]. \end{align*} The affine coordinates $(a_1, \ldots, a_n) = (x_1/x_0, \ldots, x_n/x_0)$ are by definition the image $\varphi_0([p])$. \textbf{The connection to dehomogenisation.} Substituting $X_0 = 1$ in the homogeneous polynomial $f(X_0, \ldots, X_n)$ produces a polynomial $f(1, Y_1, \ldots, Y_n)$ in $n$ variables: this is the dehomogenisation. Its value on the affine coordinates $(a_1, \ldots, a_n)$ equals the value of $f$ on the normalised representative $(1, a_1, \ldots, a_n)$ — by definition. So \begin{align*} f \text{ vanishes at } [p] \iff f(1, a_1, \ldots, a_n) = 0 \iff f^{\mathrm{deh}}(\varphi_0([p])) = 0. \end{align*} The chart and the dehomogenisation transport the vanishing condition from $\mathbb{P}^n_k$ to $\mathbb{A}^n_k$ verbatim.[/guided]

[step:Apply the equivalence to every homogeneous element of $I$] Recall that $V = V(I)$ is by definition the set of $[p] \in \mathbb{P}^n_k$ such that $f(p) = 0$ for every $f \in I$. Since $I$ is a homogeneous ideal, $V(I)$ is determined by the homogeneous elements of $I$ — these are the only ones with well-defined vanishing on $\mathbb{P}^n_k$, and $I$ is generated by them. Equivalently, by the [Vanishing is Ideal-Independent](/theorems/2133) theorem, $V(I)$ equals $V$ of any homogeneous generating set of $I$. For $[p] \in U_0$ with $\varphi_0([p]) = (a_1, \ldots, a_n)$: \begin{align*} [p] \in V \cap U_0 &\iff \text{for all homogeneous } f \in I,\ f(p) = 0 \\ &\iff \text{for all homogeneous } f \in I,\ f^{\mathrm{deh}}(a_1, \ldots, a_n) = 0 \\ &\iff \text{for all } f \in I,\ f(1, a_1, \ldots, a_n) = 0, \end{align*} where the second equivalence is the equivalence proved in Step 2 (applied to each homogeneous $f \in I$), and the third equivalence uses that $f(1, Y_1, \ldots, Y_n)$ depends only on the homogeneous components of $f$ (substituting $X_0 = 1$ commutes with taking homogeneous components, and the substitution is well-defined for any $f$, not just homogeneous ones; if $f = \sum_d f_d$ with $f_d$ homogeneous of degree $d$, then $f(1, Y) = \sum_d f_d(1, Y)$). Hence \begin{align*} \varphi_0(V \cap U_0) = \{(a_1, \ldots, a_n) \in \mathbb{A}^n_k : f(1, a_1, \ldots, a_n) = 0 \text{ for all } f \in I\} = V\!\left(\{f(1, Y_1, \ldots, Y_n) : f \in I\}\right). \end{align*} [/step]

[step:Note the analogous statement for the other charts $U_i$] The argument transports verbatim to any other standard chart \begin{align*} U_i &= \{[x_0 : \cdots : x_n] \in \mathbb{P}^n_k : x_i \neq 0\} \\ \varphi_i : U_i &\to \mathbb{A}^n_k, \qquad [x_0 : \cdots : x_n] \mapsto (x_0/x_i, \ldots, \widehat{x_i/x_i}, \ldots, x_n/x_i) \end{align*} (omitting the $i$th entry, which would be the constant $1$). Replacing the dehomogenisation $f \mapsto f(1, Y_1, \ldots, Y_n)$ with $f \mapsto f(Y_0, \ldots, Y_{i-1}, 1, Y_{i+1}, \ldots, Y_n)$ (substitute $X_i = 1$), the same argument gives \begin{align*} \varphi_i(V \cap U_i) = V\!\left(\{f(Y_0, \ldots, Y_{i-1}, 1, Y_{i+1}, \ldots, Y_n) : f \in I\}\right) \subset \mathbb{A}^n_k. \end{align*} This completes the proof: each affine piece $V \cap U_i$ is, under the chart $\varphi_i$, an affine variety in $\mathbb{A}^n_k$, with defining ideal generated by the dehomogenisations with respect to $X_i$ of the elements of $I$. [/step]