\documentclass{minimal}

\usepackage{xcolor}
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\newcommand{\A}[1]{\mathbf{A}\sp{#1}}
\newcommand{\Proj}[1]{\mathbf{P}\sp{#1}}
\newcommand{\An}{\A{n}}
\newcommand{\Projn}{\Proj{n}}
\newcommand{\Ano}{\A{n+1}}
\newcommand{\ZProj}{Z\sb{\mathbf{P}}}
\newcommand{\ZA}{Z\sb{\mathbf{A}}}
\newcommand{\rootalpha}{\sqrt{\alpha}}
\newcommand{\cl}{\mbox{cl}}
\begin{document}
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{
\includegraphics[width=\textwidth]{exIp2p7.png}\\
For part (a), just use \href{../Ip2p6/Ip2p6.html}{yesterday's mess}:
\begin{align*}
n + 1 &= \dim S(\Projn)\\
	&= \dim \Projn + 1
\end{align*}
Subtracting $1$ on both sides finishes it off.\\\\
That was easy. Hey, the hint says to use yesterday's mess for part (b) too!\\
1.10 is this guy:\\
\includegraphics[width=\textwidth]{propIp1p10.png}
(So, it turns out this is another "projectivization of affine" thing)\\
So how do we hook up 2.6 and 1.10?\\
Well, $Y$ being a quasi-projective variety means its an open subset of some projective variety. Actually, by \href{../1p6/Ip1p6.html}{1.6}, $\overline Y$ \emph{is} precisely that variety (or at least the smallest such variety). Now, the bar notation for closure is going to be a little too ambiguous in this exercise, so let's move everything to the $\cl$ notation, and we'll be using }\href{https://en.wikipedia.org/wiki/Closure_(topology)#Facts_about_closures}{this property} \textcolor{hotpink}{a lot:\\
\\
"If $A$ is a subspace of $X$ containing $S$, then the closure of $S$ computed in $A$ is equal to the intersection of $A$ and the closure of $S$ computed in $X$: $\cl_{A}S=A \cap \cl_{X}S$"\\\\
I.e. Let's call $X = \overline Y$, so we can instead write $X = \cl_{\Projn} Y$. And note that $\cl_X(Y) = \cl_{\Projn}(Y) \cap X = X \cap X = X$.\\\\. 
(So our goal in the new notation is to show $\dim X = \dim Y$)\\\\
Okay. Now let's use the homeomorphism from last time $\phi: X \cap U_0 \rightarrow Y_0$. We saw last time that 
\begin{equation}\label{eq1}
\dim X = \dim Y_0
\end{equation}
So what happens to the closure of $Y$ when intersected with $U_0$? Well:\\
\begin{align*}
\cl_{X\cap U_0}(Y\cap U_0) &= \cl_X (Y \cap U_0) \cap (X \cap U_0)\\
&= X \cap X \cap U_0 & \mbox{1.6 again: since $Y$ and $U_0$ are both open in $X$} \\
&= X \cap U_0
\end{align*}
\\
And since $\phi$ is a homeomorphism, closures commute:\\
\begin{align*}
\cl_{Y_0}(\phi(Y \cap U_0)) &= \phi(\cl_{X\cap U_0}(Y \cap U_0))\\
&= \phi(X \cap U_0)\\
&= Y_0
\end{align*}
\\
I.e. in "bar" notation:\\
\begin{equation}\label{eq2}
\overline{ \phi(Y \cap U_0) } = Y_0
\end{equation}
\\
And hence, by 1.10:\\
\begin{align*}
\dim Y_0 &= \dim \phi(Y \cap U_0)\\
&= \dim Y \cap U_0 & \mbox{($\phi$ is a homeomorphism)}
\end{align*}
combining this with \eqref{eq1}...
\begin{equation}
\dim X = \dim Y \cap U_0
\end{equation}\\
Holy shit this is getting messy. Let's just apply \href{../1p10a/Ip1p10a}{1.10a} to finish it off.\\
\begin{align*}
\dim X &= \dim Y \cap U_0\\
&\leq \dim Y & \mbox{1.10a}
\end{align*}
and also by 1.10a, $\dim Y \leq \dim X$.\\\\
(ACTUALLY, I just realized I could have just used 1.6 again to say $\dim Y = \dim Y \cap U_0$. WHATEVS)
%\includegraphics[width=\textwidth]{exIp2p4.png}
%\href{blah}{blah}
}
\end{document}