This concerns Example 2.11.4 in chapter V of Hartshorne's Algebraic Geometry.
- $\mathscr{E} := \mathcal{O} \oplus \mathcal{O}(1)$ on $\mathbb{P}^n$.
- $P_0$ is the point $[1 : 0 : \ldots : 0]$ in $\mathbb{P}^{n+1}$.
- $V$ is the blowup of $P_0$ on $\mathbb{P}^{n+1}$ ; $V$ is isomorphic to $\mathbb{P}(\mathscr{E})$.
- $Y$ is any subvariety of $\mathbb{P}^n$.
- $X_0$ is the cone over $Y$ in $\mathbb{P}^{n+1}$ with vertex $P_0$.
- $X$ is the blowup of $P_0$ on $X_0$.
Hartshorne writes "This variety $X$ is clearly the inverse image of $Y$ under the projection $\pi : V \cong \mathbb{P}(\mathscr{E}) \to \mathbb{P}^n$." How do I see this?
$\newcommand{\PP}{{\mathbb P}}$ $\newcommand{\Bl}{{\mathrm Bl}}$ Let $\pi_1:V \to \PP^n$ and $\pi_2:V \to \PP^{n+1}$ be the two projections Hartshorne introduces. As he demonstrates $\pi_1$ is the map $\PP({\mathcal E}) \to \PP^n$ and $\pi_2$ is the map $\Bl_{P_0} \PP^{n+1} = V \to \PP^{n+1}$.
Now let $p:\PP^{n+1} - P_0 \to \PP^n$ be the projection from a point and let $i:V - E \to V$ be the canonical inclusion and $E$ the exceptional divisor of the blow-up $\pi_2$ (so that $\pi_2(E) = P_0$). Furthermore call the restriction of $\pi_2$ by the name of $\pi_2':V - E \to \PP^{n+1}-P_0$ (it is an isomorphism).
Now we have the commutation relation
$$\pi_1 \circ i = p \circ \pi_2'$$
and the chain of equalities
$$ \begin{multline} \Bl_{P_0} X_0 = \overline{\pi_2'^{-1}(X_0 - P_0)} = \overline{\pi_2'^{-1} p^{-1}(Y)} = \overline{i^{-1}\pi_1^{-1}(Y)} = \pi_1^{-1}(Y) \end{multline} $$
This is the statement to be explained ("..is clearly...") Note that I have used the relation $X_0 - P_0 = p^{-1}(Y)$, which is, I think, really obvious from the definition of projective cone. (All the above reasoning would probably a bit easier to grasp, if I would have written an appropriate diagram with slanted arrows, which does not seem so easily with Mathjax).
Addendum: Let me explain $\pi_1 \circ i = p \circ \pi_2'$ more thoroughly:
$V \subseteq \PP^n \times \PP^{n+1}$ consists of pairs (warning: pairs written in inverted order of definition (by mistake)) $[x_0:\ldots:x_{n+1}] \times [y_1:\ldots:y_{n+1}]$. The map $\pi_1([x_0:\ldots]\times [y_1:\ldots]) = [y_1:\ldots:y_{n+1}]$ and $\pi_2([x_0:\ldots]\times [y_1:\ldots]) = [x_0:\ldots:x_{n+1}]$. Furthermore $p([x_0:x_1:\ldots:x_n]) = [x_1:\ldots:x_{n+1}]$. $E$ is the subset of $V$ with $x_1=\cdots=x_{n+1} = 0$. So $\pi_1(V-E)$ maps a point $Q = [x_0:\ldots] \times [y_1:\ldots]$ for which at least one $x_s$ with $s > 0$ is nonzero to $[y_1:\ldots:y_{n+1}]$. The other map $p \circ \pi_2$ maps the $Q$ to $[x_1:\ldots:x_{n+1}]$. As $V=V(x_i y_j - x_j y_i)$ these images are identical.