Morphism of projective varieties So our professor gave us this problem and I'm not sure I understand what he wants us to do.
"Given this projective variety
$$V=\left\{(x_0 : x_1 : x_2 : x_3 : x_4 ) \mid rank
\begin{pmatrix}x_0&x_1&x_2\\x_3&x_2&x_4\end{pmatrix}
≤ 1 \space \right\}$$
Show that there is a morphism $ϕ : V → P^2 (k)$ which on the open subset of V where
$(x_0 ,x_1 ,x_2 ) \neq 0 $ is given by the projection
$(x_0 : x_1 : x_2 : x_3 : x_4 ) → (x_0 : x_1 : x_2 )$."
Why can't I just say that dropping $x_3$ and $x_4$ (ie. the projection itself) is already a morphism?
 A: Your variety is isomorphic to $\mathbb P^2$ blown up in one point, say $(0:0:1)$.
To see this, note that the blowup of this point is given by the equation $rx-sz=0$ in $\mathbb P_{x,y,z}^2 \times \mathbb P_{r,s}^1$. Using the Segre embedding, we can put this in $\mathbb P^5$, with coordinates $x_0,\ldots,x_5$, with equations
$$
\begin{pmatrix}
x_0 & x_1 & x_2 \\
x_3 & x_4 & x_5 
\end{pmatrix} \leq 1.
$$
However, the equation $rx-sz$ transforms under Segre to $x_0-x_5=0$, which allows us to eliminate $x_5$ from the equations. Hence the blowup can be realized in $\mathbb P^4$ with equations 
$$
\begin{pmatrix}
x_0 & x_1 & x_2 \\
x_3 & x_4 & x_0 
\end{pmatrix} \leq 1,
$$
which after a change of coordinate is equal to your variety $V$. Now, the blowup has a natural projection to $\mathbb P^2$.
Under this description of $V$, one sees that any point on $V$ has the form $(xr:yr:zr:xs:ys) \in \mathbb P^4$, where $(x:y:z) \in \mathbb P^2$ and $(r,s) \in \mathbb P^1$ (this is because they lie in the image of the Segre embedding).
If $(x_0,x_1,x_2) \neq 0$, then $r \neq 0$ (since $x_0=xr$, etc). Hence the map forgetting the last two coordinates is a well-defined map to $\mathbb P^2$ as long as $(x_0,x_1,x_2) \neq 0$. 
