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Let $V,U$ vector subspaces of $\mathbb{C}^{n+1}$ of dimension $r+1$, $s+1$.

Let $\pi : \mathbb{C}^{n+1}/\{0\} \longrightarrow \mathbb{CP}^n$ the projection map to complex projective space defined by $\pi:(x_0, ... ,x_n)\mapsto(x_0:...:x_n)$.

This is a question that came up while doing homework and I wanted to check some reasoning. It is useful to have that $\pi(V)\cap\pi(U) = \pi(V\cap U)$ for a homework question. Is this true? It seemed to be true based on my currently poor intuition about projective space, so I tried to prove it.

My reasoning for a simple proof was to prove both inclusions. Specifically, if $u\in \pi(U\cap V)$ then there is some $x\in U\cap V$ st $u=\pi(x)$. But then $x$ is in both $U$ and $V$ so $u\in \pi(U)\cap\pi(V)$.

Going in the other direction, if $u\in \pi(U)\cap\pi(V)$ then there is $x\in U$ and $y\in V$ such that $\pi(x)=\pi(y) = u$. But if this is true then $x$ is a scalar multiple of $y$ and vis versa (by the properties of vector subspaces), so $x\in U\cap V$. Thus $u\in\pi(U\cap V)$ and we have proved the sets are equivalent.

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Looks good. Notice the first part only depends on the fact that $\pi$ is a surjective function. – Mariano Suárez-Alvarez Jan 18 '11 at 6:26
I agree with Mariano. Just a little nitpicking: the projection map is defined on $\mathbb{C}^{n+1} \smallsetminus \{0\}$, not on all of $\mathbb{C}^{n+1}$. – t.b. Jan 18 '11 at 6:28
Yes, I guess i can simplify my argument that way. Thank you. – AnonymousCoward Jan 18 '11 at 6:28
@Theo of course. I forgot to tex that but its in my notes. – AnonymousCoward Jan 18 '11 at 6:29
up vote 2 down vote accepted

Yes, this is true, and your reasons are correct. Both describe the set of one dimensional subspaces minus the origin contained in both $U$ and $V$. Since fibers of $\pi$ are one dimensional subspaces minus the origin, any fiber that intersects a subspace must be contained in that subspace. Note that $f(A\cap B)\subseteq f(A)\cap f(B)$ is always true for the same reason you gave that $\pi(U\cap V)\subseteq\pi(U)\cap\pi(V)$.

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