I am trying to solve the following problem, but I don't have any clue about how to solve it, since in class we haven't even seen the "affine geometry" part of the Michèle Audin's book: Geometry.
The problem is this:
Let $E$ a vector space and let $\mathcal{F} \subset E$ be an affine subspace of $E$ not containing $0.$ Prove that the projection $$p:E-\{0\} \to P(E)$$ restricts to an injective map from $\mathcal{F}$ to $P(E).$
According to the definition of the book, a set $\mathcal{F}$ is an affine subspace of a vector space if there is an element $A \in \mathcal{F}$ s.t. $\Theta_A(\mathcal{F})$ is a vector subspace of $E.$ One can prove that you can choose any point $x \in \mathcal{F}$ and you'll obtain $\Theta_A(\mathcal{F})=\Theta_x(\mathcal{F}).$
In the other hand, I am assuming that the projection map is s.t. $v \mapsto [v] \in P(E).$
Now, let $v,w \in \mathcal{F}$ s.t. $p(v)=p(w).$ That is, $[v]=[w]$, that is, $v=\lambda w$ for some nonzero $\lambda.$ We have that $\Theta_v(\mathcal{F})$ is a vector subspace of $E.$
But I can't find information here to tell that $v=w$. How can I do this? Any help will be appreciated.
As @Greg Martin said you already did it. The difference $w-v$ always belongs to the subspace, so if $v=\lambda w$ with $\lambda \neq 1$, you have that $w-v=(1-\lambda)w$ is in the subspace. This is a contradiction, since dividing out $1-\lambda$ you would obtain that $w$ is in the subspace.