# projective geometry hyperplane

For $j=0,\ldots,n$ consider the affine hyperplane $A_j:=e_j+\langle e_0,\ldots,e_{j-1},e_{j+1},\ldots,e_n\rangle$ in $\mathbb K^{n+1}$ and the associated embedding $\tau_j:\mathbb K^n\rightarrow\mathbb KP^n, \tau_j(x_1,\ldots,x_n):=[x_1:\ldots:x_j:1:x_{j+1}:\ldots:x_n]$, where $e_j\in\mathbb K^{n+1}$ the $j'$th unit vector is.

Now I come to my question; How can I show that the images of $\tau_j$ overlay whole $\mathbb KP^n$ or mathematically spoken: $\mathbb KP^n=\bigcup_{j=0}^n \tau_j(\mathbb K^n)$

I think it should be a short proof but since I am a newbie in projective geometry I do not really have an idea.

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I changed $<e_0,...,e_n>$ to $\langle e_0,\ldots,e_n\rangle$. That is standard TeX usage. – Michael Hardy Jan 6 '13 at 21:03

Given $[x_0:\ldots:x_n]$, there exists $j$ such that $x_j\neq0$; therefore $$\left[\frac{x_0}{x_j}:\ldots:\frac{x_n}{x_j}\right]=[x_0:\ldots:x_n]$$ has a $1$ in the $(j+1)$-th place, therefore sits in the image of $\tau_{j}$. As the point was generic, this shows that the images of these maps cover the whole projective space.
Thanks four your help, but did you use also the fact that $A_j$ is an affine hyperplane? Or because this is an affine hyplerplane it follows that there exists a $x_j$ such that $x_j\not=0$ ?? – Montaigne Jan 6 '13 at 21:19
No, I used the definition of $\mathbb{KP}^{n+1}$: it is the quotient of $\mathbb{K}^{n+2}\setminus\{0\}$ by the equivalence relation $v\sim w$ iff $v=\lambda w$ with $\lambda\in\mathbb{K}$. Therefore, every element of $\mathbb{KP}^{n+1}$ is a homogeneous (n+2)-tuple with at least one non vanishing element. – wisefool Jan 6 '13 at 21:50