Pointed Convex cone: one-to-one correspondence extreme rays - extreme points Hoi, let $V$ be a finite dimensional real vector space with inner product $\left\langle .\right\rangle$. Let $\Gamma\subset V$ and $\Gamma \neq \left\{0\right\}$ a pointed convex cone. (Pointed means $\Gamma \cap -\Gamma = \left\{0\right\}$)
I want to show there exists a hyperplane $H$ through $0$ and $x_0\in \Gamma, x_0\neq 0$ such that the extreme points of the convex set $C:=\Gamma\cap (x_0+H)$ are in one-to-one correspondence with the extreme rays of $\Gamma$. 
I know there exists $z\in \Gamma\setminus \left\{0\right\}$ such that  $\left\langle z,w\right\rangle >0$ for all $w\in \Gamma\setminus  \left\{0\right\}$. Then if we consider the hyperplane $H$ with normal $\textbf{n} =z$, then $H$ is a hyperplane through 0 such that $H\cap \Gamma = \left\{0\right\}$
Then i suppose for $x_0$ we can take some $x_0 = \lambda z$ with $\lambda >0$ and then the rest seems intuitively clear. (but this is my Euclidean intuition speaking). But how do I make this precise? 
Thanks for any help, or suggestions. 
 A: Although a $z\in\Gamma\backslash\{0\}$ such that $\langle z,w\rangle>0$ for all $w\in\Gamma\backslash\{0\}$ does not exist for general pointed cones, your result follows if we assume it does.
Let $H=z^\perp$ and let $x_0=z$.
The set of rays in $\Gamma$ is $R=(\Gamma\backslash\{0\})/\sim$, where $x\sim y$ iff $x=ay$ for some $a\in(0,\infty)$. We will denote the ray containing $x$ by $[x]=\{ax|a\in(0,\infty)\}$.
Now define the map $\phi\colon R\to (H+z)\cap\Gamma\colon[x]\mapsto\frac{\langle z,z\rangle}{\langle x,z\rangle}x$.
First of all, $\frac{\langle z,z\rangle}{\langle x,z\rangle}x\in [x]\subset\Gamma$ since $\frac{\langle z,z\rangle}{\langle x,z\rangle}>0$. Secondly $\frac{\langle z,z\rangle}{\langle x,z\rangle}x\in H+z$ since
$$\left\langle\frac{\langle z,z\rangle}{\langle x,z\rangle}x-z,z\right\rangle = \frac{\langle z,z\rangle}{\langle x,z\rangle}\langle x,z\rangle-\langle z,z\rangle =0.$$
Thirdly, $\phi$ is well defined since $\phi([ax]) = \frac{\langle z,z\rangle}{\langle ax,z\rangle} ax = \frac{\langle z,z\rangle}{\langle x,z\rangle}x = \phi([x])$.
Now I claim that $\phi$ is a bijection. First we prove injectivity: Suppose $\phi([x])=\phi([y])$, then $\frac{\langle z,z\rangle}{\langle x,z\rangle} x = \frac{\langle z,z\rangle}{\langle y,z\rangle}y$ and hence $x = \frac{\langle x,z\rangle}{\langle y,z\rangle}y$. Since $\frac{\langle x,z\rangle}{\langle y,z\rangle}>0$, this proves $x\sim y$ and hence $[x]=[y]$.
Secondly we prove surjectivity: Let $x\in(H+z)\cap\Gamma$. Since $x\in H+z$, $0=\langle x-z,z\rangle = \langle x,z\rangle -\langle z,z\rangle$, thus $\langle x,z\rangle = \langle z,z\rangle$ and we have $\phi([x]) = \frac{\langle z,z\rangle}{\langle x,z\rangle}x = x$.
Now all that remains to show is that $[x]$ is extremal if and only if $\phi([x])$ is. We do this by contraposition. Suppose $\phi([x]) = t v+(1-t)w$ for some $v\neq w\in (H+z)\cap\Gamma$ and some $t\in(0,1)$. Then $[x] = [\phi([x])] = [tv+(1-t)w] = t[v]+(1-t)[w]$. Furthermore $[v]\neq[w]$ since $\phi([v])=v\neq w=\phi([w])$.
Now suppose $[x]=t[v]+(1-t)[w]$ for some $[v]\neq[w]\in R$ and some $t\in(0,1)$.
Then $x = av+bw$ for some $a,b> 0$. Hence $$\phi([x]) = \frac{\langle z,z\rangle}{\langle x,z\rangle}x = \frac{\langle z,z\rangle}{\langle x,z\rangle}(av+bw) = a\frac{\langle z,z\rangle}{\langle x,z\rangle}v + b\frac{\langle z,z\rangle}{\langle x,z\rangle}w$$
which
$$=a\frac{\langle v,z\rangle}{\langle x,z\rangle}\phi([v])+b\frac{\langle w,z\rangle}{\langle x,z\rangle}\phi([w]).$$
Since $a\frac{\langle v,z\rangle}{\langle x,z\rangle}+b\frac{\langle w,z\rangle}{\langle x,z\rangle} = \frac{\langle av+bw,z\rangle}{\langle x,z\rangle} = \frac{\langle x,z\rangle}{\langle x,z\rangle}=1$ and since both $a\frac{\langle v,z\rangle}{\langle x,z\rangle}$ and $b\frac{\langle w,z\rangle}{\langle x,z\rangle}$ are positive this concludes the argument.
