If $U = \{(u,v)\mid 0 < u,\; 0 < v < 2\pi\}, $ and $ x(u,v) = (u\cos v,u \sin v,u+v). $
Then how can we show that $x$ is a simple surface?
Is it necessary true that for being simple implies that we need:
smoothness 1-1 and class C^k or is there anything I have overlooked?
For being simple, we need to show that it is smooth, 1-1, and of class $C^k$. How does the geometry help? I see that $x$ is a helix and to show 1-1, I'm thinking if we let $x(u,v) = x(a,b)$ then this implies $u = a$ and $v = b$, but this seems just wrong
Thanks