Local surface theory 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
 A: We can see that $x(u,v)$ is smooth, because $u, v, \cos v$, and $\sin v$ are all smooth functions. 
To see that $x(u,v)$ is one-to-one, suppose $u,u,>0$ and $0<v,v'<2\pi$ such that 
$x(u,v)=x(u',v,)$. This implies by definition that 
$$(u\cos v,u \sin v,u+v)=(u'\cos v',u' \sin v',u'+v')$$
or equivalently, 
$$\tag{1} u\cos v=u'\cos v', u \sin v=u' \sin v', u+v=u'+v'.$$
By the first two equations in $(1)$, we have
$$(u\cos v)^2+(u\sin v)^2=(u'\cos v')^2+(u'\sin v')^2$$
which implies that $u^2=u'^2$. Since $u,u,>0$, we have $u=u'$. Now by the third equation in $(1)$, we have $v=v'$. This shows that $x(u,v)$ is one-to-one.
To show that $x(u,v)$ is regular, note that its differential is given by 
$$dx=\left[
  \begin{array}{cc}
    \cos v & -u\sin v \\
    \sin v & u\cos v \\
    1 & 1 \\
  \end{array}
\right].$$
To show that its rank is 2, note that the minor given by the first and second rows is
$$\left[
  \begin{array}{cc}
    \cos v & -u\sin v \\
    \sin v & u\cos v \\
  \end{array}
\right],$$
which is nonsingular, because it has determinant $u\cos^2v+u\sin^2v=u>0$. 
A: For showing $1$-$1$. Try to think of an inverse of $x(u,v)$.
HINT: $\cos(v)^2+\sin(v)^2=1$
