I have a question about the proof that if the differential $dT$ of a transformation has rank 1 (2) at each point in a domain, then the image will be a curve (surface).
Stated more precisely (in $\mathbb{R}^3$ for convenience), suppose $T: \mathbb{R}^3\to \mathbb{R}^3$ is $C^1$ in an open set $D$ and has rank $1$ at each point $p\in D$. Then $T$ maps onto a curve in the target...in any point of $D$ there is a neighborhood in which one of the variables in the target can be used to express the others.
The proof I have read (in Buck's Advanced Calculus, p. 369-371) goes as follows: write $T$ as
$$T: \begin{cases}u=f(x,y,z)\\v=g(x,y,z) \\ w = h(x,y,z).\end{cases}$$
Take $p_0 \in D$. The matrix for $dT$ has at least one entry nonzero near $p_0\in D$ since it is rank 1, suppose for convenience it is $f_1$. Define $F(x,y,z,u) := f(x,y,z) - u = 0$, and by the implicit function theorem since $F_1 = f_1 \neq 0$, we can solve $x = K(y,z,u)$ in a neighborhood of $p_0$. Now $$v = g(K(y,z,u), y, z) = G(y,z,u)\\ w = h(K(y,z,u),y,z)=H(y,z,u).$$
We can show that $v$ and $w$ do not depend on $y$ or $z$. For instance $\frac{\partial v}{\partial y}=0$ as follows: we can regard $y,z,u$ as independent near $p_0$ by what we've just shown. Therefore, differentiating the equations for $u$ and $v$ and solving, we get $$\frac{\partial v}{\partial y}= \frac{\left|\begin{matrix} f_1 & f_2 \\ g_1 & g_2 \end{matrix}\right|}{f_1},$$ which is zero since the numerator is a $2 \times 2$ submatrix of $dT$. $\blacksquare$
My question is: we have shown that $v=G(u)$, $w = H(u)$ in a neighborhood of $p_0$, but we have not shown that these functions are $C^1$, for instance, so how do we know that the curve is not degenerate? Intuitively, $\operatorname{rank}(dT)=1$ should guarantee that the image is not a point (perhaps even that it is a smooth curve), but how have we shown this?
If we try to take the derivatives of $G$ and $H$, we get $G'(u) = g_1 \cdot K_3$, $H'(u) = h_1 \cdot K_3$, and I see no reason why $g_1$ or $h_1$ should be nonzero near $p_0$. ($K_3$ should be $\frac{1}{f_1}$ near $p_0$, so it should be nonzero.) Can someone clarify for me?
Around each point $u$ you have that $G$ and $H$ are compositions of $C^1$ functions and thus $C^1$, hence your image is a $C^1$ curve. Note that your $K$ changes as we move along the curve but different $K$s will agree on intersection (implicit function theorem).
If the function is constant in a neighbourhood around $p_0$ then the rank at $p_0$ would be 0.