Proof for showing that a set of space curves form a manifold I basically have a smooth space curve $\alpha$,with curvature $\kappa$ and $\tau$, both non-zero, and I generate a family of curves $M_{\alpha} = \{\dfrac{\alpha}{\mu} : \mu \in (0, \infty) \}$ . The new curvature and torsion functions are now $\mu\kappa$ and $\mu\tau$. I wish to show that $M_{\alpha}$ is a smooth manifold. 
I can get a multiplicative group structure on $M_{\alpha}$ through the multiplicative group structure on $(0, \infty)$. But will properties like local compactness of $(0, \infty)$ transfer onto $M_{\alpha}$. What kind of topological and geometric structure can $M_{\alpha}$ have??
Edit :> The curve $\alpha$ initially satisfies a functional relation of the form $f(\kappa, \mu) = \lambda$, $\lambda$ being a positive constant. The family $M_{\alpha} $ then arises out of curvature and torsion functions being modified as $\kappa_{\mu} = \mu\kappa$ and $\tau_{\mu} = \mu\tau$. Then $f(\kappa_{\mu}, \tau_{\mu}) = \lambda .\mu$, so I just defined a mapping $\psi :M_{\alpha} \to (0, \infty)$ as $$\psi(\dfrac{\alpha}{\mu}) = \mu$$ and ergo, $$\dfrac{\alpha}{\mu_1} * \dfrac{\alpha}{\mu_2} = \dfrac{\alpha}{\mu_1 . \mu_2}$$ is the multiplicative binary operation. Does what I have done seem sensible?
 A: I'll just make a few remarks.
First: Let $I$ be the domain of your smooth curve $\alpha \colon I \to \mathbb{R}^3$, which I'll assume to be an open interval.  The map
$$\begin{align}
X \colon I \times (0,\infty) & \to \mathbb{R}^3 \\
X(t,\mu) & = \mu \alpha(t)
\end{align}$$
is a smooth surjection onto $M_\alpha$.  The map $X$ will be an immersion at points $(t_0, \mu_0)$ for which the tangent vector $\alpha'(t_0)$ is not a scalar multiple of the position vector $\alpha(t_0)$.
That is, if the tangent vector $\alpha'(t_0)$ does not lie on the (radial) ray from the origin to the point $\alpha(t_0)$, then the map $X$ will be an immersion at $(t_0, \mu_0)$.
Second: Note that if the curve $\alpha$ ever intersects a ray emanating from the origin in two or more points, then the map $X$ will not be injective.  In particular, if $\alpha$ intersects itself, then $X$ will not be injective.

Conclusion: If $\alpha$ is not tangent to any ray through the origin, and if $\alpha$ never intersects a ray through the origin in more than one point, then $M_\alpha$ is the image of an injective immersion.

Thus, in this case, $M_\alpha$ has a unique topology and smooth manifold structure for which $X \colon I \times (0,\infty) \to M_\alpha$ is a diffeomorphism.  In short: $M_\alpha$ is diffeomorphic to a plane in this case.

Here are two examples to illustrate the extent to which the above can fail if hypotheses are ignored.
(1) If $\alpha(t) = (t,0,0)$ is a segment of the $x$-axis, then $M_\alpha$ is literally just that same exact segment.  In other words, $M_\alpha$ is only $1$-dimensional in this example.
(2) Here is a more interesting example.  Consider a smooth curve $\alpha$ which has the following behavior:


*

*$\alpha(t)$ lies on the positive $x$-axis for $t \in (-\infty, 0)$

*$\alpha(t)$ lies on the quarter-plane $\{(x,y,0) \colon x,y > 0\}$ for $t \in (0, 1)$

*$\alpha(t)$ lies on the positive $y$-axis for $t \in (1, 2)$

*$\alpha(t)$ lies on the quarter-plane $\{(0,y,z) \colon y,z > 0\}$ for $t \in (2, 3)$

*$\alpha(t)$ lies on the vertical line $\{\alpha(3) + s(0,0,1)\}$ for $t \in (3, +\infty)$.


Then $M_\alpha$ is the union of the quadrant $\{(x,y,0) \colon x,y > 0\}$ in the $xy$-plane and the quadrant $\{(0,y,z) \colon y,z > 0\}$ in the $yz$-plane.
