How does parametrization of the intersection of two surfaces induce a space curve? Given a two surfaces say: $z=1-y$ and $ x^2+y^2+z^2=1$, we find that they intersect at: $$x^2-2yz=0$$
How is the above a space curve? Is it not just another surface?
And why do we need to introduce a parameter, say $y=t$ , for it to be a space curve? I really don't understand the purpose of parametrization.
 A: When "geometric objects" intersect transversely, or in "general position" ,the dimension of their intersection is the dimension of the ambient space (which I am assuming here is $\mathbb R^3$) minus the sum of their respective codimensions, say "Cod" i.e., 
$$Dim (S_1 \cap S_2)_{\mathbb R^3}=Dim(\mathbb R^3)-Cod(S_1)-Cd(S_2) $$
Where $S_1, S_2$ are the two surfaces , giving $$3-1-1=1 $$
So the dimension of the set of points in the intersecting surfaces is $1$, i.e., the intersection is a curve.
A: You are right. $$x^2-2yz=0$$ is not a space curve. Is it not just another surface but one that shares a commonalty. It is new a cone, a surface similar to the plane and sphere given, with which it shares the same curved intersection. It is one among a set of possible infinitely many surfaces with a common track or locus.
$ z=1-y $ and $ x^2+y^2+z^2=1$, do not intersect at $$x^2-2yz=0.$$
In fact no two surfaces intersect along another surface!
Just as three curved lines can be  concurrent, three surfaces can have a ( I don't know if a proper word exists .."concur-line" :) for lack of one).
If any two of the surfaces is given, the third and indefinitely many more can be set up or found.
$$ (x,y,z) = (\pm \sqrt {2 t ( 1-t)} , t , (1-t) ) \tag{1}$$ is an easy parameterization of the common curved intersection. It has fixed or unique position in 3-space. Here also we can have several possibilities of changing the velocity vector. It is like a race-track. The track is fixed but so many speed / gradient  variations are possible prescribed at any point on common line. The purpose of such parameterization is to find embedding in 3-space for this common line of cutting.
This line gives (x,y,z) coordinates as a function of a needed single parameter.
A simpler example. You can imagine all spheres passing through $ x^2 + y^2 = 1$ with their centers on z-axis and varying radii.
EDIT1:
2 errors corrected. $ x^2- 2 y z = 0$ is a cone, not a hyperboloid. But my remarks do not suffer. Other is typo about $\sqrt (..) $ in parameterization.
The red circle is the "concur-line" of the cone, plane and sphere of parameterization (1).

EDIT2:
Picturization of all possible coniciods passing through this line (including the above three) is indicated in code given (please make the correction): 
https://mathematica.stackexchange.com/questions/87639/manipulated-surfaces-to-include-parametric3d-plot-of-concurrent-line
The animation shows successive changes, as morphing parameter $lambda$ ( zero for circular cone) is varied: 
Double plane $\rightarrow$ oblate ellipsoid $\rightarrow$  hyperboloid of 1 sheet $\rightarrow$ cone $\rightarrow$ sphere $\rightarrow$ hyperboloid of 2 sheets $\rightarrow$  prolate ellipsoid $\rightarrow$ Double plane
This I hope completes an answer to a question  " How is the parameterization of all surfaces inducing a common single space curve of intersection "?
A: The intersection of two surfaces will be a curve.  Here, one surface is the sphere defined by 
$$x^2+y^2+z^2=1 \tag 1$$
while the second surface is the plane defined by 
$$z=1-y \tag 2$$.

Now, the curve of intersection can be defined by a parametric curve 
$$\vec r(t)=\hat xx(t)+\hat yy(t)+\hat zz(t)$$
where $t\in [t_1,t_2]$.
And since this curve lies on both the sphere defined by $(1)$ and the plane defined by $(2)$, we have simultaneously
$$z(t)=1-y(t) \tag 3$$
and 
$$x^2(t)+y^2(t)+z^2(t)=1 \tag 4$$
Now, $(3)$ and $(4)$ have a common solution
$$x^2(t)+2y^2(t)-2y(t)=0 \tag 5$$
and 
$$z(t)=1-y(t) \tag 6$$

We rewrite $(5)$ as 
$$x^2(t)+2(y(t)-1/2)^2=1/2 \tag 7$$
from which we can see that a parametric curve describing the intersection of $(1)$ and $(2)$ can be expressed as
$$x(t)=\frac{1}{\sqrt{2}}\cos t$$
$$y(t)=\frac12+\frac12 \sin t$$
$$z(t)=\frac12-\frac12 \sin t$$
for $0\le t\le 2\pi$.

NOTE:
The parametric description is not unique and in fact, we need not use a parametric description to describe the curve of intersection.  Here, we could have as easily described the curve in terms of one of the coordinates.   For example, we can write
$$x=\pm\sqrt{y(1-y)}$$
$$y=y$$
$$z=1-y$$
for $0\le y\le 1$.
