Gluing manifolds

Is it true that gluing can be realized two ways? In the example of gluing two cylinders to obtain a torus one can imbed each cylinders into the torus like this:

A point $(\cos \phi , \sin \phi, \theta)$ in $C_1 = S^1 \times [0,1]$ maps to $x = \cos(2 \theta)(R+r\cos(\phi))$, $y=\sin(2 \theta)(R+r\cos(\phi))$, $z=r\sin(\phi)$ and $(\cos \phi , \sin \phi, \theta)$ in $C_2 = S^1 \times [0,1]$ maps to $x = \cos(\pi + 2 \theta)(R+r\cos(\phi))$, $y=\sin(\pi + 2 \theta)(R+r\cos(\phi))$, $z=r\sin(\phi)$. These maps define an imbedding into the torus in $\mathbb R^3$. This is a gluing where the boudaries of the cylinders are identified.

A different way of realizing the gluing: Set $X = C_1 \times \{1\} \cup C_2 \times \{2\}$. Define $f: \partial C_1 \times \{1\} \to \partial C_2 \times \{2\}$ as $(s,1,1) \to (s,1,2)$ and $(s,0,1) \to (s,0,2)$. Then define $x \sim y \iff f(x) = y$ and $T = X / \sim$.

Is the first the same as the second but with an imbedding i.e.: Does one have to make a quotient space in the first also (before imbedding)? Does gluing always mean take quotient?

-
Is gluing the same with a homeomorphism between a quotient space? Then imbedding into $\mathbb R^3$ is not a gluing. – goobie Jan 16 '13 at 10:26
(I have not read your question thoroughly as I'm in a hurry.) Typically gluing is easier if you are dealing only with the topological structure. In this case you can basically do whatever you want. But if you want your glued manifold to have differential structure, you need to be more careful as you will need to define an atlas on the glued manifold. Maybe this applies to the case at hand: one of the gluings is topological only, the other is apt to receive a differential structure. – Giuseppe Negro Jan 16 '13 at 11:01
@GiuseppeNegro Thank you for your comment. Isn't it the case that for the first approach with the imbedding one must also define a quotient space? Otherwise the two maps cannot be glued together into a homeomorphism. And the result must be a space together with a homeomorphism. Therefore the first approach must also do what the second approach does and therefore the first approach is the second approach but with additionally added an imbedding into $\mathbb R^3$? – goobie Jan 16 '13 at 12:55