# How does gluing work?

If one "glues" together a cylinder $C$ with a cylinder $C'$ the resulting space should be a torus as a subspace of $\mathbb R^3$. Both $C$ and $C'$ are $S^1 \times [0,1]$. If I understand it correctly the identification requires two maps $f: C \to \mathbb R^3$ and $g: C' \to \mathbb R^3$ with the property that $f(x)=g(x) \in S^1 \subseteq \mathbb R^3$ on $S^1 \times \{0\}$ and $f(x)=g(x) \in S^1 \subseteq \mathbb R^3$ on $S^1 \times \{1\}$ such that $f,g$ are continuous and one-to-one.

What I'm struggling with is how to define $f$ and $g$. I tried as follows: My idea is to define $f$ and $g$ such that they are inclusion maps into the torus at the origin of $\mathbb R^3$. The torus can be parameterized as $x = \cos(s)(R+r\cos(t))$, $y=\sin(s)(R+r\cos(t))$, $z=r\sin(t)$ for $s,t \in [0,2\pi)$.

For a point $(x,y)$ on $C$ how to define $f:C \to \mathbb R^3$? I can not do it. Thank you for any help!!

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Try visualizing it - what happens (visually) if you 'slice' your torus by a plane through the $z$-axis, say the plane $y=0$? What would this correspond to in terms of the torus's parametrization? –  Steven Stadnicki Jan 15 '13 at 16:05
Also, note that after the first gluing you again obtain a cylinder, just bigger. Therefore, it's not a strictly necessary step here. It's enough to glue together two ends of a single cylinder together. –  Marek Jan 15 '13 at 16:28

Instead of $(x,y)$, let's write $(\cos(\theta), \sin(\theta), \phi)$. Then $f$ should send this to $(\cos(\theta)(R + r\cos(\pi\phi)),\sin(\theta)(R + r\cos(\pi\phi)),r\sin(\pi\phi))$. Thus, the $\theta$ coordinate is sent to the $s$ coordinate, and the $\phi$ coordinate is sent to the first $[0,\pi]$ of the $t$ coordinate. $g$ should do likewise, but covering the $[\pi,2\pi]$ of the $t$ coordinate.
If you continue to study topology, you'll probably see gluings like this treated more abstractly, in arguments like the following. Both cylinders are product spaces $S^1 \times [0,1]$, and the gluing ignores the $S^1$ coordinate. Thus, the glued space should be $S^1$ times the space you get by gluing two intervals along their endpoints -- that is, it's $S^1 \times S^1$, a torus.
Is it possible that it is the other way around and $f$ should send $\Phi \to s = 2 \Phi$ and $\Theta \to t= \Theta$? The parametrization of $S^\times [0,1]$ you suggest is such that $\Theta \in [0, 2 \pi]$ and $\Phi \in [0 , \pi / 2]$. –  goobie Jan 16 '13 at 8:01
I don't think so... $\phi$ takes values in $[0,1]$, so $\pi\phi$ takes values in $[0,\pi]$. It shouldn't matter whether we send $\phi$ to $s$ and $\theta$ to $t$ or the other way around -- these are two distinct ways of gluing the two cylinders to give the torus! –  Paul VanKoughnett Jan 17 '13 at 1:50
I used $\phi \in [0, \pi / 2]$ as the angle in the middle of the cylinder. It also seems to work. Thank you for answering! –  goobie Jan 17 '13 at 13:55