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Question Define an infinitely long cylinder.

Definition: Quotient topology

Let $\left ( X,\tau \right )$ be a topological space. Let ~ be an equivalence relation on X and let $X^{*}$ be he collection of equivalence classes. Let $\pi: X\rightarrow X^{*}$ be the map taking each point to its equivalence class, i.e. $\pi\left ( x \right )=\left [ x \right ]$.

The quotient topology $\tau^{*}$ on $X^{*}$ is the collection of sets $U \subseteq X^{*}$ such that $\pi^{-1}\left ( U \right ) \in \tau$.

I am having a bit of problem beginning this question. A cylinder has a disc at the end. So perhaps there should be an $S^{1}=\left [ 0,1 \right ]$ we could work from?

However, any useful hints to kickstart me would be utmost helpful.

Thanks in advance.

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  • $\begingroup$ What is the question, exactly? (Also, what does "has a 1-sphere area" mean?) $\endgroup$ Commented Sep 14, 2016 at 14:10
  • $\begingroup$ @NoahSchweber I have edited the question. $\endgroup$ Commented Sep 14, 2016 at 14:15
  • $\begingroup$ An infinitely long cylinder doesn't have any ends! It is defined by $\mathbb{R}\times S_1$. $\endgroup$
    – zooby
    Commented Sep 14, 2016 at 14:18
  • $\begingroup$ I apologise. But that's what the question on my notes asked for . I typed word for word. $\endgroup$ Commented Sep 14, 2016 at 14:18
  • $\begingroup$ I thought his question was the title. "Define an infinitely long cylinder." $\endgroup$
    – zooby
    Commented Sep 14, 2016 at 14:21

1 Answer 1

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Your definition of "cylinder" is incorrect: a cylinder does not have "caps" at the end. It should just be the "tube" part. (As a side note, you shouldn't write e.g. "$S^1=[0, 1]$" - "$S^1$" is the standard notation for the $1$-sphere, that is, the unit circle. In your previous edit, what you described as "$1$-sphere" is actually called a "$2$-disc" or "$2$-ball" - an $n$-sphere is the boundary of a $(n+1)$-ball. The "$n$" refers to the dimension of the object - so a circle is one-dimensional, even though it "lives" most naturally in two-dimensional space, and the cap at the end of a cylinder is two-dimensional.)

Intuitively, you can form an infinite cylinder by taking $\mathbb{R}^2$ and "rolling it up". The point of the question is to make this precise by formalizing it as a quotient space construction. HINT: let's say I roll the plane along the $x$-axis, and wind up with a cylinder with circumference $1$. Note that the points $(0, 0)$ and $(1, 0)$ wind up "overlapping." Do you see, in general, which points overlap with which other points?

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  • $\begingroup$ Let me give this some thought. $\endgroup$ Commented Sep 14, 2016 at 14:42
  • $\begingroup$ Yes I do see it geometrically. Suppose the glueing occurs at all points along the boundary (0,y) and (1,y) for y in the reals. This produces a long cylinder and stretches as far as the y-values goes. So, (0,0)~(1,0), (0,1)~(1,1),... However, I understand the need to formalise this but am still familiar with using the definition of quotient space to do so. $\endgroup$ Commented Sep 15, 2016 at 4:43

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