I'm self-studying Mendelson's Introduction to Topology. There is an example in the identification topology section that I cannot understand:

Let $X$ and $Y$ be topological spaces and let $A$ be a non-empty closed subset of $X$. Assume that $X$ and $Y$ are disjoint and that a continuous function $f : A \to Y$ is given. Form the set $(X - A) \cup Y$ and define a function $\varphi: X \cup Y \to (X - A) \cup Y$ by $\varphi(x) = f(x)$ for $x \in A$, $\varphi(x) = x$ for $x \in X - A$, and $\varphi(y) = y$ for $y \in Y$. Give $X \cup Y$ the topology in which a set is open (or closed) if and only if its intersections with $X$ and $Y$ are both open (or closed). $\varphi$ is onto. Let $X \cup_f Y$ be the set $(X - A) \cup Y$ with the identification topology defined by $\varphi$.

Let $I^2$ be the unit square in $\mathbb{R}^2$ and let $A$ be the union of its two vertical edges. Let $Y = [0, 1]$ be the unit interval. Define $f : I^2 \to Y$ by $f(x, y) = y$. Then $I^2 \cup_f Y$ is a cylinder formed by identifying the two vertical edges of $I^2$.

I don't understand how $I^2 \cup_f Y$ can be a cylinder. The set is equal to $(I^2 - A) \cup Y$. Which is a union of a subset of $\mathbb{R}^2$ and $[0, 1]$. How can this be a cylinder?

The book has an exercise that constructs a torus in a similar manner. I'm hoping to be able to solve it once I understand this example.

I looked up some examples online. While I understand the definitions and theorems of identification topologies, I have no clue how geometric objects are constructed.

  • 2
    $\begingroup$ I'm not quite sure where your confusion comes from. Why can that set not be a cylinder? Look at it from the other side: Take a cylinder (i.e. tube of finite length). Draw a vertical line on it. Now your cylinder is a union of a square and a line - but there are now small open neighbourhoods that 'reach around' the square - that's what the identification topology does. (The is bent/curved now, but that doesn't matter: Topological structure doesn't capture curvature in general.) $\endgroup$ – us2012 Aug 16 '12 at 0:16
  • $\begingroup$ @us2012 - Thanks for your comment. How do I rigorously prove it's a cylinder? Why not a torus? Is it by finding a homeomorphism between the two sets? $\endgroup$ – PeterM Aug 16 '12 at 0:18
  • 1
    $\begingroup$ @Peter, you should not worry about the geometry of the spaces, you should only take care of the topological properties. $\endgroup$ – Sigur Aug 16 '12 at 0:30
  • $\begingroup$ @Sigur - I understand the topological properties. I know what the open sets are in the constructed space for example. The problem is, the book instructs me to see how a space is a geometric object. $\endgroup$ – PeterM Aug 16 '12 at 8:24

Yeah so this is one of those things where imho intuition (given excellently by us2012) is way more important than the details.

For the details, fix the cylinder as $[0,1] \times S^1$, with coordinates $(h, \theta), \theta \in [0, 2\pi)$ (how high on the cylinder you are, and where you are on the circle at that height).

For ease of notation, describe points in $I^2 \cup_f A$ as just $(x,y) \in I^2$, modulo $(0,y) \sim (1, y)$.

Then we have (inverse, continuous) maps $I^2 \cup_f A \rightarrow [0,1] \times S^1$ given by $(x,y) \mapsto (x, 2\pi y)$, and $[0,1] \times S^1 \rightarrow I^2 \cup_f A$ given by $(h, \theta) \mapsto (h, \frac{\theta}{2\pi})$.

I'll leave checking these are inverse and continuous as an exercise.

  • $\begingroup$ Thank you. This is exactly what I was looking for. When my intuition fails, I'll look for a homeomorphism to prove the correspondence. $\endgroup$ – PeterM Aug 16 '12 at 8:25

I think it's good that you ask this question, plus one. Your intuition will eventually develop, don't worry. I had trouble understanding identification topologies too when I saw them first. It just takes some time to get used to, don't worry. The way I think about it now, is as follows:

You have two spaces $X,Y$ and you know how you want to "glue" them together, namely, you take all the points in $A \subset X$ and stick them on $Y$. The map $f$ tells you where on $Y$ you stick them. In pictures it looks something like this:

enter image description here

In the example, this looks like this:

enter image description here

  • $\begingroup$ Thanks for the drawings. Very helpful. :) $\endgroup$ – PeterM Aug 16 '12 at 8:27
  • $\begingroup$ @Peter Glad to hear : ) $\endgroup$ – Rudy the Reindeer Aug 16 '12 at 9:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.