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I am a bit confused with the definition of a suspension. This the definition.

For a space $X$, denote $SX$ the suspension of $X$ in which this is the quotient space $$\frac{X \times I}{\sim}$$ where $\sim$ is the equivalence of relation of $X \times \{0\}$ and $X \times \{1\}$ collapsed to a point.

The typical example is to set $X = S^n$. For $n = 1$, this is a "cylinder". What I don't understand is that why when we collapsed the top and end point of the cylinder our quotient space immediately becomes a "double-cone"? For example let's say $X \times \{1\} \to \{x_1 \} \times \{1 \}$ and $X \times \{0\} \to \{x_2\} \times \{0\}$. I don't understand why suddenly points close to $0$ and $1$ "shrink". For example, at $X \times \{3/4\}$, the "cone' picture depicts $S^1$ with a smaller radius.

To clarify what the problem is when we "shrink", at $\{3/4\}$ $X$ is no longer $S^n$

If the above example is too diffuclt to explain, we can work with $X = I$, so that $(X \times I) / {\sim}$ is a "diamond" on $\mathbb{R}^2$

Michael Hardy
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Lemon
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  • The intuition might be that, if $\epsilon$ is close to zero, then $(x, \epsilon) \in X \times I$ is close to $(x, 0)$, so when you collapse the 0 end to a single point, then in the quotient, $(x, \epsilon)$ is close to that point. Given $\epsilon$, this is true for every $x$, so the subspace $X \times {\epsilon}$ is in some sense smaller than the subspace $X \times {1/2}$. – John Palmieri Mar 20 '18 at 23:26
  • @JohnPalmieri yeah that is actually what confuses me right now. When the definition of suspension first was shown to me, I didn't have to argue why it would be a cylinder, but the more I thought about it, these questions "corrected" me. – Lemon Mar 20 '18 at 23:31
  • Note MathJax usage, as in my edit to this question. If you write $A\sim B,$ coded as A\sim B, you see a certain amount of space to the left and right of the binary relation symbol. But in $A/{\sim},$ that symbol is just a noun and that spacing is not appropriate. So write A/{\sim} instead of A/\sim and then you'll see $A/{\sim}$ instead of $A/\sim.$ I edited the question accordingly. Contrast the two here: $$ \begin{align} & A/{\sim} \ \ & A/\sim \end{align} $$ – Michael Hardy Mar 20 '18 at 23:48

4 Answers4

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You are right that when people draw it as a cone, they're taking a bit of license. Identifying the circle at the end of the cylinder to a single point technically does not change the "slope" of the cylinder. A more accurate depiction would be to keep the cylinder with same slope (say, horizontal), no shrinking or tapering, and just color the boundary circle at $0$ to notate that all points on the circle are the same.

However, this is hard to understand if you're new to quotients (infinitely many colored points represent a single point??), while a cone is quite easy to understand. And as far as the topology is concerned, the two options are homeomorphic.

And the tapering or sloped cone picture shows you something: points near the apex are near each other. On the cylinder (with colored end circle) that's harder to see.

And while the tapered cone picture is not isometric to a straight cylinder with its boundary circle smashed (which is why I say they are "taking license" with this picture), they are homeomorphic. If all you care is about the topology, then it is harmless to use the easier to draw and easier to understand cone.

ziggurism
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  • I am not too "new" into quotients, but I definitely would say these quotient structures in algebraic topology has helped me understand quotients in other contexts; but like u said, I don't have it 10000% handed down. You also mentioned this "nearness" near the apex, well that's thing, where is this coming from? Even if $\epsilon \to {0}$ or $\epsilon \to {1}$, we aren't quotienting out those spaces right? – Lemon Mar 20 '18 at 23:12
  • And this may have been in your post, but I missed it, but why they do fix the "collapsation" of the end points to the centre? I m gonna try to provide a picture in a bit, but taking $X$ to be $S^1$, why can't we fix the $X \times {0}$ to say "south pole" and $X \times {1}$ to say "north pole" (basically I want them to be on opposite end, so it isn't a "perfect cylinder". – Lemon Mar 20 '18 at 23:13
  • http://prntscr.com/iu1qzi Okay this is not the best picture, but it is the best i can draw to illustrate what i meant. – Lemon Mar 20 '18 at 23:27
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    @Hawk: It's just conventional to draw the endpoints as being over the center; this has no deep meaning whatsoever. If $X\subset\mathbb{R}^n$ and you actually realize the suspension as a subset of $\mathbb{R}^{n+1}$ geometrically as in my answer, moving around the endpoints a bit will not change the space up to homeomorphism. – Eric Wofsey Mar 20 '18 at 23:36
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    More broadly, I feel like you are forgetting that we are dealing with a topological space here. We don't care exactly what it looks like geometrically; we just care about it up to homeomorphism! – Eric Wofsey Mar 20 '18 at 23:37
  • @EricWofsey if all we care about is topological spaces up to homeomorphism, then yeah, there's no point getting upset about the difference between the cone and the smushed cylinder. But for example as Riemannian manifolds they are not isometric (I think). The cone picture is wrong. – ziggurism Mar 21 '18 at 16:22
  • (a) Why should the suspension have the structure of a Riemannian manifold? (b) Your question starts off with a "space" $X$, which places everything in the context of topology, not geometry. In the context of topology, the cone picture is just fine. – John Palmieri Mar 21 '18 at 18:05
  • @JohnPalmieri I don't disagree. But OP wants to understand where the cone picture came from. While it is allowable and equivalent in a topological context, it is an addition added for simplicity, which is not correct in all contexts. This what I describe in my answer as "taking license". – ziggurism Mar 21 '18 at 18:34
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Here is a very precise statement that you may find useful. Let $X$ be a compact subspace of $\mathbb{R}^n$. Let $$cX=\{((1-t)x,t)\in\mathbb{R}^n\times\mathbb{R}:x\in X,t\in[0,1]\}$$ and let $CX$ be the quotient $X\times [0,1]/\sim$ where $\sim$ collapses $X\times\{1\}$ to a point. The idea here is that $cX$ is the geometric cone picture that everyone draws (where the copies of $X$ shrink as you approach the tip) while $CX$ is the cone defined as a quotient space. I claim then that actually $CX$ and $cX$ are homeomorphic, via the map $f:CX\to cX$ that sends $[x,t]$ to $((1-t)x,t)$ (where $[x,t]$ is the equivalence class of $(x,t)\in X\times[0,1]$). Indeed, $f$ is clearly continuous and a bijection. Since $CX$ is compact and $cX$ is Hausdorff, that implies $f$ is a homeomorphism.

You can do a similar construction for the suspension, to show that when $X$ is a compact subspace of $\mathbb{R}^n$, $SX$ is homeomorphic to the "geometric" suspension of $X$ formed by taking two cones on $X$ in $\mathbb{R}^{n+1}$ on opposite sides.

Eric Wofsey
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  • So from what I am understanding, you are saying this "cylinder" and "cone" business in the context of $CX$ (same as when you say "$CX$ is the cone...") is via the map $f$ and these geometric names given to $CX$ is only because of $cX$? And $CX$ is just an "algebraic" object with no geometry until we realize we can identify this object with your map $f$? – Lemon Mar 20 '18 at 23:23
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    No, I'm not saying anything that drastic. I'm just making precise the notion that "shrinking" the copies of $X$ as you approach the point of the cone doesn't actually change your space up to homeomorphism. As I said, $cX$ is the space that everyone is actually drawing when they draw a picture of a cone like that. On the other hand, $CX$ is the way "cone" is usually actually defined. So I'm proving that the pictures actually are accurate and these two spaces are homeomorphic, at least for nice $X$. – Eric Wofsey Mar 20 '18 at 23:28
  • Another way to say essentially the same thing is that a topology on a space provides, for each point $p$, a notion of "$x$ is close to $p$" for varying $x$ (by considering neighborhoods of $p$), but it does not provide a notion of "$x$ is close to $y$" when both $x$ and $y$ are variable. That's why there's no notion of "Cauchy sequence" or "uniform continuity" in general topological spaces. These notions depend on having a metric (or a uniform structure). For suspensions, it makes sense to say that points $(x,t)$ with $t\to1$ approach the "top" point but not that they approach each other. – Andreas Blass Mar 20 '18 at 23:57
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I expect the picture helps to see that the suspension $\frac{S^1\times I}{\sim}$ is a sphere: enter image description here

janmarqz
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There is a fundamental problem whenever one draws a picture of a topological space: a picture has lots of properties that topological spaces don't have - length, volume, distance, thickness, etc. The standard picture of the suspension as a double cone is perfectly correct, but it includes a lot of structure that the suspension doesn't have, such as the fibers over $\epsilon$ getting "smaller" as $\epsilon$ approaches $0$ or $1$. The fibers don't have a size, so you can draw them however you want!

The key properties that the suspension has in common with the double cone are:

  1. The image of $\{0\} \times X$ in $SX$ is a single point $x_0$ (and similarly for $\{1\} \times X$)
  2. Every neighborhood of $x_0$ in $SX$ contains an open set $V$ with the property that $V - \{x_0\} \cong X \times (0, \epsilon)$.

In particular, if $X$ happened to be a metric space then you could show that any compatible metric on $SX$ would have to have the property that the diameter of the fibers $\epsilon \times X$ must approach $0$ as $\epsilon \to 0$, but this is a computation in metric geometry rather than topology. Also, there is nothing forcing the metric to look like what you would traditionally think of as a cone; for instance you could use:

$$\cos d_{SX}((x,t),(y,s)) = \cos(t)\sin(s) + \sin(t) \sin(s) \cos(d_{X}(x,y))$$

In the case where $X = S^n$ with the round metric, this recovers the round metric on $SX \cong S^{n+1}$.

Paul Siegel
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