Definition of Suspension (Topology).

Question

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 $\mathbb{R}^2$

Answer 1

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.

Answer 2

I expect the picture helps to see that the suspension $\frac{S^1\times I}{\sim}$ is a sphere:

Answer 3

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.

Answer 4

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 $SX \cong S^{n+1}$.