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Definition $:$ For a given topological space $X$ we define the cone of $X$ denoted by $C(X)$ as follows $:$

$$C(X) : = (X \times I)/ (X \times 0)$$ where $I = [0,1]$ and we define the suspension of $X$ denoted by $S(X)$ as follows $:$

$$S(X) : = C(X) / X$$ where $X$ is viewed as a subspace of $C(X)$ by means of the embedding $x \longmapsto (x,1).$

Now with this definition in mind I am able to show that $C(\Bbb S^n) \cong \Bbb D^{n+1}.$ The proof is as follows $:$

Consider the map $g : \Bbb S^n \times I \longrightarrow \Bbb D^{n+1}$ defined by $(x,t) \longmapsto tx.$ Then it is easy to see that $g$ is a continuous surjection. Hence by universal property of quotient topology $g$ induces a bijective continuous map $f : (\Bbb S^n \times I)/ (\Bbb S^n \times 0) \longrightarrow \Bbb D^{n+1}.$ But now the domain of $f$ is compact and the range of $f$ is Hausdorff and hence $f$ is a homeomorphism, which is what we wanted to show.

Now what is $S(\Bbb S^n)\ $? By definition $S(\Bbb S^n) = C(\Bbb S^n)/ \Bbb S^n \cong \Bbb D^{n+1} / \Bbb S^n.$ But what is $\Bbb D^{n+1} / \Bbb S^n\ $? In order to understand this concept I tried to draw pictures for the suspension of $\Bbb S^1$ and I saw that $S(\Bbb S^1)$ is a double cone which can be equivalently viewed as two cones of $\Bbb S^1$ with their bases identified. Now cone of $\Bbb S^1$ is $\Bbb D^2.$ So $S(\Bbb S^1)$ is nothing but two copies of $\Bbb D^2$ with their boundaries identified which is homeomorphic to the $2$-sphere $\Bbb S^2.$ So what I intuitively feel is that $S(\Bbb S^n) \cong \Bbb S^{n+1}.$ But I am unable to show that analytically. Would anybody please help me in this regard?

Thanks in advance.

Anil Bagchi.
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  • Think of $X$ as being embedded as $X \times {1/2}$. The suspension is then exactly the double cone as you say, where they are glued together at the $1/2$ coordinate. It might be easier if you write suspension like $S(X) = (X \times [0,1]/(X \times {0}))/(X \times {1})$, as that looks a bit more symmetric (and it should be clear that the order of identifying the 'ends' doesn't matter). – Dan Rust Apr 15 '21 at 16:27
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    @DanRust how do I get an explicit homeomorphism? As I am a new to this topic it will be better if you illustrate it a bit more. I want to somehow use the universal property of quotient topology which I am familiar with from the course I had in the previous semester in point set topology. – Anil Bagchi. Apr 15 '21 at 16:32
  • Call the two cones $C_1(X)$ and $C_2(X)$ and then define the map $f \colon C_1(X) \sqcup C_2(X) \to S(X)$ by $f(x,i) = (x,i/2)$ if $(x,i) \in C_1(X)$ and $f(x,i) = (x,1-i/2)$ if $(x,i) \in C_2(X)$. It's not too hard to check this is a well defined map. Now use the gluing lemma, because you'll notice that $(x,1)$ in $C_1(X)$ and $(x,1)$ in $C_2(X)$ both get mapped to the same point in $S(X)$. – Dan Rust Apr 15 '21 at 16:39
  • Does this answer your question? $B^n/S^{n-1}$ is homeomorphic to $S^n$ – Paul Frost Apr 15 '21 at 21:20

3 Answers3

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Let $C_1 = X \times [0,1]/(X \times \{0\}) = C_2$ be two cones and define $f_j \colon C_j \to S(X)$ by

$$f_j(x,s) = \begin{cases}(x,s/2)& \text{if } j=1\\(x,1-s/2) &\text{if } j=2\end{cases}.$$

Each of these maps is well-defined and continuous, and so by the gluing lemma, the induced map $f$ on the space $Y = \left(C_1 \cup C_2\right) /(x,1)_1\sim (x,1)_2$ to $S(X)$ is also well defined and continuous. It's easy to see this map $f$ is surjective, and only a little more work to show it has a well defined inverse.

I guess the most difficult part is in showing this inverse is also continuous. You can do this I guess by applying the gluing lemma again by cutting $S(X)$ along the line $X\times \{1/2\}$ and then each half is a local inverse of $f_j$ for the corresponding $j$.

Dan Rust
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  • In this case the continuity of the inverse of the map from $Y$ to $S(\Bbb S^n)$ is forced (as long as we know that this map is a continuous bijection) since $Y$ is compact and $S(\Bbb S^n)$ is Hausdorff as it is the quotient of a compact Hausdorff space (and hence normal and in particular regular) $\Bbb D^{n+1}$ by a closed subset $\Bbb S^n.$ – Anil Bagchi. Apr 15 '21 at 17:29
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I will instead model the suspension as $X \times [-1,1]$, then collapsing the two sides.

Picture this for the circle and the 2-sphere. The picture is that the great circle on the 2-sphere is $S^1 \times \{0\}$, and that as you go to the extremities the great circle shrinks to a point. I think of the time parameter as measuring the height.

So consider the map $f: S^1 \times [-1,1] \to S^2$ given by $$f(x,y,t) = (x\sqrt{1-t^2}, y\sqrt{1-t^2}, t).$$

I will leave the rest to you (argue that this is a quotient map, whose induced equivalence relation is exactly that of the suspension; you should have to do very little other than set theory).

This is the only map it could possibly be; the square root factor is forced on me.

user915325
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Your definition can be also written as $S(X)=X\times I/(X\times 0\cup X\times 1)$. For the purpose of that answer I replace $I$ with $[-1,1]$.

Write down $S^n=\{(x_0,\ldots, x_n)\in\mathbb{R}^{n+1}\ |\ x_0^2+\cdots+x_n^2=1\}$ and then define

$$f:S^n\times [-1,1]\to S^{n+1}$$ $$f((x_0,\ldots, x_n),t)=(x_0\cdot c,\ldots, x_n\cdot c,t)$$

Where $c$ is such that the right side belongs to $S^{n+1}$. With a simple computation this gives us

$$c=\sqrt{\frac{1-t^2}{\sum x_i^2}}=\sqrt{1-t^2}$$

The intuition here is as follows: for $S^1$ our $S^1\times[-1,1]$ is a tube in $\mathbb{R}^3$ and then we collapse the tube towards $S^2$ by doing planar contraction at each level. Only levels $1$ and $-1$ are fully contracted to a point, the contraction is injective everywhere else. The construction above generalizes this idea to any dimension.

With that I leave as an exercise that $f$ is continuous and surjective. You can also check that $f(v,t)=f(w,t')$ if and only if "$v=w$ and $t=t'$" or "$t=t'=0$" or "$t=t'=1$". Therefore $f$ induces continuous bijection $F:S(S^n)\to S^{n+1}$, $F([x])=f(x)$ which is a homeomorphism since $S(S^n)$ is compact.

freakish
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