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Let $S(X) = (X \times [-1,1])/R$ denote the suspension of $X$, where the classes of $R$ are $X \times \{1\}, X \times \{-1\}$ and all singletons $\{a\}$, where $a \in X \times (-1,1)$. Show that $S(S^{n-1})$ is homeomorphic to $S^n$. (Don’t confuse $S$ and $S^n$, $S$ is the suspension and $S^n$ is the unit sphere).

So I want to find an cont bijection with continuous inverse $f:S(S^{n-1}) \to S^n$ and since every output of $f$ must be of length one I’m thinking about $f(x) = \frac{x}{\|x\|}$ which would satisfy the conditions of homeomorphism, but I’m not sure if this works? Also should is it neccessarily defined in all of $S(S^{n-1})$? I don’t think $S(S^{n-1})$ contains $\vec{0}$?

Wondera
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The map $f(\vec{x}, t) = (\sqrt{1-t^2}\cdot \vec{x}, t)$ is well-defined from $\Bbb S^{n-1} \times [-1,1]$ to $S^n$, where $\vec{x}$ is an $n$-dimensional real vector, and clearly continuous.

$f(\vec{x},t)=f(\vec{y},t')$ iff $t=1$ or $t=-1$ or else $t \in (-1,1)$ and $\vec{x}=\vec{y}$, as can be easily verified. So the unique map $g: S(S^{n-1}) \to S^n$ that obeys $g \circ q= f$ (where $q: \Bbb S^{n-1} \times [-1,1] \to S(S^{n-1})$ is the suspension quotient) is bijective and continuous and by compactness is a homeomorphism.

Henno Brandsma
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  • Thanks! Not sure I entirely understood where the first coordinate of the output of $f$ came from? Also is there a reason why $x \mapsto \frac{x}{|x|}$ doesn’t work? – Wondera Jul 23 '21 at 10:53
  • @Wondera compute the norm of the image vector (in $\Bbb R^{n+1}$) and see why we need the square root factor, there is no choice... Your map idea doesn't work well with the quotient construction. – Henno Brandsma Jul 23 '21 at 11:00
  • can you please clarify what is the equivalence relation of the map q? – Intuition Jan 29 '24 at 08:52
  • can you please explain how the injectivity of the map f is still valid $t \neq t'$? Or this case can not happen? ..... Also when you said at $t =1$ you mean automatically that $t'$ will be equal 1? – Intuition Jan 29 '24 at 08:56
  • can you please clarify what the square root of $1 - t^2$ is doing to the first nth coordinates? Why you multiplied by it? – Intuition Jan 29 '24 at 08:59