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The reduced suspension of the pointed space $(X,x_0)$ is the smash product $(\mathbb S^1\wedge I, *)$ of $(X,x_0)$ with the $(\mathbb S^1,s_0)$ and is denoted by $\operatorname{S}X$.

My problem is to show that $\operatorname{S}X$ is homeomorphic to $ (I\times X)/{\sim}$, where $\sim \; =\{0\}\times X\cup I\times \{x_0\}\cup \{1\}\times X$. Consider the following diagram

$\require{AMScd} \begin{CD} I\times X @>\pi_\sim>> (I\times X)/{\sim} \\ @V\exp\times 1_X VV @VVh V \\ \mathbb S^1 \times X @>>\pi_\wedge> \operatorname{S}X \end{CD}$

where $\pi_\sim$ and $\pi_\wedge$ are quotient maps and $\exp:I\to \mathbb S^1$ is the exponential map which maps each $t\in I$ to $e^{2\pi i t}\in\mathbb S^1$. The universal property of the quotient implies that there exists $f:(I\times X)/\sim \; \to \mathbb S^1 \times X$ such that $f\circ \pi_\sim=\exp\times 1_X$. now we define $h:=\pi_\wedge\circ f$.

$h$ is continuous (by The universal property of the quotient, $f$ is continuous) and one-to-one correspondence. Now how we can show that $h$ is open or maybe closed? Is it true that $\pi_\wedge$ is a closed map?

Stefan Hamcke
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M.A.
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  • Are you sure there is such an $f$ ? The quotient map $\pi_\sim$ identifies more than exp$\times 1_X$. – Stefan Hamcke Apr 29 '16 at 16:14
  • @StefanHamcke The universal property of the quotient implies existence of $f$ but maybe this $f$ isn't useful to define a homeomorphism. – M.A. Apr 29 '16 at 16:44
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    No, the universal property implies existence only if $\exp\times 1_X$ identifies points whenever they are identified by $\pi_\sim$. Consider $(0,x_0)$ and $(0.5, x_0)$. These are identified by $\pi_\sim$, so they would have the same image under $\text{exp}\times 1_X=f\pi_\sim$ if such an $f$ existed. – Stefan Hamcke Apr 29 '16 at 16:59
  • @StefanHamcke Thx Stefan.That is, such $f$ exists if and only if $(t_1,x_2)\sim (t_2,x_2)$ implies tha $\exp\times 1_X (t_1,x_2)= \exp\times 1_X (t_2,x_2)$ – M.A. Apr 29 '16 at 17:18
  • Yes, that is the universal property, and it is why the map $h$ indeed does exist. – Stefan Hamcke Apr 29 '16 at 17:52

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$$ \require{AMScd} \begin{CD} I\times X @>\pi_\sim>>(I\times X)/{\sim} \\ @V\exp\times 1_X VV @VVh V \\ S^1 \times X @>>\pi_\wedge> \operatorname{S}X \end{CD} $$

The map $\exp$ is a perfect map, that is a closed map whose fibers are compact. For any space $X$, the map $\exp\times 1_X$ is therefore a closed map.

Now $\pi_\sim$ identifies all of $(\{0,1\}\times X) \cup (I\times\{x_0\})$ to a point. The map $\exp\times 1_X$ identifies $(0,x)$ with $(1,x)$ for any $x\in X$, and the equivalence class is $(s_0,x)$. Then the map $\pi_\wedge$ collapses $(\{s_0\}\times X) \cup (S^1\times\{x_0\})$ to a point, so when two point are identified via $\pi_\sim$, they have the same image under $\pi_\wedge \circ \exp\times 1_X$. Since the latter map is a quotient map, so is the induced map $h$. Now it only remains to show that $h$ is injective ...

Stefan Hamcke
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  • if $(t_1,x_1)\sim (t_2,x_2)$ then $\pi_\wedge \circ \exp\times 1_X (t_1,x_1)=\pi_\wedge \circ \exp\times 1_X (t_2,x_2)$ NOW The universal property of the quotient implies existence of $h$. The composition of an identification and a quotient map is an identification map then $\pi_\wedge \circ \exp\times 1_X$ is an identification map (I must show that $h$ is an identification too) and use this theorem: An injective identification map is a homeomorphism; its enough to show that $h$ is injective. This thought is true? – M.A. Apr 29 '16 at 18:41
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    @M.A. Yes, this is true. I can also give a reference for "product of perfect map and identity is closed map", if you like. And note that for injectivity of $h$, it suffices to show that $\pi_\sim$ identifies two elements whenever $\pi_\wedge\circ\text{exp}\times 1_X$ does. – Stefan Hamcke Apr 29 '16 at 20:16
  • I‎ would be very grateful if you ‎could ‎provide me that reference. – M.A. Apr 30 '16 at 01:59
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    @M.A. See the section about proper maps in chapter 3.6 of Brown's Topology and Groupoids. Or have a look at this post for the far more general result that the product of perfect maps is perfect. – Stefan Hamcke Apr 30 '16 at 17:10