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Do join and suspension of topological spaces always commute, i.e. is it true that $\sum(A\star B)=A\star(\sum B)$?

I suppose that it is not true in general (but, for example, everything works in the case of two spheres), but perhaps there is an epimorphism from one space to another?

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I'm dealing with the unreduced suspension $SX=X\times[-1,1]/((x,t)\sim(x',t)\forall x,x'\in X,t=\pm1)$

Let's first look at the case $A=B=\{*\}$. Then the join of $A$ and $B$ is an interval $I$, and its suspension is the space $SI$, a square with each the top and the bottom line identified to a point. One may think of this as the space $$Y=\{(s,t)\mid |t|\le s\le 1\}$$ and a homeomorphism $SI\to Y$ is given by the map $$(s,t)\mapsto(1-(1-s)(1-|t|),t)$$ On the other hand, The suspension of $B$ is an interval $[-1,1]$, and its join with $A$ is a cone $C[-1,1]$ over that interval. A homeomorphism $C[-1,1]\to Y$ is given by $$(t,s)\mapsto(s,st)$$

So there is a homeomorphism $h:SI\to C[-1,1]$. Its composition with the quotient map $I\times[-1,1]\to SI$ factors as $$I\times[-1,1]\xrightarrow f [-1,1]\times I\to C[-1,1]$$ with the discontinuous $f$ being defined by $$f(s,t)=(f_t(s,t),f_s(s,t))=\begin{cases} \left(\frac t{1-(1-s)(1-|t|)},1-(1-s)(1-|t|)\right) &\text{if }s\ne0\text{ or }t\ne 0 \\ (0,0) &\text{if }s=t=0 \end{cases}$$ Conversely, the composition of $h^{-1}$ with the quotient map $[-1,1]\times I\to C[-1,1]$ factors as $$[-1,1]\times I\xrightarrow g I\times [-1,1]\to SI$$ with the discontinuous $g$ being defined by $$g(t,s)=(g_s(t,s),g_t(t,s))=\begin{cases} \left(1-\frac{1-s}{1-s|t|},st\right) &\text{if }s\ne 1\text{ or }|t|\ne 1 \\ (1,t) &\text{if }s=1\text{ and }t=\pm 1 \end{cases}$$

For arbitrary $A$ and $B$, the suspension of their join is a quotient of $A\times B\times I\times[-1,1]$, and the quotient map is the composite $$A\times B\times I\times[-1,1]\to (A*B)\times[-1,1]\to S(A*B)$$ The first map is the product of a quotient map and the identity on the locally compact space $[-1,1]$, thus again a quotient map. If we send a point $(a,b,s,t)$ to $[a,\,[b,f_t(s,t)],\,f_s(s,t)]\in A*SB$, then this induces a function from $S(A*B)$ to $A*SB$ as two point have the same image whenever they get identified under $p$.
Sending $(a,b,t,s)$ to $[[a,b,g_s(s,t)],\,g_t(s,t)]\in S(A*B)$ induces a function $\beta$ in the other direction since two points have the same image whenever they get identified by the continuous map $q$ which is the composite $$A\times B\times[-1,1]\times I\to A\times SB\times I\to A*SB$$
One may now check that these functions are inverse to each other. That means we have a bijection $$\alpha:S(A*B)\to A*SB$$

In order to show the continuity of $\alpha$, it suffices to show that $\alpha p$ is continuous. So take an open set $U\subseteq A*SB$. Its preimage under $q$ is then an open $q$-saturated subset of $A\times B\times[-1,1]\times I$. Note that $\alpha p= q(\mathbf 1_A\times\mathbf 1_B\times f)$ and $\beta q= p(\mathbf 1_A\times\mathbf 1_B\times g)$. Since $q^{-1}(U)$ is open, each point $x=(a,b,t,s)$ in this set is covered by an open box $\cal W_x=\cal A_x\times B_x\times T_x\times S_x$ contained in $q^{-1}(U)$.
If $s>0$, then we may choose $\cal S_x$ to not contain $0$, and that means that $\cal T_x\times\cal S_x$ is $d$ saturated where $d:[-1,1]\times I\to C[-1,1]$ is the quotient map shrinking $[-1,1]\times\{0\}$ to a point. Note that preimages of $d$-saturated open sets under $f$ are $c$-saturated open sets, where $c:I\times[-1,1]\to SI$ is the suspension map. So in this case, $(\mathbf 1_A\times\mathbf 1_B\times f)^{-1}(\cal W_x)$ is open.
If $s=0$, then $q^{-1}(U)$ contains $\{a\}\times\{b\}\times[-1,1]\times\{0\}$, which is compact. We can thus choose $\cal T_x$ to be $[-1,1]$. But this makes $\cal T_x\times S_x$ $d$-saturated, hence $(\mathbf 1_A\times\mathbf 1_B\times f)^{-1}(\cal W_x)$ is open.
This shows that $(\alpha p)^{-1}(U)$ is an open set, so $\alpha$ is a continuous bijection.

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  • $\begingroup$ For the openness of $\alpha$, I would simply assume that $A$ and $B$ should be spaces such that $A\times B\times[-1,1]\times I\to A\times SB\times I$ is a quotient map. This holds for example when $A$ is locally compact or when $B$ is compact. $\endgroup$ – Stefan Hamcke Apr 17 '15 at 14:24
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When working with "nice" spaces, you can also use the associativity of the join and the observation that $\Sigma X = S^0 * X$.

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