Do join and suspension commute?

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?

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.

• 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. – Stefan Hamcke Apr 17 '15 at 14:24

When working with "nice" spaces, you can also use the associativity of the join and the observation that $\Sigma X = S^0 * X$.