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