Let $SX$ denote the reduced suspension of $X$. If $A\subset X$, what is the relationship between $S(X/A)$ and $SX/SA$?
In commutivity of suspension and quotient space, a comment says these spaces are homeomorphic, but I don't follow the argument. Can someone outline a proof of this result that requires less category theory than the comment from the link, or explain what the commenter means?
Thanks in advance.
For topological spaces taking quotient spaces commutes with taking products with compact spaces in the sense that we may first quotient our space by the relation and then take the product, or we may take the product and then take a levelwise quotient (i.e. $(a,b) \sim (a',b)$ if $a \sim a'$).
Since suspension is the same as taking a product with the interval and then quotienting by $\sim$ which identifies all coordinates that involve the basepoint or endpoints of the interval, we have that $\Sigma(X/A)= (X/A \times I )/ \sim \: = ((X \times I)/((a,t) = (*,t))/ \sim )$ where $a \in A$. It is true for any space $Y$ that if I have two equivalence relations $Q,R$, quotienting by the equivalence relation $Q$ and then by the relation generated by $R$ on $Y/Q$ is canonically homeomorphic to quotienting by $Q$ on $Y/R$. We see that in this instance letting $Y=X \times I$, $Q=((a,t) = (*,t))$ and $R=\sim$ tells us that we can first quotient by $\sim$ which gives us $X \times I / \sim=\Sigma X$ and then quotient by $((a,t) = (*,t))$ which is now $((a,t)=(*,*))$ which is the same as $\Sigma (X)/\Sigma(A)$ since all the basepoints in each copy of $A$ have been collapsed to a single point.