Let $X$ be a topological space, $A\subset X$ a closed subspace. $CA$ means the cone of $A$, and by $SA$ I'll denote the suspension of $A$.
I need to prove that
$$ \left( \left( (X \cup CA) \cup CX \right) \cup C(X \cup CA) \right)/ C(X \cup CA) \approx SX$$
I figured out that $((X \cup CA) \cup CX )/ CX \approx SA$ but for the first I can't even write down a sketch to help understanding the situation.
Can someone gives some advice about how to visualize the space in exam? I made some sketches for two easy cases, (where X and A are two concentric circles, or X is a disk and A a circle) but I can't generalize it an a satisfying way
Let's first clear up the notation. Let $Y := (((X ∪ CA) ∪ CX) ∪ C(X ∪ CA))$. I assume that the meaning is such that the following hold: $X ∩ CA = A$, $CA ∩ CX = A$, $(X ∪ CA ∪ CX) ∩ C(X ∪ CA) = X ∪ CA$.
Now let's prove the following claim.
Proof: Let $q: A ∪ B \to (A ∪ B) / B$ be the quotient and $i: A \to A ∪ B$ be the inclusion. Then $q \circ i$ is onto $(A ∪ B) / B$, is injective on $A \setminus B$ whereas $A ∩ B$ is contracted to a point. So to prove the claim it remains to see that $q \circ i$ is a quotient mapping. Let $(q \circ i)^{-1}[U]$ be open in $A$. Then either is is disjoint with $B$ and so is open in $A ∪ B$ or is joint with $B$ and so $q^{-1}[U] ⊇ B$ and $q^{-1}[U] ∩ A$ is open so again $q^{-1}[U]$ is open in $A ∪ B$. In both cases by $q$ being quotient, we have that $U$ is open which we wanted.
Now we can apply the claim to get the proposition: $(X ∪ CA ∪ CX ∪ C(X ∪ CA)) / C(X ∪ CA) \approx (X ∪ CA ∪ CX) / (X ∪ CA) \approx CX / X \approx SX$.
Note that we don't need that $A$ is closed, we just need that $X$ is non-empty.