It is natural to ask if it is possible for the mapping cone $X\cup_\alpha CA$ to be homeomorphic to the mapping cone $X\cup_\beta CB$ with $A$ and $B$ nonhomeomorphic. Is there a standard go-to example for this?
I have vague memories that there are manifolds $M$ and $N$ that are not homeomorphic, but $M\times \mathbb{R} \cong N \times \mathbb{R}$, and it seems like it might be a mere hop, skip, and a jump from there to an example.
The double suspension theorem says that if $Y$ is a homology $3$-sphere, then its double suspension $\Sigma^2 Y$ is homeomorphic to $S^5$. If we take $Y$ to be the Poincaré sphere, then $\Sigma Y$ is not a topological manifold, since the suspension points are not manifold points, and in particular $\Sigma Y$ is not homeomorphic to $S^4$. Taking these two spaces as $A$ and $B$ and maps to a point as $\alpha$ and $\beta$ gives a fairly well-known example.
Let $X$ be a line with countably many whiskers, i.e., the subset of the plane given by $$ X=(\mathbb R\times\{0\})\cup\bigcup_{n\in\mathbb N}(\{n\}\times[0,1]). $$ Then adding one more whisker produces the same (meaning homeomorphic) result as adding two more whiskers, even though $1\neq2$. That is, let $A=\{(-1,0)\}$ and $B=\{(-1,0),(-2,0)\}$, with $\alpha$ and $\beta$ being the inclusion maps.
