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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.

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  • $\begingroup$ Regarding your manifolds, the Whitehead manifold $W$ and $\mathbb{R}^3$ are not homeomorphic, but $W\times\mathbb{R}$ is homeomorphic to $\mathbb{R}^4$. $\endgroup$
    – Steve D
    Aug 26, 2020 at 4:14

2 Answers 2

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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.

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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.

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  • $\begingroup$ But adding the cone on $B$ will make a closed loop, right? But this seems a promising approach. $\endgroup$
    – Jeff Strom
    Aug 26, 2020 at 2:58
  • $\begingroup$ How about instead a line with semicircles joining $0$ (the basepoint) to $n$ for $n\in \mathbb{N}$. Then attaching a cone on a finite set does not change the homeomorphism type. $\endgroup$
    – Jeff Strom
    Aug 26, 2020 at 3:00
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    $\begingroup$ @JeffStrom Oops, you're right. Start with infinitely many whiskers and infinitely many loops (or arches). That seems to fix the problem. $\endgroup$ Aug 26, 2020 at 3:01
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    $\begingroup$ Another example to describe the same phenomenon: a graph with two vertices $a,b$, infinitely many edges connecting $a$ and $b$, plus infinitely many edges connected just to $a$ should do also the trick if $A=\{a\}$ and $B=\{a,b\}$ with $\alpha,\beta$ inclusions. $\endgroup$ Aug 26, 2020 at 14:19

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