For any topological space $A$, the cone $C(A)$ is defined to be $A \times [0,\infty)$ with $A \times 0$ identified to a point (cone point).


Let $X$ and $Y$ be two compact Hausdorff spaces such that there is a homeomorphism between $C(X)$ and $C(Y)$ which preserves the cone points. Can we prove that $X$ and $Y$ are homeomorphic?



Does homeomorphism between cones imply homeomorphism between sections
The answer is no. Let $X$, $Y$ be any two smoothly h-cobordant closed manifolds of dimension $\ge 4$ that are non-homeomorphic. For example, we can take $X=S^2\times L(7,1)$ and $Y=S^2\times L(7,2)$ where as usual $L(p,q)$ is a 3-dimensional lens space. By the weak h-cobordism theorem the interior of the h-cobordism is diffeomorphic to $X\times\mathbb R$ and also to $Y\times \mathbb R$. The interior is two-ended, and we can compactify it by adding two points (= the ends). The "identity" homeomorphism $X\times\mathbb R\to Y\times \mathbb R$ extends to
a homeomorphism of the compactifications that takes ends to ends. Removing the points corresponding to one end gives a homeomorphism from the cone over $X$ to the cone over $Y$ that preserves the cone point.


Conversely, if $X$, $Y$ are closed manifolds and $X\times\mathbb R$ is diffeomorphic to $Y\times \mathbb R$, then $X$, $Y$ are h-cobordant.


The same trick of course works when $X$, $Y$ are compact non-homeomorphic spaces and $X\times\mathbb R$ is homeomorphic to $Y\times \mathbb R$.


