Let $X$ be a manifold without boundary and let $Y$ and $Z$ be two manifolds with boundary such that $X$ is homeomorphic to their interiors: $X \cong Y^\circ \cong Z^\circ$. Does it follow that $Y \cong Z$ as well? That is, does there exist a homeomorphism that extends to the boundaries as well? If not, can we at least say that $\partial Y \cong \partial Z$ even if perhaps they are "attached" to their respective interiors differently?


I rarely work with manifolds with boundary so I apologize if this is a stupid question. It seems like it should be true but I haven't been able to find a reference for this. It is certainly obvious that we have a homotopy equivalence $Y \simeq Z$ since a manifold with boundary is homotopy equivalent to its interior.



Is the boundary of a manifold topologically unique?
Since you say in a comment that you might be satisfied with a homotopy equivalence, let me sketch a proof that the homotopy type of the boundary depends only on the interior. 


Let $Y$ be the interior of $X$ and let $DY$ be the space of all proper maps $[0,1)\to Y$, suitably topologized. Then $DY$ must be homotopy equivalent to $\partial X$. 


To see this, first observe that $DY$ is homotopy equivalent to the space $D'Y$ of germs of such maps, where two maps $[0,1)\to Y$ have the same germ if they agree on $[a,1)$ for some $0\le a<1$. 


Then observe that $D'Y$ depends only a neighborhood of $\partial X$, and that by using a collar neighborhood $C\cong I\times\partial X$ we find $D'Y$ to be homeomorphic to the space of all germs of proper maps $[0,1)\to [0,1)\times\partial X
$. 


The latter space of germs is homotopy equivalent to the space of all proper maps $[0,1)\to [0,1)\times\partial X$. 


Finally, this last space is the product of 


(1) the space of all proper maps $[0,1)\to [0,1)$ and 


(2) the space of all continuous maps $[0,1)\to \partial X$.


(1) is contractible. (2) is homotopy equivalent to $\partial X$.


