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.

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    $\begingroup$ I expect that the homotopy type of the boundary is uniquely determined from the interior along the lines of a "fundamental group at infinity"-type construction. But I would be surprised if you could do much more than that. Maybe try total spaces of rank 4 vector bundles over $S^4$ whose sphere bundles are homotopy equivalent but not homeomorphic. $\endgroup$ – Mike Miller Aug 7 '18 at 0:44
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    $\begingroup$ This topic was discussed previously at least in mathoverflow.net/q/22441, mathoverflow.net/q/34602, mathoverflow.net/q/81714, mathoverflow.net/q/83356. $\endgroup$ – Igor Khavkine Aug 7 '18 at 1:04
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    $\begingroup$ It seems that this is a duplicate of the third question you link (your own). A one-line answer: the manifolds are $h$-cobordant, so homotopy equivalent, and homeomorphic if simply connected. $\endgroup$ – Mike Miller Aug 7 '18 at 1:08
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    $\begingroup$ Without other assumptions it is false: a manifold without boundary is a manifold with boundary. So one may have the trivial situation $X=Y=Y^\circ=Z^\circ\not \cong Z$; and more situations, if $\partial Z$ is not connected $\endgroup$ – Pietro Majer Aug 7 '18 at 6:21
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    $\begingroup$ An additional assumption to exclude these obvious counterexamples might be compact. $\endgroup$ – Michael Bächtold Aug 7 '18 at 8:40

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 $\partial X$.


I don't think this is true, by a classical `swindle' that I think is due to Stallings. You can find variations on this argument eg in Milnor's paper, Two complexes which are homeomorphic but combinatorially distinct (Annals 1961). He refers to some notes of Stallings.

Here is how I recall the argument, as applied to your question. I might be off by a sign, depending on the dimension.

Consider a manifold $M$ for which there is an h-cobordism $Y$ from $M$ to itself with non-zero Whitehead torsion; then your $X$ will be two copies of $M$, or (taking orientations into account) $X = -M \cup M$. Then obviously $\partial Y = X$. On the other hand, you could take $Z = M \times I$.

Claim: $Y^0 \cong Z^0 = M \times R$.

Proof: Consider an h-cobordism $Y'$ from $M$ to itself whose torsion is the negative of that of $(Y,M,M)$. Then $$Y \cup_M Y' \cong M \times I \cong Y' \cup_M Y.\quad (*)$$ Now consider $$ \cdots (Y \cup Y') \cup (Y \cup Y') \cup Y \cup (Y'\cup Y) \cup (Y'\cup Y) \cdots $$ By (*), this is $(-\infty,0] \times M \cup Y \cup [0,\infty)\times M = Y^0$.

On the other hand, you can put the $Y$ in the middle onto one side, and redistribute the parentheses to see that this is an infinite union of copies of $Y \cup Y' = M \times I$, and hence is $Z^0$.

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    $\begingroup$ It is perhaps easier to apply the weak h-cobordism theorem: all h-cobordisms are diffeomorphic after removing one boundary component (in higher dimensions). This is done by engulfing. $\endgroup$ – Igor Belegradek Aug 7 '18 at 2:08

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