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Let $X$ be a compact topological manifold with boundary. Suppose its interior is homeomorphic to $\mathbb{R}^n$. Is $X$ homeomorphic to a (closed) ball?

Context: I want to show that a certain compactification of moduli of convex $n$-polygons (up to scaling and rotations) is a $(2n-4)$-cell. I can degenerate all sides (but $2$, of course), keeping records of the slopes, and all angles to $\pi$ (again except $2$). If more structure is needed, like smoothness on the interior, PL at the boundary - you have it.

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    $\begingroup$ Is this really trivial? See also this question for the smooth category: mathoverflow.net/questions/158391/… (Edit: it has been pointed out to me that the preceding sentence might be misread as saying "isn't this really trivial?" What I meant to suggest is that I thought the question had merit. Sorry if there was any confusion about this.) $\endgroup$
    – Todd Trimble
    Commented Nov 11, 2015 at 19:02
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    $\begingroup$ Meta discussion: meta.mathoverflow.net/a/2564/2926 One guess for why the question was really closed is that it lacks context/motivation, so you might consider adding some. (My own opinion is that the question is a "natural" one to ask, and natural questions are often self-motivating, but any information about context or why you ask would probably improve the question. It need not be a long story.) $\endgroup$
    – Todd Trimble
    Commented Nov 11, 2015 at 21:59
  • $\begingroup$ Thank you for providing the context. I added some formatting. I cast a vote to reopen. $\endgroup$
    – user9072
    Commented Nov 12, 2015 at 16:24
  • $\begingroup$ By a standard argument if you remove a small disk from the interior of $X$, you get a topological $h$-cobordism of simply-connected manifolds. One of them is homeomorphic to a sphere, and hence by Poincare conjecture so is the other one. The $h$-cobordism is certainly trivial if $n\neq 4$ by the $h$-cobordism theorem in those dimensions. I hesitate slightly if $n=4$ but this case should be okay too. $\endgroup$
    – Igor Belegradek
    Commented Nov 12, 2015 at 20:14

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For $n \geq 6$, this follows from the topological h-cobordism theorem (due to Kirby-Siebenmann). Indeed, removing a topological ball from the interiour of $X$ yields an h-cobordism. If this is trivial, i.e., homeomorphic to $S^{n-1} \times [0,1]$ relative boundary, you glue the ball back in to see that $X$ must be a topological ball.

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    $\begingroup$ What about n=4? $\endgroup$
    – Pupkin
    Commented Nov 12, 2015 at 18:02
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    $\begingroup$ We're only working topologically, yes? Topological h-cobordism is true in all dimensions. $\endgroup$
    – mme
    Commented Nov 12, 2015 at 23:45
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    $\begingroup$ Don't you need to know that $\partial X$ is simply connected, in order to apply the $h$-cobordism theorem? $\endgroup$
    – Marco Golla
    Commented Nov 13, 2015 at 1:12
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    $\begingroup$ @MarcoGolla. $\mathbb R^n$ is simply connected at infinity, this implies that $\delta X$ is simply connected. $\endgroup$
    – Andreas Thom
    Commented Nov 13, 2015 at 4:59
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    $\begingroup$ @IgorBelegradek: I see your point, thank you. This is the 4-dimensional annulus theorem which Quinn proved in 1982. $\endgroup$
    – mme
    Commented Nov 13, 2015 at 13:49
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I wish to address the case $n=4$. This would follow if any $4$-dimensional h-cobordism between $3$-spheres were trivial but at the moment I am not sure how to prove this. There is however a different argument.

Brown proved in [A proof of the generalized Schoenflies theorem, Bull. Amer. Math. Soc. 66 (1960), 74-76], http://www.ams.org/journals/bull/1960-66-02/S0002-9904-1960-10400-4/home.html that any topological embedding of $S^{n-1}\times [-1,1]$ into $S^n$, then the closures of either component of the complement of $S^{n-1}\times \{0\}$ is a topological disk.

Now let's show that $X$ is homeomorphic to $D^4$. Since $\partial X$ is a simply-connected $3$-manifold it is homeomorphic to $S^3$ (by Perelman). By the topological collar theorem (also due to Brown) there is a collar neighborhood $C$ of $\partial X$ in $X$. Attaching a $4$-disk along $\partial X$ gives a $4$-sphere. Thus $C$ is a copy of $S^3\times I$ in $S^4$, hence by Brown the closure of $X-C$ is homeomorphic to $D^4$, and hence the same is true about $X$.

Perhaps a variation of the above shows that any h-coborsims between 3-spheres is topologically a product but I do not see this.

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Igor's answer is spot-on, avoiding the h-cobordism theorem, which is fragile in low dimensions. Here is a proposal to tweak his answer (which works in all dimensions), so as to use the Poincare Conjecture only once, just where it is required.

Step 1. Boundary X is a homotopy sphere (noted already), and hence a sphere, by the PC. (Big machinery here, which varies with the dimension, but there's no way to avoid it. If you know in advance that boundary X is a sphere, so much the better.)

The next two steps use two classic theorems of Morton Brown (both mentioned already), and they work uniformly in all dimensions.

Step 2. Boundary X is collared in X (Brown, beautifully redone by Connelly). And so, by pulling X inward into its interior, we can assume that X lies in R^n, hence in S^n.

Step 3. Now Brown's Schoenflies Theorem (pure magic) implies that boundary X bounds a closed ball. Done.

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  • $\begingroup$ It took me a while to see why bdry X is a homotopy sphere, so let me record an argument in case it's useful for anyone else. First, we show pi_k(bdry X) = 0 for 0 <= k <= n-2. Choose U to be a collar neighborhood of the boundary. Consider a map f: S^k -> bdry X. Choose R such that R^n \ B_R(0) is contained in U. We can homotope f to a map into R^n \ B_R(0), so it is nullhomotopic. Since bdry X is simply-connected, H_{n-1}(bdry X; Z) = Z. Choose f to be a map from bdry X to S^{n-1} that induces an isomorphism on H_{n-1}. Homology Whitehead now implies that f is a homotopy equivalence. $\endgroup$
    – Nathaniel Bottman
    Commented Mar 9, 2019 at 18:10

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