let $M$ be a smooth $n$-manifold with boundary $\partial M$; I denote by $M^o$ the internal part of $M$, that is $M \smallsetminus \partial M$. The question is the same as in the title: let $M$ and $N$ be two compact orientable topological manifolds such that $M^o$ is homeomorphic to $N^o$. Does this imply that $M$ is homeomorphic to $N$? Can we say something about the connected components of the boundary?

I think that this question should have a really easy answer, but I cannot find it. One thing that I could prove is that the Euler characteristic of $\partial M$ must be the same as the one of $\partial N$, this being an easy consequence of the fact that $M^o$ is homotopically equivalent to $M$. The next step i tried is to prove that the sequence of Euler characteristics of the boundary components must agree: $$ ((\partial M)_1, \ldots , (\partial M)_r) = ((\partial N)_1, \ldots , (\partial N)_r) ;$$ but I do not see a way to prove this. I am specially interested in the case of 3-manifolds, and from this one could conclude that $\partial M$ is homeomorphic to $\partial N$, because Euler characteristic identifies closed surfaces. Do you have any tip or any simple proof?

  • $\begingroup$ This is a duplicate of an old question: mathoverflow.net/questions/95129/…. $\endgroup$ Sep 16, 2018 at 21:47
  • $\begingroup$ You are right, I tried to search for it but i didn't found that question. Anyway, here a positive answer for dimension 3 was exhibited, so I think it was useful to ask it again. $\endgroup$
    – P. Tolo
    Sep 22, 2018 at 18:44
  • $\begingroup$ Your title asks whether $M^o \cong N^o \implies \partial M \cong \partial N$, but your body asks whether $M^o \cong N^o \implies M \cong N$. Are those the same question? $\endgroup$
    – LSpice
    Jan 14, 2020 at 22:29
  • $\begingroup$ Hi LSpice, I just read this comment. The second one obviously imply the first one, and when I asked the question I was sure that the second one was true. I was just searching for a proof of that. Maybe they are equivalent, I do not know. $\endgroup$
    – P. Tolo
    Jan 23, 2020 at 14:06

4 Answers 4


In general, this is wrong.

Take $M=L(7,1)\times S^{2n}\times[0,1]$ and $N=L(7,2)\times S^{2n}\times[0,1]$.

Their boundaries $L(7,1)\times S^{2n}$ and $L(7,2)\times S^{2n}$ are not homeomorphic but $h$-cobordant, as proven by Milnor in

J. Milnor, Two complexes which are homeomorphic but combinatorially distinct. Ann. of Math. (2) 74 (1961) 575–590.

It follows that $\mathring{M}=L(7,1)\times S^{2n}\times \mathbb R=L(7,2)\times S^{2n}\times \mathbb R=\mathring{N}$.

However, in dimension $3$ this should be true, following from work of Edwards:

C. Edwards, Concentricity in 3-manifolds. Trans. Amer. Math. Soc. 113 (1964) 406–423.

He showed that two compact $3$-manifolds are homeomorphic if and only if their interiors are homeomorphic. A main ingerdient in his proof is (as you said) that oriented $2$-manifolds are determined by their Euler characteristic.

  • $\begingroup$ It seems that one of two L(7,1) in the third line should be L(7,2). $\endgroup$ Sep 14, 2018 at 16:24
  • $\begingroup$ @ Taras Banakh: Thanks, I have changed it. $\endgroup$
    – Marc Kegel
    Sep 16, 2018 at 10:37

(Marc Kegel posted his answer just before I posted this. I will leave it because perhaps this helps elaborate some of the points)

No, this is not true in general.

Here is an example of what can happen in high dimensions. Let $N_1$ and $N_2$ be two manifolds which are h-cobordant but not homeomorphic*. We let M be the h-cobordism, which we view as a bordism from $N_1$ to $N_2$. The s-cobordism theorem (assuming $\dim M > 4$) tells us that there is an inverse bordism $M'$ from $N_2$ to $N_1$.

This is a bordism with the properties $$M \circ M' = M \cup_{N_2} M' = N_1 \times [0,1]$$ and $$M' \circ M = M' \cup_{N_1} M = N_2 \times [0,1]$$

Now we do the swindle: We consider the infinite series of composites $M \circ M' \circ M \circ M' \circ \cdots$

On the one hand we get $$(M \circ M') \circ (M \circ M') \circ \cdots = N_1 \times [0,1] \circ N_1 \times [0,1] \circ \dots = N_1 \times [0,1)$$ One the other hand we get $$M \circ (M' \circ M) \circ (M' \circ M) \circ \cdots = M \circ N_2 \times [0,1] \circ N_2 \times [0,1] \circ \cdots$$ $$ = M \circ N_2 \times [0,1) = M \setminus N_2$$

Removing the remaining $N_1$ boundary we see that $M^0 = N_1 \times (0,1)$.

So $M$ and $N_1 \times [0,1]$ have the same interior, but different boundaries.

*such manifolds are constructed, for example, in "How different can $h$-cobordant manifolds be ?" by Bjorn Jahren and Slawomir Kwasik


As the two answers already given prove, this is not true. On the other hand, one can show that these spaces cannot be distinguished by homology or homotopy groups. This greatly strengthens your statement that they have the same Euler characteristic. Considering $\pi_0$ in particular, we can say that $M$ and $N$ have the same number of connected components. Considering $H^{\ast}$ as a functor to Rings-op, you should be able to show that (for $\dim M=\dim N=2$) they have the same number of connected components of each genus.

Let $F$ be some function from topological spaces to some category closed under limits, such as $H_j$ or $\pi_j$. Let $U = M^{\circ} \cong N^{\circ}$. Consider $$\lim_{\leftarrow} F(U \setminus K)$$ where the limit is taken over all compact subsets $K$ of $U$.

Now, fix metrics on $M$ and $N$. Let $(\partial M)_{\delta}$ be an open $\delta$ neighborhood of $\partial M$ in $M$, and let $(\partial N)_{\epsilon}$ be likewise. Then $M \setminus (\partial M)_{\delta}$ and $N \setminus (\partial N)_{\epsilon}$ are each cofinal in sets of the form $U \setminus K$, so we deduce that $$\lim_{0 \leftarrow \delta} F((\partial M)_{\delta}) = \lim_{0 \leftarrow \epsilon} F((\partial N)_{\epsilon})=\lim_{\leftarrow} F(U \setminus K).$$ But, for $\delta$ small enough, $\partial M$ is a deformation retract of $(\partial M)_{\delta}$ and likewise for $N$. So, if $F$ turns deformation retracts into the identity, we deduce that $$F(\partial M) = F(\partial N).$$

When $F=H_j$ or $\pi_j$, this invariant is called the "homology at infinity" or "homotopy at infinity" of $U$.

Here is a quicker proof that $\partial M$ is homotopy equivalent to $\partial N$ without homotopy limits. Choose $\delta_1$ small enough that, for all $\delta<\delta_1$, we have $(\partial M)_{\delta} \cong (\partial M) \times \mathbb{R}$. Choose $\epsilon_1$ small enough that the analogous condition is true, and also so that $(\partial N)_{\epsilon_1} \subset (\partial M)_{\delta_1}$. Choose $\delta_2$ and $\epsilon_2$ small enough that $(\partial N)_{\epsilon_2} \subset (\partial M)_{\delta_2} \subset (\partial N)_{\epsilon_1} \subset (\partial M)_{\delta_1}$.

Each of the inclusions above is a map in the homotopy category. Since $(\partial M)_{\delta_2}$, $(\partial M)_{\delta_1}$ and $\partial M$ are canonically homotopy equivalent, and likewise for $N$, we get maps $$\partial N \overset{\alpha_2}{\longrightarrow} \partial M \overset{\beta}{\longrightarrow} \partial N \overset{\alpha_1}{\longrightarrow} \partial M.$$ The composites $\beta \circ \alpha_1$ and $\alpha_2 \circ \beta$ are each homotopic to the identity. So $\alpha_2 \circ \beta \circ \alpha_1 = \alpha_1 = \alpha_2$ in the homotopy category, and we may denote both $\alpha_1$ and $\alpha_2$ by $\alpha$. We have shown that $\alpha$ and $\beta$ are inverse maps in the homotopy category.

  • 3
    $\begingroup$ In fact the same proof using homotopy limits will show that the whole homotopy type of the boundary is determined by the interior (precisely: the shape at ∞ of the interior is the constant prospace corresponding to the homotopy type of the boundary) $\endgroup$ Sep 15, 2018 at 8:18
  • $\begingroup$ @DenisNardin Thanks! I was imagining this was true, but I googled to check that the homotopy category was closed under limits and all the sources were about higher category theory that I didn't understand. $\endgroup$ Sep 15, 2018 at 12:14
  • $\begingroup$ Yes, the homotopy category is not a refined enough tool to do this kind of work (in particular a prospace is not correctly defined as a pro object for the homotopy category). Still, the theory can be made work with some effort, for example using (∞,1)-categories or model categories. $\endgroup$ Sep 15, 2018 at 13:44

An aside, related to some other answers: in fact any example arises from $h$-cobordisms.

If $e: M^\circ \to N^\circ$ is a homeomorphism and $j: M \to M^\circ$ is an embedding isotopic (through embeddings into $M$) to the identity map of $M$, then we can set $H = N \setminus ej(M^\circ)$. This compact manifold comes with a preferred diffeomorphism $\partial H = \partial M \amalg \partial N$, and there is a canonical homeomorphism $N \cong M \cup_{\partial M} H$.

In this situation $H$ must be an $h$-cobordism. Proof: first, $H \setminus \partial N$ is homeomorphic to $M^\circ \setminus j(M^\circ)$. It follows from the isotopy extension theorem that $M \setminus j(M^\circ)$ is homeomorphic to a collar $[0,1] \times \partial M \hookrightarrow M$, and hence $M^\circ \setminus j(M^\circ) \cong (0,1] \times \partial M$. In particular the inclusion $\partial M \hookrightarrow H \setminus \partial N \simeq H$ is a homotopy equivalence.

By the way, the (non)uniqueness question discussed here has an existence counterpart: given an open manifold, is it possible to find a homeomorphism to the interior of a compact manifold with boundary? Both existence and uniqueness are related to algebraic $K$-theory of the group ring of the fundamental group, as explained by Tom Goodwillie's here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.