# Does $M^o=N^o$ imply that $\partial M = \partial N$?

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?

• This is a duplicate of an old question: mathoverflow.net/questions/95129/…. – Igor Belegradek Sep 16 at 21:47
• 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. – P. Tolo Sep 22 at 18:44

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.

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

• 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) – Denis Nardin Sep 15 at 8:18
• @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. – David E Speyer Sep 15 at 12:14
• 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. – Denis Nardin Sep 15 at 13:44