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?
 A: 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.
A: (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
A: 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.
A: 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.
