Edit: According to the interesting comments of Michael Albanese and Nick L we revise the question as follows:

By manifold compactification of a manifold $M$ we mean a compact manifold $\tilde{M}$ which contains $M$ as an open dense subset.

Assume that $M$ is an open connected manifold which admits a manifold compactification. Does $M$ necessarily admit a manifold compactification with zero Euler characteristic?

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    $\begingroup$ @Neal: Once you remove the pinch point, what's left is homeomorphic to a cylinder which embeds in a torus. $\endgroup$ – Michael Albanese Nov 3 '17 at 20:32
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    $\begingroup$ Possible duplicate of Compactification of a manifold $\endgroup$ – Michael Albanese Nov 3 '17 at 20:49
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    $\begingroup$ No. The complement of a point in an oriented surface of genus 2 cannot be embedded in a torus or a Klein bottle. $\endgroup$ – Tom Goodwillie Nov 4 '17 at 12:38
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    $\begingroup$ Perhaps use the fact that the genus two surface contains a separating curve which is non-trivial in $\pi_{1}$, You just have to prove that the class of this curve doesn't die in the compactification. $\endgroup$ – Nick L Nov 5 '17 at 16:52
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    $\begingroup$ No it is trivial in homology but non-trivial in the fundamenetal group. This is discussed here math.stackexchange.com/questions/1031069/…. $\endgroup$ – Nick L Nov 12 '17 at 11:22

First note that odd dimensions the question of Euler characteristic $0$ is automatic, $M$ will embed in the orientable double cover of $\tilde{M}$, which will have $\chi = 0$ by Poincare Duality.

In even dimension = $2n$ (we assume $n > 1$), we recall the following fact. If $M_{1},M_{2}$ are compact connected manifolds then $\chi(M_{1} \# M_{2}) = \chi(M_{1}) + \chi(M_{2}) - \chi(S^{2n}) = \chi(M_{1}) + \chi(M_{2}) - 2$.

To prove that some embedding into any manifold implies embedding in to Euler characteristic $0$ manifold, it is sufficient to show that any integer is equal to the Euler characteristic of some manifold of dimension $2n$, since we can do connect sums on the complement of the embedding $M \hookrightarrow \tilde{M}$ (It is easy to see that the embedding can be changed so that this complement contains an open set) to shift the Euler characteristic of $\tilde{M}$ to the correct value.

In dimension $4$ we have $\chi(\mathbb{R}\mathbb{P}^{2} \times \mathbb{R}\mathbb{P}^{2}) =1$ and $\chi(\mathbb{C} \mathbb{P}^{2} ) = 3$. So for any $4$-manifold $N$ connect summing with $\mathbb{R}\mathbb{P}^{2} \times \mathbb{R}\mathbb{P}^{2}$ subtracts 1 from $\chi(N)$, connect summing with $\mathbb{C} \mathbb{P}^{2} $ adds one to $\chi(N)$, hence there is a $4$-manifold with Euler characteristic equal to any integer. In higher even dimension taking appropriate products with $\mathbb{R}\mathbb{P}^{2}$ will give the same result.

Edit 1 As Misha points out we can now ensure the image is dense by using the fact that every connected n-manifold is a compactification of an open n-cell.

Edit 2. Note that the above solution holds in even dimensions atleast $4$. I will give details a counterexample in the dimension $2$ case (which was pointed out Tom Goodwillie).

Let $S$ be an orientable surface of genus $g \geq 2$. I will show there is no embedding $I: S \setminus \{p\} \hookrightarrow S'$ (for some $p \in S$). Where $S'$ is a compact surface of Euler characteristic $0$ (i.e. a torus or a Klein bottle).

We argue by contractiction, suppose such an embedding $I$, exists. Let $C$ be the boundary of a small neighbourhood of $p$.

Suppose there is an embbeding $I: S \setminus \{p\} \hookrightarrow S' $, then image of $I(C)$ decomposes $S'$ as the connect sum of $S$ and compact surface $\Sigma$. In symbols: $S' = S \# \Sigma$. Hence by the formula for Euler characteristic of a connect sum given above $\chi(S') \leq -2$.

To see this note by the classification of surfaces $\chi(\Sigma) \leq 2$ and by direct computation $\chi(S) = 2-2g \leq -2$, the fact that $\chi(S') \leq -2$ now follows directly from the formula for Euler characteristic of a connect sum. This is the desired contradiction.

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    $\begingroup$ You can easily complete your solution by using the fact that every connected n-manifold is a compactification of an open n-cell. $\endgroup$ – Misha Nov 5 '17 at 2:48
  • $\begingroup$ Thank you for your answer.And I thank you @Misha for your comment. $\endgroup$ – Ali Taghavi Nov 5 '17 at 11:52
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    $\begingroup$ @Misha: will you please elaborate your comment: how does the fact you quoted help to find a dense embedding out of an open embedding? I would say one needs a more general "every connected n-manifold is a compactification of (any manifold diffeomorphic to) any nonempty open subset of its" (which I think is also true). $\endgroup$ – Pietro Majer Nov 12 '17 at 17:23

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