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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 $4$. Let me construct a counterexample in the dimension $2$ case.

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.

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.

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 $4$. Let me construct a counterexample in the dimension $2$ case.

Let $S$ be a genus $3$ orientable surface. 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. Write $S$ as $\mathbb{T}^{2} \# \mathbb{T}^{2} \#\mathbb{T}^{2}$, and let $C \subset S$ be the curve along which the first connect sum is made. We set $p$ so that is contained in the first factor. (in other words, contracting $C$ gives a wedge sum of a genus two surface and a torus, we set $p$ so that it lies in the torus)

Suppose there is an embbeding $I: S \setminus \{p\} \hookrightarrow S' $, then image of $I(C)$ decomposed $S'$ into a connect sum of a genus $2$ surface and compact surface $\Sigma$. In symbols: $S' = \mathbb{T}^{2} \# \mathbb{T}^{2} \# \Sigma$. 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 $\chi(\mathbb{T}^{2} \# \mathbb{T}^{2}) = -2$, 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.

(Note that this proof unfortunately does not apply to a puctured genus 2 surface)

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 $4$. Let me construct a counterexample in the dimension $2$ case.

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.

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: After reading the question more carefully, I realise this is not a complete solution. The embedding we obtain does not have dense image after doing the connect sums. (Although see the comment of Misha below).

Edit 2. Note that the above solution holds in even dimensions atleast $4$. Let me construct a counterexample in the dimension $2$ case.

Let $S$ be a genus $3$ orientable surface. 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. Write $S$ as $\mathbb{T}^{2} \# \mathbb{T}^{2} \#\mathbb{T}^{2}$, and let $C \subset S$ be the curve along which the first connect sum is made. We set $p$ so that is contained in the first factor. (in other words, contracting $C$ gives a wedge sum of a genus two surface and a torus, we set $p$ so that it lies in the torus)

Suppose there is an embbeding $I: S \setminus \{p\} \hookrightarrow S' $, then image of $I(C)$ decomposed $S'$ into a connect sum of a genus $2$ surface and compact surface $\Sigma$. In symbols: $S' = \mathbb{T}^{2} \# \mathbb{T}^{2} \# \Sigma$. 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 $\chi(\mathbb{T}^{2} \# \mathbb{T}^{2}) = -2$, 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.

(Note that this proof unfortunately does not apply to a puctured genus 2 surface)

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 $4$. Let me construct a counterexample in the dimension $2$ case.

Let $S$ be a genus $3$ orientable surface. 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. Write $S$ as $\mathbb{T}^{2} \# \mathbb{T}^{2} \#\mathbb{T}^{2}$, and let $C \subset S$ be the curve along which the first connect sum is made. We set $p$ so that is contained in the first factor. (in other words, contracting $C$ gives a wedge sum of a genus two surface and a torus, we set $p$ so that it lies in the torus)

Suppose there is an embbeding $I: S \setminus \{p\} \hookrightarrow S' $, then image of $I(C)$ decomposed $S'$ into a connect sum of a genus $2$ surface and compact surface $\Sigma$. In symbols: $S' = \mathbb{T}^{2} \# \mathbb{T}^{2} \# \Sigma$. 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 $\chi(\mathbb{T}^{2} \# \mathbb{T}^{2}) = -2$, 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.

(Note that this proof unfortunately does not apply to a puctured genus 2 surface)