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I can’t seem to find the answer to the question in the title despite searching for a while. In the examples I can think of it seems true, though that isn’t many. Help is appreciated.

JLA
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The connected sum of complex manifolds of dimension $\ge 2$ does not have a natural complex structure (sometimes, it does not have a complex structure at all!). A meaningful interpretation of your question then is:

Suppose that $X, Y$ are $n$-dimensional Stein manifolds. Is $X\# Y$ homeomorphic (diffeomorphic) to a Stein manifold?

The answer to this is negative whenever $n\ge 2$. More precisely, the connected sum of Stein manifolds is never even homotopy-equivalent to a Stein manifold, when $n\ge 2$.

The reason is that every $n$-dimensional Stein manifold $X$ is homotopy-equivalent to an $n$-dimensional CW complex, hence, $H_{2n-1}(X)=0$ unless $n\le 1$. But if $X, Y$ are connected open $2n$-dimensional manifold then the separating sphere in $X\# Y$ represents a nontrivial class in $H_{2n-1}(X\# Y)$.

On the other hand, the connected sum of Stein Riemann surfaces (with the natural complex structure) is again Stein. This is non-trivial and follows from the fact that a Riemann surface is Stein if and only if it is open (proven by Behnke and Stein).

Moishe Kohan
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  • Thank you for your response, but I am confused as to how it can be correct. For one, in that link you posted it is stated that $M # \overline N$ has a natural complex structure. Secondly, as I understand it is true that $H_{2n-1}(X# Y)=H_{2n-1}(X\vee Y)=H_{2n-1}(X)\oplus H_{2n-1}(X)=0,,$ since $H_{2n-1}(X)$ and $H_{2n-1}(Y)$ are $0$ for $n>1$ (see https://math.stackexchange.com/questions/187413/computing-the-homology-and-cohomology-of-connected-sum/187417). – JLA Jan 04 '20 at 23:04
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    @JLA: You should read the linked answer more closely: The horizontal lines separate the correct from the incorrect. As for the computation of $H_*$, the first equality that you wrote is false, you are incorrectly applying the MV-sequence. Observe that $H_{2n-1}(X-{x})\cong {\mathbb Z}$ since $X$ is noncompact. And the linked answer by KReiser is wrong as well: He implicitly assumes compactness of at least one of the manifolds. – Moishe Kohan Jan 04 '20 at 23:14