Let $X$ and $Y$ be finite and connected CW-complexes of dimension $n$ with exactly one $n$-cell. Then we can define their connected sum $X\#Y$ just like in the manifold case: extract an $n$-dimensional open disk from the interior of the $n$-cell of $X$, do the same for $Y$ and glue them along the boundaries. (I think one could still define the connected sum if $X$ and $Y$ had more than one $n$-cell, but the final result would depend on the chosen $n$-cell). Here is the problem:
If $X$ and $Y$ are CW-complexes satisfying the conditions above and $H_n(X) \simeq H_n(Y) \simeq \mathbb{Z}$, then $$H_k(X \# Y) = \left\{ \begin{array}{ccc} H_k(X)\oplus H_k(Y) & \mbox{if} & 0<k<n\\ \mathbb{Z} & \mbox{if} & k=0,n\\ 0 & \mbox{if} & k>n \end{array}\right.$$
The situation here is pretty similiar to the manifold case (like this question): We take open sets A slightly bigger than the "punctured" X and $B$ slightly bigger than the "puntured" $Y$ such that $A \cup B = X \# Y$, $A \cap B$ is homotopy equivalent to $S^{n-1}$ and $A$,$B$ deformation retracts to $X^{n-1}$, $Y^{n-1}$, respectively. Then we use Mayer-Vietoris.
For $k<n-1$ it works, but for the last two groups it is not that easy. We have the exact sequence $$\dots\to H_n(X^{n-1}) \oplus H_n(Y^{n-1}) \to H_n(X \# Y)\to H_{n-1}(S^{n-1}) \to H_{n-1}(X^{n-1})\oplus H_{n-1}(Y^{n-1}) \to H_{n-1}(X \# Y) \to H_{n-1}(S^{n-1}) \to \dots$$ which turns out to be $$\dots \to 0 \to H_n(X \# Y)\to \mathbb{Z}\to H_{n-1}(X) \oplus H_{n-1}(Y) \to H_{n-1}(X \# Y) \to 0 \to \dots $$
At this point, on the manifold case, one uses that $X \# Y$ is again a manifold and its top homology group is known, along with some orientability arguments. I think we can't adapt this reasoning for CW-complexes.
I tried to show that $\mathbb{Z}\to H_{n-1}(X) \oplus H_{n-1}(Y)$ is the zero homomorphism directly, using that $H_n(X) \simeq H_n(Y) \simeq \mathbb{Z}$ (which I think I have not used yet), but I couldn't. Any ideas?
