I know that for any surfaces, $K, L$ forming $K$#$L$ (where # denotes the connected sum) is done by deleting a disk from each and gluing together the boundary. I also know a few facts about the homology groups of surfaces:
1) $H_0(K)=H_0(L) \simeq \mathbb Z,$ since each is connected, the connected sum is connected.
2) $H_1(K$#$L) = H_1(K) \bigoplus H_1(L)$
3)$H_2(K$#$L)$ is either isomorphic to $\mathbb Z$ or to 0, depending on the orientability of $K$ and $L$.
I came to this conclusion (which my professor has confirmed) by considering the case of connected sum of tori, and the connected sum of projective planes, and using the fact that all surfaces are homeomorphic to one of these, or the sphere.
This seemed pretty okay when dealing with connected sums of tori and projective planes. The problem arises when dealing with classifications of surfaces of the form $T$#$P$, where $T$ is a torus and $P$ is a projective plane. To illustrate, we note that $H_1 (T)= \mathbb Z \bigoplus \mathbb Z,$ and that $H_1(P) = \mathbb Z/ 2 \mathbb Z$
So by the above, forming the connect sum would give $H_1(T$#$P)=\mathbb Z \bigoplus \mathbb Z\bigoplus (\mathbb Z /2\mathbb Z)$.
However, it a well known fact that $T$#$P$=$P$#$P$#$P$. Using statement (2) above on the right of this equation gives $H_1$$(P$#$P$#$P)$=$(\mathbb Z/ 2 \mathbb Z) \bigoplus (\mathbb Z/ 2 \mathbb Z) \bigoplus (\mathbb Z/ 2 \mathbb Z)$
But the problem is this. We computed the homology of both sides of the above equation, and together they imply that:
$\mathbb Z \bigoplus \mathbb Z\bigoplus (\mathbb Z /2\mathbb Z) = (\mathbb Z/ 2 \mathbb Z) \bigoplus (\mathbb Z/ 2 \mathbb Z) \bigoplus (\mathbb Z/ 2 \mathbb Z)$.
But this is false, since the group on the right side have exactly 8 elements, and the group on the left has an infinite number of elements.
So the question is, what went wrong in this analysis?