Determining surfaces by self-gluing after removing interiors of two disks. Given a surface $M$, after removing the interiors of two discs on the surface and then self-gluing along the boundary you obtain two surfaces: $M_+$ which is when orientation is preserved, and $M_-$ when orientation is reversed.
the problem is to determine the surfaces $M_+$ and $M_-$, when M is:


*

*Sphere

*Torus

*Projective plane


So it is obvious, in the case of the sphere, that the removal of two disks gives us a cylinder, and we have that M_+ is the Torus ($T^2$) and $M_-$ is the Klein Bottle ($K=\mathbb{RP}^2 \# \mathbb{RP}^2$  ).
In the case of the Torus, intuitively it seems that $M_+$ should be the 2-Torus ($T^2\#T^2$) and $M_-$ should be the connected sum of 3 $\mathbb{RP}^2$ ($=T^2\# \mathbb{RP}^2$). The fundamental polygon is tricky. 
For the projective plane, not sure but possibly a Klein bottle for $M_+$.
 A: Hint: For an arbitrary connected surface $M$ we have $M_\pm = M\# (S^2_\pm)$. My solution has two steps: 1) prove that this is true, and then 2) describe $S^2_\pm$.

Solution:
Just to make the construction more explicit, I will nail down some of my conventions. For a connected surface $M$ consider two embeddings $\varphi_1, \varphi_2 \colon D^2 \to M$ whose images are disjoint and contained in the interior of $M$, and such that one is orientation-preserving and the other is reversing. Let $\bar{M}$ be $M$ without the interiors of the images of $\varphi_1$ and $\varphi_2$. Now we can construct two surfaces
$$ M_+ = \bar{M}/(\varphi_1(x) \sim \varphi_2(x))\text{  and  }M_- = \bar{M}/(\varphi_1(x) \sim \varphi_2(\rho x)) $$
where $x\in S^1$ and $\rho\colon S^1 \to S^1$ is an orientation-reversing diffeomorphism. (To see why I don't want both $\varphi_1$ and $\varphi_2$ to preserve orientation, draw what happens when $M = D^2$.)
Since $M$ is connected, the images of $\varphi_1$ and $\varphi_2$ are contained in a larger disk, say $D\subset M$. By gluing a disc to the boundary of $D_+$ (and $D_-$) you can see that $S^2_+ = D_+ \cup D^2$ (and $S^2_- = D_- \cup D^2$). By the definition of connected sum it follows that 
$$ M_+ = M \# (S^2_+) \text{  and  } M_- = M \# (S^2_-). $$
(In words, the idea is that you could either just consider the two disks inside $M$ and do your operation, but it's equivalent to encompass them in a bigger disk and excise them out first, then do the operation on this cut-out disk, and then just glue the disk back in. This second procedure is equivalent to taking disks out of $M$ and $S^2_\pm$ and then gluing along the boundaries, i.e. the connected sum.)
Now it only remains to describe $S^2_\pm$, which you have already done: $S^2_+ = T^2$ and $S^2_- = K$, i.e. for every connected surface 
$$ M_+ = M\# T^2\text{   and   } M_- = M\# K$$
