2 -manifolds that can't be decomposed into two equal parts The real projective plane has the property that if you divide into two different 2-manifolds, they will not be homeomorphic (i.e. one will be orientable, and the other non-orientable).
The sphere does not have this property. It can be divided into two homeomorphic disks.
My question is, what 2-manifolds can't be divided into equal (homeomorphic) parts (where the parts are 2-manifolds)?
Note: We say that $M$ can be divided into manifolds $A$ and $B$ if $A \cup B = M$ and $\operatorname{int}(A \cap B) = \emptyset$.
(The interesting thing about these spaces is you can play a game like Projex on them. Simply choose some anti-reflexive total relation on manifolds, and a tiling on the manifold. Plays take turn coloring tiles, and when the board is filled, the relation determines the winner.)
 A: You don't want $A \cap B = \varnothing$; you want $\partial A = \partial B = A \cap B$. 
Recall that the Euler characteristic is well-behaved under decomposition: if $X = A \cup B$, then $\chi(X) = \chi(A) + \chi(B) - \chi(A \cap B)$. If a (compact without boundary) surface $\Sigma$ can be written as $\Sigma = A \cup B$ where $A \cong B$ is a (compact) surface. The above formula gives $\chi(\Sigma) = 2\chi(A) - \chi(A \cap B)$. But $A \cap B$ is a disjoint union of circles, and the Euler characteristic of a circle is zero. So $\chi(\Sigma) = 2\chi(A)$, and in particular is even.
If $\Sigma_g$ is the surface of genus $g$ (that is, the connected sum of $g$ copies of $T^2$), and $N_g$ the connected sum of $g$ copies of $\Bbb{RP}^2$, then $\chi(\Sigma_g) = 2 - 2g$. Indeed, you can decompose this into two pieces (perhaps the easiest way is to slice it in half along the $xy$-plane, in the usual presentation of these surfaces: image). $\chi(N_g) = 2-g$, however, and so only $N_{2g}$ can be a 'double'. If $N_{g,1}$ is what you get when you delete a small open disc from $N_g$, so that it has one boundary circle, $N_{g,1} \cup_{\partial} N_{g,1} = N_{2g}$ (we're just doing the connected sum operation!) 
Punchline: Only the connected sum of an odd number of copies of $\Bbb{RP}^2$ have this property.
