connected sums of closed orientable manifold is orientable a general version: connected sums of closed manifold is orientable iff both are orientable.
I think this can be prove by using homology theory, but I don't know how.Thanks.
 A: You can also have a differential eye on that matter. I will use a less precise vocabulary than in the other answers.
A manifold is orientable if and only if, when you follow a (smooth) path, you never come back to the starting point with the orientation reversed (as happens for example in the Möbius band). That can be seen using the orientation cover, if you know what this is, or as the most trivial result in obstruction theory.
In a connected sum, I can see the whole manifold as the union of the two pieces along a thickening of a sphere of codimension one. On this intersection, I can fix a compatible orientation once and for all. Now, if I had a path starting in this sphere and ending there, but where orientation is reversed, I could slightly modify that path in order to ensure that it has only finitely intersection points with the sphere. So my path is the union of finitely many paths starting from the sphere, ending there and never coming back there meanwhile. As I have a coherent choice of orientation along this sphere, at least one of those paths has changed orientation, so at least one of the pieces isn't orientable.
I believe that this argument can be made rigorous in several ways (using the orientation cover, using the first Stiefel-Whitney class, using a differential topology definition of orientation, etc.) but I think that regardless of the formalisation you choose, it really tells you the whole story.
A: If the connect sum is orientable, so are both pieces:
Proof:  We'll use the fact that an $n$-manifold is closed and orientable iff $H_n(M) = \mathbb{Z}$.  Assume $M_1$ is nonorientable and consider the connect sum $M_1\sharp M_2$.
The pair $(M_1\sharp M_2, M_2-\{p\})$ gives rise to a long exact sequence, a portion of which is $$...\rightarrow H_n(M_2-\{p\})\rightarrow H_n(M_1\sharp M_2)\rightarrow H_n(M_1\sharp M_2, M_2-\{p\})\rightarrow...$$
Now, $M_2-\{p\}$ is not closed so $H_n(M_2-\{p\}) = 0$.  Also, we can identify $H_n(M_1\sharp M_2, M_2-\{p\})$ with $H_n(M_1\sharp M_2/M_2-\{p\}) = H_n(M_1) = 0$, since $M_1$ is nonorientable.  By exactness, the middle term $H_n(M_1\sharp M_2)$ must be 0.  Since $M_1\sharp M_2$ is clearly closed, it must be nonorientable. $\square$
I don't know how to show the converse using homology, but one can see that the connect sum of orientable manifolds is orientable as follows.  Choose orientations on $M_1$ and $M_2$.  These choices induce orientations at every point of $M_1\sharp M_2$; the only issue is whether or not these orientations agree on the intersection of $M_1$ and $M_2$, i.e., on an $S^{n-1}\times (0,1)$.  Since $S^{n-1}\times (0,1) $ is connected (if $n > 1$, which we may as well assume since all $1$-manifolds are orientable), these orientations either agree on every point of $S^{n-1}\times (0,1)$ or disagree on every such point.  If they disagree, it's clear that choosing the reverse orientation on $M_2$ will make them agree.  But then this defines an orientation on $M_1\sharp M_2$, so it's orientable.
A: Consider the perspective of simplicial homology, for manifolds M,M'. Assume WOLG that M,M' are both connected: if an m-manifold M is orientable (I think that there is a result that all manifolds can be made into simplicial complexes), this means that the top cycle --call it m'-- can be assigned a coherent orientation, so that m' is a cycle --that does not bound, since m is the highest dimension--, i.e., the net boundary of m cancels out , e.g., in the simplest case of a loop with boundary a-a=0. This means m', which represents M itself, is a non-trivial cycle, which generates the top homology class. If your coefficient ring is $\mathbb Z$, then the top homology will be $\mathbb Z$; consider going n-times about the cycle. 
  Now, the key is that the two orientable manifolds can be glued so that, at the circle of gluing, the total boundary cancels out, and the resulting manifold M#M' is still orientable. As a specific example,  consider a square a,b,c,d, with arrows going all in  the same direction, so that the net boundary is : (b-a)+(c-b)+(d-c)+(a-d)=(b-b)+(a-a)+(c-c)+(d-d)=0 . Now glue a second square a',b',c',d' along a common edge, (say (b,c) with (b',c')), but reverse the orientation of the edge  (b',c') in M when gluing, and notice how the simplex resulting from the gluing also has net boundary zero. 
Now, the key general point is that , at the cycle C where we collapse M with M', we change the orientation of C in either M or M', so that, along the common cycle, where you are doing the gluing, the respective boundaries cancel each other out, and the remaining orientations of M-C and M-C' remain the same, so that M#M' is orientable.
