Construct 2-Fold Cover of M Using H^1(M;Z2)

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That's right. Just glue the two manifolds together. (But to be able to visualize this you might have turn one of the manifolds inside out.) When you cut along the nonbounding cycle, the resulting boundary is a two fold cover of the cycle. If the cycle is orientable then the cover is disconnected and if it is not orientable the cover will be connected.

Example. Start with P3 and cut along a P2-hypersurface. This will result in a ball with S2-boundary, say [itex]\{|(x,y,z)|\leq 1\}[/itex]. Take the other manifold to be the ball [itex]\{|(x,y,z)|\geq 1\}\cup \{\infty\}[/itex]. Gluing them together results in S3. (See http://sketchesoftopology.wordpress.com/2009/07/25/two-balls/" for a nice animation.)

Yet another way to think of the gluing, is to again start with two copies of the manifold and consider the nonbounding cycle to be a "magic membrane". If you pass through it you do not arrive on the other side, but instead the other side of the membrane in the other manifold.
 
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Oh bah -- I confused myself by trying to picture things, didn't I? One only needs to specify the gluing locally, so the fact that the cut locally splits the manifold into two pieces is all you need. Alas, there's a few technical details I would to work out (e.g. that the cut really can be covered by open balls that are cut in half -- I'm worried about the possibility of some pathological behavior).

That's why I really liked the universal covering space approach -- I really want to take advantage of something whose details have already been worked out (and is in the domain of things I know a little about). But I suppose I ought to be able to understand this way too...
 
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OrderOfThings said:
That's right. Just glue the two manifolds together. (But to be able to visualize this you might have turn one of the manifolds inside out.) When you cut along the nonbounding cycle, the resulting boundary is a two fold cover of the cycle. If the cycle is orientable then the cover is disconnected and if it is not orientable the cover will be connected.

Example. Start with P3 and cut along a P2-hypersurface. This will result in a ball with S2-boundary, say [itex]\{|(x,y,z)|\leq 1\}[/itex]. Take the other manifold to be the ball [itex]\{|(x,y,z)|\geq 1\}\cup \{\infty\}[/itex]. Gluing them together results in S3. (See http://sketchesoftopology.wordpress.com/2009/07/25/two-balls/" for a nice animation.)

Yet another way to think of the gluing, is to again start with two copies of the manifold and consider the nonbounding cycle to be a "magic membrane". If you pass through it you do not arrive on the other side, but instead the other side of the membrane in the other manifold.
Now I see what your are saying. So it is like the classical thing of cutting a Moebius band down the middle and getting a twisted cylinder of twice the length.

This makes me think that the general situation is: start with a manifold with boundary and a fixed point free involution of the boundary. The quotient space by the involution contains a hypersurface that is covered twice by the boundary. the hypersurface's Poincare dual is the corresponding element of H^1(quotient;Z2).

It would be interesting to do some examples. If you know of any good ones I'd like to see them.

For starters I am wondering what you get from involutions of a torus as the boundary of a solid torus. Suppose I rotate in 1 direction by 180 degrees and reflect in the other. The seam manifold will be a Klein bottle. What if I just rotate 180 degrees in one direction?

What if I take a solid Klein bottle and take the involution that rotates 180 degrees along its fiber direction?
 
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