Möbius strips with 3 twists to make a Klein bottle I've been looking into Klein bottles and Möbius strips. What would happen if you took two "Möbius" strips with three twists in them, each oriented opposite eachother, and then connected the edges. Much like doing it with a 1 twist Möbius strip for a Klein bottle, is there any indication what a 3 twist would look like? Or if theres a specific term I need to search for to find out? I've done a bit of digging and can't find anything. 
 A: There's a couple things going on with your question. First, you have to understand that there is no such thing as orientation on a Mobius band. If your three-times-twisted band is not an orientable surface, there is no way to say that they are oriented opposite to one another. You should be able to convince yourself readily that this is the case for your surface. You go around the band on the top, twist three times, which gets you on to the bottom, and the loop is closed, so the surface is not orientable.
Next, you have to convince yourself that a twice-twisted Mobius band is topologically equivalent to a cylinder. There is a homeomorphism between the twice twisted band and the cylinder which is most easily seen by representing them both as quotients of a square. In this setting, one can see easily that both are formed by just gluing the left and right edges of the square - no flipping!
Finally, if your Mobius band has more than two twists, you can slice it into parts where each has at most two twists, and apply the above reasoning. The most natural way to do it is that every time you find a pair of twists, you consider them separately as a twice-twisted band, use the above homeomorphism to eliminate them, and continue.
In this way, you see that there are only, up to homeomorphism, two ways to make any sort of 'band.' There are the usual cylinders, and Mobius bands, and that's it!
The upshot to your question then is that your three-times-twisted bands are actually ordinary bands, and so you can glue them together to create a really funky looking Klein bottle, but it is in fact topologically a Klein bottle, rather than something more exotic.  When you learn the classification of surfaces, you will see that in a sense, there are no non-orientable surfaces more exotic than Klein bottles, and Klein bottles that have been surgically conjoined.
A: A two-dimensional creature living inside the strip wouldn't be able to tell the difference between the one-twist and three-twist constructions, because the strip's "local neighborhoods" all fit together in the same way in both cases. (From the creature's perspective, the world is a long strip in which if you go far enough, you get back to where you started but find yourself flipped over.) Formally, we say that the one-twist and three-twist constructions are different embeddings of the same topological space (the Mobius strip) into 3-dimensional space ($\Bbb R^3$). You can think of a topological space as an abstract description of how the neighborhoods fit together, independent of any embedding.
Similarly, when we say that gluing the Mobius strip's edge to itself produces the Klein bottle, we're not talking about embeddings - we're just abstractly "joining" certain points of a topological space to each other to define a quotient space. This is like installing teleports in the creature's 2D world (so that when you cross what used to be the edge of the strip, you “come back in on the other side”), and then looking at the new way that all the neighborhoods fit together. There's no 3D space involved in this operation, so it doesn't make sense to ask about the one-twist and three-twist versions of it. That's the thrust of Alfred's answer.
However, it sounds like you’re visualizing the gluing as an operation on embeddings, which makes sense. The one-twist and three-twist embeddings are indeed distinct (i.e. they’re not isotopic), so it makes sense to ask which versions of the Klein bottle we can get from each of them.
There’s a problem, though. If you start with any embedding of the Mobius strip in $\Bbb R^3$ and try to stretch/bend it to make an embedding of the Klein bottle (by lining the edge up with itself in the right way), you'll be forced to make the strip intersect itself, because the Klein bottle can be immersed, but not embedded, in $\Bbb R^3$.
To figure out which immersions you can get, and whether this depends on the number of twists you start with, we need to choose what sorts of stretching/bending are allowed, and we can’t choose isotopy because it doesn’t allow self-intersections, so the natural choice is regular homotopy instead. But it turns out that many distinct (i.e. non-isotopic) embeddings of the Mobius strip are regular-homotopic to each other, because you can add and remove twists just by letting the strip pass through itself. This maneuver (essentially the belt trick) lets you add any multiple of 4 to twist count, so in particular, you can turn a 1-twist embedding into a (-3)-twist embedding (i.e. a 3-twist embedding with the opposite twist orientation). So as part of morphing your Mobius strip into the Klein bottle, you’ll be able to switch between these two forms, which means that the set of immersions you can get is the same regardless of which one you start with.
