Intuitive Question about Klein Bottles and Torsion in Surface Groups This is just a geometric sort of question about how to picture/intuit on homology, hope it's considered attractive and not just "vague" haha
So if you take a typical generator for the torsion element in $H_1(K)$, where $K$ is the Klein Bottle, you can picture deforming the loop down the 'length of the tube' and getting the orientation reversed when you get back around.
However, I'm wondering if there is another way to visualize the torsion.  If you take a stretchy piece of string and wind it around twice the torsion element, then tie the ends off, is there a way to deform the string so that it simply "falls off" the Klein Bottle?  Or is this asking too much?
Obviously "falls off" is sort of a 3-d notion.  I guess what I'm asking is that if your Klein bottle is sitting in a high-dimensional Euclidean Space (I guess 5+ should suffice to avoid any complications at all, but maybe 4 is fine as well), is the map that winds a circle around (a tubular neighborhood around) the torsion element twice homotopy-equivalent to a free loop rel $K$?  Or is this not quite the right question?
Thanks!
 A: What you're asking sounds more like torsion in the fundamental group. The intuition works for instance for $\mathbb{RP}^2$. Wind your loop around twice and suddenly you can continuously deform your loop into a point. Since $H_1$ is the abelianisation of $\pi_1$, part of this intuition is reflected onto $H_1$ as well.
The case of the Klein bottle would be a bit more subtle because although H1 has torsion, it doesn't come from torsion in $\pi_1$ (I think $\pi_1$ is torsionfree). Somewhat tautologically, you should still be able to say that, upon going around your loop twice, it becomes the boundary of some $2$-cycle, and perhaps if you squint hard enough you can see this geometrically.


OK so I said that if you squint you get to see the torsion so let's squint. We have two $2$-simplices $A$ and $B$, and we can make sense of their boundaries. We have $\partial A = a + b - c$, and $\partial B = -a + b + c$. That means that $\partial(A + B) = 2b$, which means that by adding $b$ up to itself we suddenly get to realise it as the boundary of a $2$-simplex.
This makes sense from the picture. $A$ and $B$ together can be thought of as the entire square, and tracing along its boundary we see that we pass through $b$ twice with the correct orientation.
