1-cycles of $T^2$ distinguished by embedding The first fundamental group of a $T^2$ is $\mathbb{Z} \times \mathbb{Z}$, generated by two cycles typically referred to as the "a" and "b" cycles of the torus. As an abstract mathematical object, nothing distinguishes these two cycles: it is arbitrary which cycle we call the "a" cycle and which one we call the "b" cycle. However, upon embedding in $R^3$, there is a difference between the "a" and "b" cycle: in particular, as can be seen in the attached image, one may smoothly shrink the b-cycle to zero size, leaving the a-cycle as a remaining $S^1$. The a-cycle, however, does not seem to admit a continuous deformation to zero size that leaves the b-cycle as a remaining $S^1$ embedded in $\mathbb{R}^3$.
My question is: what is the underlying mathematics that explains this distinction between the shrinking of the a-cycle and the b-cycle of a $T^2$ upon embedding in $\mathbb{R}^3$? I am particularly interested in how this distinction (which I can visualize in the case at hand) extends to embeddings of more general Riemann surfaces in more general d-manifolds (where things are not so easy to visualize).
Torus shrinking image
 A: I would say that what is worth remembering here is that you are making a particular choice of embedding of the topological space $S^1 \times S^1$ into $\mathbb{R}^3$. Your remarks about cycles admitting certain deformations don't make much mathematical sense to me, but I do see what you mean intuitively, in this setting. Unfortunately, here you do indeed need to make your intuition precise, and that's where your question stops making sense.
Due to the embedding which we've chosen of $T^2$ into $\mathbb{R}^3$, it may appear as though cycle $b$ can be "deformed" in a way that cycle $a$ cannot, but in fact this isn't quite right in any real sense. Whatever you've done to $b$, you can do to $a$.
More precisely, you've said that we can "shrink" the $b$ cycle. What does this mean? Without knowing (although I would guess that you require a word beginning with "h"), I assert that whatever precise mathematical meaning you choose to attach to this statement - whatever that meaning turns out to be - you can apply it in exactly the same way to $a$.
