Relationship between the $2$-plane bundles over $S^2$ and $\mathbb{Z}$ I want to follow up on this answer by asking a few more questions (posting directly on the question didn't seem to "bump" the thread). I was trying to read the referenced text (Husemoller's Fiber Bundles), but I couldn't understand that much of it since the "Clutching Constructions" chapter was a bit terse.
Given that you know there's a one-to-one correspondence between the isomorphism classes of $2$-dimensional bundles on $S^2$ and $\mathbb{Z}$, how do you know which bundle corresponds to which integer? For example, consider something like the trivial, tangent, or cotangent bundles.
Or to expand my question a bit more generally, given a map $f:S^1\rightarrow GL(n,\mathbb{R})$, what does the bundle represented by $f$, $E_f$, look like?
I think once I see an example, I'll be able to understand clutching constructions much better. I appreciate any help!
 A: The bundle corresponding to $0\in \mathbb{Z}$ is, not surprisingly, the trivial bundle.
For $z\in \mathbb{Z}$, the bundle corresponding to $-z$ is the same bundle with opposite orientation on the fibers, so we may as well just try to understand $z = 1, 2, 3,...$.  Let $E_z$ be the rank 2 oriented vector bundle corresponding to the positive integer $z$.
Now, stick an inner product on $E_z$ and let $M_z\subseteq E_z$ be the set of all unit length vectors.  Note that $M_z = f^{-1}(1)$ where $f:E_z\rightarrow\mathbb{R}$ is $f(p) = |p| $ and that $1$ is clearly a regular value.  Hence, $M_z$ is actually an embedded submanifold.  $M_z$ is called the unit sphere bundle over $S^2$.
The projection map, when restricted to $M_z$, gives a bundle $S^1\rightarrow M_z\rightarrow S^2$.
Via some abstract nonsense, the number $z$ can be identified with the Euler class of this bundle.  The Euler class is technically an element of $H^2(S^2;\mathbb{Z})$, but since we're bluring the distinction between positive and negative numbers, we can take it to be a well defined integer.
There is something called the Gysin sequence which is an exact sequence involving $S^2$ and $M_z$ which can be computed with knowledge of the Euler class.  Suffice it to say that we can show, using the Gysin sequence, that $H^2(M_z) = \mathbb{Z}/z\mathbb{Z}$.  This is how we tell the bundles apart.  In particular, if we can construct a bundle where $H^2(M_z) = \mathbb{Z}/z\mathbb{Z}$, then we must have found them all.
Now, to your questions.


*

*The Euler class of the tangent bundle of an oriented manifold can be identified with the Euler characteristic times the fundamental class.  Since the Euler characteristic of $S^2$ is $2$, the tangent bundle is $E_2$ (or $E_{-2}$).  The cotangent bundle of a manifold is always bundle isomorhpic to the tangent bundle via the flat/sharp maps (assuming a background Riemannian metric).

*As to what the bundle "looks like", here's how I think about it.
Start with the Hopf bundle $S^1\rightarrow S^3\rightarrow S^2$.  This is a principal bundle, so there is a "preferred" direction on the $S^1$s.  Now, think of each $S^1$ as sitting inside of an $\mathbb{R}^2$, with each $\mathbb{R}^2$ disjoint from every other $\mathbb{R}^2$.  Give each $\mathbb{R}^2$ the orientation which makes the "preferred" direction on the $S^1$, say, counter clockwise.  This is $E_1$.  (Notice that in this case, $M_1 = S^3$ and we have $H^2(S^3) = 0$ as claimed.)
Now, there is a canonical $\mathbb{Z}/z\mathbb{Z}$ inside of $S^1$, the $z$th roots of unity.  Dividing by this action, we get a bundle $S^1/(\mathbb{Z}/z\mathbb{Z})\rightarrow S^3/(\mathbb{Z}/z\mathbb{Z})\rightarrow S^2$.
The fiber is still a circle, but the total space becomes a lens space (or $\mathbb{R}P^3$ when $z=2$).  Do the same trick of filling in each circle fiber with an oriented $\mathbb{R}^2$ to make the bundle $E_z$.  In this case, the sphere bundle is the lens space.
Incidentally, this provides one of the more complicated proofs I know of that the unit tangent bundle of $S^2$ (i.e., $M_2$) is diffeomorphic to $\mathbb{R}P^3$.
A: This doesn't directly answer your question, but the computations become simpler I think if you restrict to oriented 2-plane bundles.  Now these are classified by maps (up to homotopy) $S^1 \to GL_+(2)$.  But by putting a metric on any bundle and making sure our transition functions respect the metric, such bundles are actually classified by maps $S^1 \to SO(2) \simeq S^1$ (another way to see this is that $GL_+(2)$ deformation retracts onto $SO(2)$).  But any map to a sphere to itself is determined up to homotopy by its degree.  For maps $S^1 \to S^1$ this means they all look like $e^{i\theta}\mapsto e^{in\theta}$, with $n$ being the degree of the map.
Further since $SO(2) \simeq U(1) \subset GL(1,\mathbb C)$, we actually see that any oriented 2-plane bundle is a complex line bundle over $S^2 \simeq \mathbb C P^1$.  These are classified by their first Chern classes, which lie in $H^2(S^2;\mathbb Z)$.  Since the Chern class of the tautological line bundle $H \to \mathbb CP^1$ generates $H^2(S^2)$ and since the Chern class is a group isomorphism from line bundles (with tensor product as the operation) to $H^2(S^2;\mathbb Z)$, it follows that any complex line bundle is a tensor product of copies of $H$ and its dual $H^*$.
Under these correspondences the trivial bundle corresponds to the integer 0 since its transition functions can be made constant.  $TS^2$ and $T^* S^2$ are isomorphic (as is always the case for tangent and cotangent bundles, which can be seen by putting a metric on your manifold) and it can be shown that $T \mathbb CP^1$ direct sum with the trivial complex line bundle is $H^* \oplus H^*$.  From properties of the total chern class it follows that the first chern class of $T\mathbb CP^1$ is twice that of $H^*$. 
I find Husemoller to be a little difficult in many places.  I would recommend Hatcher's book on Vecotor bundles and K-theory (available freely on his website) and Milnor and Stasheff's Characteristic classes.
