I have just started to learn about characteristic classes and before learning more about the ways to compute them it would be nice to compute some examples using tools I already know. I only started learning homology and cohomogy a few weeks ago so the ideas here will be very basic, the cost being a very long thought process for something that could be done in a paragraph.
Consider the projective space $P^1(\mathbb{C})$. It is well known that the line bundles $E$ over a topological space $X$ are classified up to isomorphsim by their Chern classes $c_1(E) \in H^2(X;\mathbb{Z})$. In the case of $P^1(\mathbb{C}) \cong S^2$ then $H^2(P^1(\mathbb{C});\mathbb{Z})=\mathbb{Z}$. The classes of complex line bundles over $S^2$ are also classified by homotopy maps $S^1 \rightarrow U(1) \cong \pi_1(S^1) \cong \mathbb{Z}$. These two statements are clearly the same in some way, which I would like to show.
In this case $c_1(E)$ is the top chern class and so is the Euler class of the vector bundle. Define the euler class via the Thom isomophism as folows: there exists an element of the relative cohomology group $u \in H^2(E, E \setminus E_0; \mathbf{Z})$ so the the restriction of $U$ to each fiber of $E$ is the orientation class on $F$. The Euler class is the image of this form Under the canonical maps $H^r(E, E \setminus E_0; \mathbf{Z}) \to H^r(E; \mathbf{Z}) \to H^r(X; \mathbf{Z})$.
Now $P^1(\mathbb{C})$ can be covered by two open sets, both homeomorphic to $\mathbb{C}$, and the vector bundle over these will be the trivial bundle so given coordinites $(z, w)$ on $\mathbb{C} \times \mathbb{C}-0$ the thom class on these is going to (I guess) like $\frac{1}{w} dw \wedge d \bar{w}$. I now try to glue these two back together to get a thom class over the whole bundle. Glue the two copies together by $(z , w) = (z^{-1} , \frac{z^n}{\|z\|^n}w)$. This corressponds to the vector bundle with hopotopy class $n \in \mathbb{Z} \cong \pi_1 (U(1))$. Now we want to see which thom class this gluing procedure produces. At this stage I've kind of lost the wood for the trees but I went ahead and did a lot of transforming into new coordinites and wedging and pulling back anyway. The results were confusing at best, but I got something of the form $n z^{-1} dz \wedge d \bar{z} $ as the class $c_1(E) \in H^2(P^1(\mathbb{C});\mathbb{Z})$ corresponding to the vector bundle with clutching function $(z , w) = (z^{-1} , \frac{z^n}{\|z\|^n}w)$.
I was intially quite excited that everything had come together so well but I think I must have made a mistake doing hopeful cancellations or just blatently making up what I was doing based on the answer I hoped to get or something. I think the class $n=2$ should corresspond to the case of the tangent sphere and here we expect the Euler class to be twice the generating class of $H^2(P^1(\mathbb{C});\mathbb{Z})$ and it looks like this is the case. Likewise $n=0$ seems to give the trivial case. On the other hand I though $n=1$ respresented the tautogical bundle? In which case I should get $-1$ times the generating class of $H^2(P^1(\mathbb{C});\mathbb{Z})$, but I don't get that. I would like to head back and try and get everything more exactly, but Id like to know my search isnt fruitless and I'm a little out of my depth. Does the line of reasoning, at least, look correct?