Confusion between sheaf cohomology of the cotangent bundle of complex projective space and the geometric picture The main aim in all of this is to find the singular cohomology of $T^*\mathbb{P}^1$ (as a manifold).
I know that for this space the $k$-th cohomology coincides with the $k$-th sheaf cohomology, and then Hartshorne Theorem III.5.1 says that $H^0(T^*\mathbb{P}^1) = H^1(T^*\mathbb{P}^1) = \mathbb{C}$, if I take for granted that $T^*\mathbb{P}^1 = \mathcal{O}(-2)$.
On the other hand, I can draw a crude picture of the sphere $S^2$ and some tangent planes to model $T^*\mathbb{P}^1$, and it seems that $H^4 = H^0 = \mathbb{C}$ whilst the other degrees have zero cohomology. This is because I can have 0-cochains defined on discrete points and 4-cochains on the entire bundle, whilst the 1, 2, 3-cochains can be 'contracted' to a point. The 2-cochain in particular is either a cochain on a fibre or a cochain on the base space, both of which should be zero.
This isn't at all a rigorous argument, and if the first paragraph is correct, it should be wrong. I'm vaguely aware of Riemann-Roch but don't know how to apply it, so an answer using relatively simpler techniques would be ideal, even to correct my answer.
In particular, I get the feeling that the cohomology should be either $\mathbb{C}$ in degrees 0 and 4 or degrees 2 and 4, but I can't figure out why it should be one or the other, and this isn't given by the first paragraph.
What am I misunderstanding?
 A: Question: "In particular, I get the feeling that the cohomology should be either C in degrees 0 and 4 or degrees 2 and 4, but I can't figure out why it should be one or the other, and this isn't given by the first paragraph. What am I misunderstanding?"
Answer: @mi.f.zh - If you view $S:=T^*(\mathbb{P}^1_{\mathbb{C}})$ as a complex manifold with the strong topology, you may consider the constant sheaf $F:=\mathbb{Z}_S$. For any open subset $U \subseteq S$ it follows $F(U):=\mathbb{Z}$. It follows $F$ is a sheaf of abelian groups on $S$ (restriction maps are the identity maps) and you may define the cohomology of $F$ in the sense of Hartshorne, Chapter III.
It follows (...) singular cohomology of $S$ may be calculated using the sheaf cohomology of $F$: There is an isomorphism
$$H_{sing}^i(S,\mathbb{Z}) \cong H^i(S,F)$$
where $H^i(S,F)$ is the cohomology introduced in Hartshorne. The constant sheaf $F$ is not a quasi coherent $\mathcal{O}_S$-module, hence singular cohomology differs much from the cohomology of quasi coherent sheaves.
Note: For a complex projective manifold $X$ (with $X^{alg}$ its "corresponding" algebraic variety) and for any coherent $\mathcal{O}_X$-module $E$ (with $E^{alg}$ its corresponding coherent $\mathcal{O}_{X^{alg}}$-module) it follows there is an isomorphism of complex vector spaces
$$H^i(X,E) \cong H^i(X^{alg},E^{alg}).$$
for all $i\geq 0$. Here $E$ is a coherent sheaf on $X$ in the strong topology and $E^{alg}$ is a coherent sheaf on $X^{alg}$ in the Zariski topology.
The singular cohomology groups are (finitely generated) abelian groups in general, sheaf cohomology groups are quasi coherent sheaves: If $E$ is a finite rank locally trivial sheaf on $X:=Spec(A)$, it follows  its 0'th sheaf cohomology group $H^0(X,E)\cong P$ is a finite rank projective $A$-module.
Note:  (...) means you will find this proved in a course on complex manifolds. Maybe Griffiths-Harris "Principles of algebraic geometry" proves it or gives references. Constant sheaves in the Zariski topology are not that useful. They are useful in the etale topology (in the definition of various characteristic $p$ cohomology theories such as l-adic etale cohomology).
In Milne ("Etale cohomology", Theorem 3.12) you find the following result: Let $X \subseteq \mathbb{P}^d_{\mathbb{C}}$ be a smooth projective scheme and let $X(\mathbb{C})$ be the "associated" complex projective manifold. Let $X_{et}$ be $X$ equipped with the etale topology and let $M$ be a finite abelian group. It follows
$$H^i_{sing}(X(\mathbb{C}), M) \cong H^i_{et}(X_{et},M_{X_{et}})$$
hence the etale topology calculates singular cohomology with finite coefficients $M$.  Here $M_{X_{et}}$ is the constant sheaf (in the etale topology) defined by $M$ and $H^i_{et}$ is "etale cohomology". Hence you may use the etale topology on $X$ to calculate singular cohomology with finite coefficients. In the "etale topology" on $X$ your open sets are no longer subsets of $X$. The open sets $U$ are etale morphisms $\phi_U:U \rightarrow X$. Any open immersion $i:U \rightarrow X$ is etale, hence Zariski open sets are open in the etale topology. The fact that you may give an algebraic construction of singular cohomology on $X(\mathbb{C})$ shows that the etale topology is useful.
