I’ve a problem understanding why the action of the second cohomology group (integer coefficients) of an oriented smooth manifold $M$ is free and transitive on the set of $Spin^c$. I’m following these notes on the nLab (Prop D.43) but they dont prove that the action is transitive.
For every two $Spin^c$-structure on $M$, say $P,P’$ the author constructs a chern class $\delta$ (Distinguishing chern class) and then claim that $$P’\cong P\times_{K} U(\Bbb L_{\delta})$$ where $K=\ker \pi\colon Spin^c(n)\to SO(n)$ and $U(\Bbb L_{\delta})$ is the $S^1$-bundle over $M$ associated to the chern class $\delta$.
how can I start building a map between $P’$ and $P\times_{K} U(\Bbb L_{\delta})$ for example?
I'm aware that this question is somehow technical, but i can’t find any resource spelling out the details of transitiveness and freenes of such action.
You consider the fiber sequence $\Bbb{CP}^\infty \to B\text{Spin}^c(n) \to BSO(n).$ Things are easiest if you choose a model of this fibration in which $\Bbb{CP}^\infty$ is a group, and this is a principal $G$-bundle.
A brief point on principal bundles: given a principal $G$-bundle $P \to B$, there is an associated bundle of groups, $\text{Aut}(P)$; there is an isomorphism of groups $\text{Aut}(P)_x \cong G$ (an automorphism of the $G$-set $P_x$ is given by right-multiplication by some element of $G$). This bundle is not necessarily trivial, but if $P$ had transition functions $\rho_{\alpha \beta}: U_{\alpha \beta} \to G$ acting on elements of $G$ by left-multiplication, then the transition functions for $\text{Aut}(P)$ act on elements by conjugation by $\rho_{\alpha \beta}$ instead of left multiplication. In particular, if $G$ is abelian, this bundle is trivial, and sections of $\text{Aut}(P)$ are just maps $B \to G$.
Now, given two sections $\sigma_1$ and $\sigma_2$ of $P$, there is an automorphism $g$ of $P$ determined by sending $\sigma_1(x)g(x) = \sigma_2(x)$. Because both $\sigma_1$ and $\sigma_2$ are continuous sections, as is the inversion operation in $G$, so too is $g: B \to \text{Aut}(P)$ continuous. We see that $\Gamma(P)$ is acted on simply transitively by $\Gamma(\text{Aut}(P))$ - provided $\Gamma(P)$ is nonempty. When $G$ is abelian, $\Gamma(\text{Aut}(P)) = \text{Map}(M, G)$, as above.
Suppose you have an oriented vector bundle $E$ (coded by a map $f: X \to BSO(n)$), $\text{spin}^c$ structures on $E$ are the same thing as lifts to $\tilde f: X \to B\text{Spin}^c(n)$. This is the same as the space of sections of a principal $\Bbb{CP}^\infty$-bundle written $\text{Spin}^c(E)$, the bundle whose fiber above $x$ is the space of $\text{spin}^c$ structures on $E_x$ (this is the pullback to $X$ of the fiber sequence in the first line of this answer). What was written above is that picking an element of $\Gamma(\text{Spin}^c(E))$ gives a canonical bijection with $\text{Map}(X, \Bbb{CP}^\infty)$. (Of course, this is assuming there exists a $\text{spin}^c$ structure on $E$!)
Passing to homotopy classes, we see that the space of $\text{spin}^c$ structures on $E$ up to isomorphism is acted on simply transitively by $[X, \Bbb{CP}^\infty] = [X, K(\Bbb Z,2)] = H^2(X;\Bbb Z)$, assuming there was a $\text{spin}^c$ structure to begin with.
Another argument (essentially the same as this) runs through obstruction theory, which determines whether or not lifts exist and how many they are; the point being that $H^n(X;\pi_n F)$ is zero only for $n = 2$, where we get $H^2(X;\Bbb Z)$. The existence of a lift is determined by a class in $H^3(X;\Bbb Z)$, which turns out to be $\beta w_2(E)$, where $\beta$ is the Bockstein map.