$C^\infty$ involution of Manifold and surjective local diffeomorphism give rise to an isomorphism? this is a question from a textbook that I was looking at volume 1 of Werner Greub's "Connections, Curvature and Cohomology", where I found the following problem. 
Let $M$ be a manifold and $A:M \to M$ a $C^\infty$ involution such that $A^2(x) = x$ for all points in $M$. Let $\pi : M \to N$ be a surjective local diffeomorphism between $M$ and another manifold $N$ such that for each $y \in N$, there is some $x\in M$ such that $\pi^{-1}(y)=\{x,A(x)\}$. We want to show that the pullback $\pi^*: \Omega^*(N) \to \Omega^*_+(M)$ is an isomorphism, where:
$$\Omega^*_+(M)=  \{ \omega \in \Omega^*(M)\: |\: A^*\omega = +\omega\}$$
Now, the first part of the setup required is trivial, one can easily show that:
$$\Omega^*(M) = \Omega^*_-(M) \oplus \Omega^*_+(M) $$
And that $\pi^*$ is injective, by noting that $A^* \circ \pi^* = \pi^*$, so:
$$\ker(\pi^*) = \{\tau \in \pi^*(\Omega^*(N)) \subseteq \Omega^*_+(M)\: |\: A^*(\omega) = 0  \}  = \{0\}$$
But, showing it is surjective has me a bit lost. Locally, we know that $\pi_*$ is an isomorphism, but I don't really know how to proceed from this to being able to write a form on $\Omega^*_+(M)$ as the pullback of a form in $\Omega^*(N)$. Thanks.
 A: Rather than  jumping at solving the problem, let us first understand what it says. We shall do it with the aid of an example. Let $M = S^n \subset \mathbb R^{n+1}$, the sphere of center $0$ and radius $1$. Let $A : S^n \to S^n$ be $A(x) = -x$. Let $N = \mathbb P^n$, the real projective space of dimension $n$. Remember that it is defined as $S^n / \sim$, where $x \sim y \iff y = -x$. It is natural, then, to take $\pi : M \to N$ to be $\pi (x) = \{x, -x\}$. If $V_p \subset N$ is a small neighbourhood of $p \in N$, and if $x \in M$ is such that $\pi (x) = p$, there exists a small neighbourhood $U_x$ of $x$ that is diffeomorphic to $V_p$ through $\pi \big| _{U_x}$. Furthermore, $A(x) = -x$ is the other point the image of which is $p$, and $U_{-x} = -U_x = \{ -y \mid y \in U_x \}$ is a neighbourhood of it, diffeomorphic to $V_p$ through $\pi \big| _{U_{-x}}$. In other words, $\pi ^{-1} (V_p) = U_x \cup U_{-x}$, and $U_x \cap U_{-x} = \emptyset$, so $V_p$ is "covered" by two disjoint "sheets", each being diffeomorphic to $V_p$.

Returning to your problem, let us first show that $\pi \circ A = \pi$. If $x \in M$ and $p = \pi (x)$ then, according to the hypotheses, $\pi ^{-1} (p) = \{ x, A(x) \}$, whence $\pi (A(x)) = p = \pi (x)$, so $\pi \circ A = \pi$. It follows that $\pi ^* = (\pi \circ A) ^* = A^* \circ \pi ^*$. This tells us that if $\omega \in \Omega(N)$ then $A^* (\pi ^* \omega) = \pi ^* \omega$, so $\pi^* (\Omega (N)) \subseteq \Omega _+ (M)$. (This was a simple but necessary step, because in principle we only know that $\pi^* (\Omega (N)) \subseteq \Omega (M)$.)
Next, let us show injectivity: if $\omega \in \Omega (N)$ with $\pi^* \omega = 0$, and if $x \in M$, then let $U_x \subset M$ be some open neighbourhood of $x$ such that $V_p = \pi (U_x)$ is an open neighbourhood of $p = \pi (x) \in N$ and is diffeomorphic to $U_x$ (this exists because $\pi$ is a local diffeomorphism). If $\pi^* \omega = 0$, then in particular $0 = (\pi^* \omega) \big| _{U_x} = (\pi \big| _{U_x})^* (\omega \big| _{V_p})$, and since $\pi \big| _{U_x}$ is a diffeomorphism onto $V_p$ we deduce that $\omega \big| _{V_p} = 0$. Since the restriction of $\omega$ around every point of $N$ is $0$, it follows that $\omega = 0$.
Finally, surjectivity. Let $\eta \in \Omega_+ (M)$; we want to construct $\omega \in \Omega (N)$ such that $\pi^* \omega = \eta$. Pick $p \in N$ and let $x \in \pi^{-1} (p)$. Let $U_x$ and $V_p$ be diffeomorphic neighbourhoods, like above. Remember that unlike the pull-back, the push-forward cannot be defined for arbitrary maps, but only for diffeomorphisms. Fortunately, $\pi \big| _{U_x}$ is a diffeomorphism onto $V_p$, so we may speak about the push-forward of $\eta \big| _{U_x}$ through $\pi \big| _{U_x}$ - call it $\omega_p$. Clearly, $\omega_p$ is a form on $V_p$ such that $(\pi \big| _{U_x}) ^* \omega_p = \eta \big| _{U_x}$, by the very definition of $\omega_p$. The problem is that, so far, $\omega_p$ only lives on $V_p$, whereas we need a global $\omega$, living on $N$. We would like to "patch together" all these $\omega_p$ into a single object, so let us see if we may do it.
Let $q \in N$ such that $V_p \cap V_q \ne \emptyset$. Let $y \in \pi^{-1} (q)$, let $U_y$ be the open neighbourhood of $y$ diffeomorphic to $V_q$. On $V_p \cap V_q$ we have two local forms: $\omega_p$ (the local push-forward of $\omega \big| _{U_x}$) and $\omega_q$ (the local push-forward of $\omega \big| _{U_y}$). If they are equal on $V_p \cap V_q$, we are done. There are two possibilities:


*

*if $U_x \cap U_y \ne \emptyset$, then we have $[ \eta \big| _{U_x} ] \big| _{U_x \cap U_y} = \eta \big| _{U_x \cap U_y} = [ \eta \big| _{U_y} ] \big| _{U_x \cap U_y}$, so these two forms coincide on $U_x \cap U_y$, and so will their respective push-forwards $\omega_p$ and $\omega_q$ on $V_p \cap V_q$;

*if $U_x \cap U_y = \emptyset$, then $A(U_x) \cap U_y \ne \emptyset$; we then have $[ \eta \big| _{A (U_x)} ] \big| _{A (U_x) \cap U_y} = \eta \big| _{A (U_x) \cap U_y} = [ \eta \big| _{U_y} ] \big| _{A (U_x) \cap U_y}$, so these two forms coincide on $A(U_x) \cap U_y$, and so will their push-forwards on $V_p \cap V_q$ (because $V_p = \pi (U_x) = \pi (A (U_x))$, because $\pi \circ A = \pi$, as already shown above); but, remembering that $A^* \eta = \eta$, we have $\eta \big| _{A(U_x)} = (A^* \eta) \big| _{U_x} = \eta \big| _{U_x}$, so the push-forward of $\eta \big| _{A(U_x)}$ is the same as the push-forward $\omega_p$ of $\eta \big| _{U_x}$, hence $\omega_p$ and $\omega_q$ coincide on $V_p \cap V_q$.
In both cases, we have deduced that $\omega_p = \omega_q$ on $V_p \cap V_q$. Things are easy now: on every $V_p$ define $\omega$ to be equal to $\omega_p$. If $V_p \cap V_q \ne \emptyset$, we have just shown that $\omega_p = \omega_q$ on $V_p \cap V_q$, so $\omega$ is well-defined.
(Notice that all our arguments were local, i.e. done on some small neighbourhood around every point, despite the conclusion being global. This is a proof technique to keep in mind in differential geometry.)
