Where is a closed form also exact? I'm not very familiar with exterior derivatives. I've some trouble following argument (which is a part of a proof that if the Riemann tensor vanishes, $R^{\,\rho}_{\;\,\sigma \mu \nu}=0$, iff there exists a coordinate system where the components of the metric are constant.)
One finally get to the equation:
$$\nabla_\mu \omega_\nu =0$$
Taking the anti-symmetrization of the last equation:
$$\nabla_{[\mu} \omega_{\nu]} =\text{d}\omega=0 $$
Which means that $\omega$ is closed. However in general this does not mean that $\omega$ is exact, i.e. $\omega=\text{d}\alpha$, for some scalar function $\alpha$. Then 'since we have restricted the topology of the region in which we are working' the one-form must also be exact. 
I don't see how the vanishing of the Riemann tensor directly leads to the bold-texted conclusion. More generally in what space is a closed form also always exact ? 
 A: I) When Ref. 1 writes 

[...] since we have restricted the topology of the region in which we are working,

it is referring to a previous comment: 

Technically, these statements should be restricted to a region of the manifold that is simply-connected (all loops in the region can be smoothly deformed to a point without leaving the region); we will implicitly assume this condition below.

II) Ref. 1 is apparently alluding to the Poincaré lemma in the text. The Poincaré lemma comes in various versions and generalizations, e.g., 


*

*In a contractible manifold, every closed form is exact (except for zero-forms).

*In a star-shaped neighborhood, every closed form is exact (except for zero-forms).

*In a sufficiently small neighborhood, every closed form is exact (except for zero-forms).

*In a connected and simply-connected manifold, every closed one-form is exact.

*In a manifold $M$ where the homotopy groups $\pi_0(M)=\pi_1(M)=\ldots=\pi_r(M)=0$ vanish, every closed $r$-form is exact.   
Specifically, Ref. 1. is using version 4. 
III) Example: Magnetic monopole. Let the manifold be $M=\mathbb{R}^3\backslash\{0\}$, which is connected and simply-connected, but not contractible. The two-form
$$B~:=~\sum_{i,j,k\in\{1,2,3\}}\epsilon_{ijk}\frac{x^i}{r^3}\mathrm{d}x^j\wedge \mathrm{d}x^k, \qquad r~:=~\sqrt{\sum_{i\in\{1,2,3\}}(x^i)^2}~>~0, $$
of the dualized magnetic field is a closed two-form but not an exact two-form on $M$. (The magnetic potential one-form $A$ cannot be defined on a so-called Dirac-string.)
References:


*

*S. Carroll, Spacetime and Geometry, Section 3.6, page 124-125.

