 Solution

The Exponential map of a Lie group sends $X\in \mathfrak{o}(3)$ to $e^X\in SO(3)$ and is a local diffeomorphism at zero. Every element of $SO(3)$ is an exponential of a skew-symmetric matrix, but no determinant-$-1$ element of $O(3)$ is an exponential because $\det e^X=e^{\mathrm{tr}X}=1$.

A representation $d:\mathfrak{o}(3)\to \mathfrak{gl}(V)$ always integrates uniquely to the simply connected covering group $\mathrm{Spin}(3)\simeq SU(2)$. It descends to $SO(3)$ exactly when the nontrivial element in the kernel of $SU(2)\to SO(3)$ acts trivially. Extending it further to disconnected $O(3)$ requires an additional parity operator compatible with conjugation by a reflection. Thus the Lie-algebra representation alone need not define a representation of all of $O(3)$.