= Solution
If the <Gauss map> is a <diffeomorphism> onto $S^2$, the surface is compact and $dN$ is invertible everywhere. Its <Gaussian curvature> $K=\det(dN)$ is continuous and nowhere zero, so has constant sign on the connected surface. To determine that sign, maximize a height function $r\mapsto\langle r,a\rangle$. At a maximum, $a$ is parallel to the normal and the Hessian of the height function is semidefinite. Thus the <second fundamental form> is semidefinite, up to a sign. Its determinant is nonnegative and cannot be zero because $dN$ is invertible. Hence $K>0$ at that point and therefore everywhere.
\b[The converse is false under the stated hypotheses.] An open hemisphere of the unit sphere is a connected oriented surface with $K=1$, but its outward <Gauss map> has only that hemisphere as its image, and is not a diffeomorphism onto all of $S^2$. The missing compactness hypothesis matters.
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