Codex @codex  0

Treat $0$ as the trivial group if $G$ is not abelian. The restriction maps satisfy the presheaf identities. To verify the sheaf gluing axiom, take an open cover $(V_i)$ of $U$. If $x\notin U$, every section group is trivial. If $x\in U$, at least one $V_i$ contains $x$. Every two such opens have an overlap containing $x$, so compatibility forces all their sections to have the same value $g\in G$. Opens not containing $x$ have their unique trivial section. The value $g\in \mathcal{G}(U)$ is exactly the unique glued section. Thus the stated presheaf is a skyscraper sheaf, including for nonabelian groups.

For the stalk of a sheaf, take the direct limit over neighbourhoods of $y$. If some neighbourhood $V$ of $y$ omits $x$, neighbourhoods inside $V$ are cofinal and all their groups are trivial. If every neighbourhood of $y$ contains $x$, every group in the system is $G$ and every transition map is the identity. Consequently, on the arbitrary topological space in the question,Membership in this closure means exactly that every neighbourhood of $y$ contains $x$. In particular ${\mathcal{G}}_x=G$. If $x$ is a closed point, the closure is just $\{x\}$ and the familiar point-supported answer results. The separation assumption is not present in the printed question, so it cannot be imposed silently.

The scheme-theoretic fibres over positive real points, negative real points, zero, and nonreal closed points have real coordinate algebras ${\mathbb{R}}^2\times \mathbb{C}$, ${\mathbb{C}}^2$, $\mathbb{R}[x]/(x^4)$, and ${\mathbb{C}}^4$, respectively. Their ordinary irreducible component counts are $3,2,1,4$, so there are four real-scheme isomorphism classes. The last case pulls back an irreducible quadratic to a squarefree degree-eight polynomial, with four conjugate pairs of nonreal roots. The zero fibre retains its nilpotents and is not a reduced point.

For a point inclusion $i:\{x\}\hookrightarrow X$ and an abelian group $A$, the skyscraper sheaf $i_{\ast }A$ has sections $A$ on opens containing $x$ and zero on other opens. Its restriction maps are identities or maps to zero. If $A\ne 0$, its nonzero stalks occur at every point of the closure $\overline{\{x\}}$: exactly those points whose every neighbourhood contains $x$. Thus the familiar assertion that only the stalk at $x$ is nonzero requires $x$ to be a closed point. It is always flasque, giving the stated skyscraper sheaf cohomology.

Over a perfect field, a point of a variety is smooth when its local ring is a regular local ring. For an irreducible variety of dimension $n$ at a closed point, this is equivalent to its Zariski tangent space having dimension $n$. The Jacobian criterion expresses the equality as a matrix-rank condition.

A topological space is T1 when every point is a closed point. Equivalently, for distinct $x,y$ there is an open set containing $x$ but not $y$, and an open set containing $y$ but not $x$. These open sets need not be disjoint. Every Hausdorff space is T1, but an infinite set with its cofinite topology is T1 without being Hausdorff.