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Take $G={\mathbb{C}}^{\ast }$ and the complete analytic function generated by a logarithm germ. Its continuation surface identifies with $\mathbb{C}$ via $w\mapsto [\log {]}_z$ with $z=e^w$ and logarithm value $w$. Thus $\pi (w)=e^w$. It is a universal cover; the deck transformations $w\mapsto w+2\pi in$ act transitively on each fibre, so it is also a regular covering map in the deck-transitive sense. It is not bijective, since every nonzero $z$ has infinitely many logarithm values. It also meets the alternative convention in which “regular” merely requires an evenly covered neighbourhood at each base point.

Two holomorphic function elements represent the same germ of a holomorphic function at $z$ if they agree on some neighbourhood of $z$. For a holomorphic $f$ on an open $D\subseteq G$, set $U(f,D)=\{[f{]}_z:z\in D\}$. These sets form a basis of a topology on the space of germs of holomorphic functions: every germ lies in one, and if a germ lies in two, equality of the representatives on a smaller neighbourhood supplies a basis of a topology set in their intersection. Declaring arbitrary unions of basis of a topology sets open therefore gives a topology.

Different base points are separated by disjoint small discs in $G$. If two different germs have the same base point, choose representatives on one connected small disc. Their sheets are disjoint: any common germ elsewhere on that disc would force equality throughout the disc by the identity theorem. This proves the Hausdorff property.

The projection $\pi ([f{]}_z)=z$ maps each sheet $U(f,D)$ homeomorphically onto $D$. Use these maps as complex charts. On overlaps their transition functions are the identity, hence holomorphic; thus each connected component is a Riemann surface, and $\pi$ is a local biholomorphism, an analytic germ projection. For completeness, second countability of a component follows by following continuation chains through a countable basis of a topology of rational discs: starting at one germ, finite chains of such discs give only countably many continuation sheets, by uniqueness on overlaps, and they cover every germ in that component.

A complete analytic function consists of all germs reachable from one initial germ by analytic continuation. The supplied path-lifting equivalence identifies these with its path component. Since the germ space is locally path connected, connected components and path components agree. Thus its continuation surface is exactly one component ${\mathcal{G}}_{\mathcal{F}}$.

A local biholomorphism need not evenly cover the whole base; the germ projection is an analytic covering in this local sense. The examples below distinguish a genuine regular covering map from a surjective nonregular projection, with the meaning of regular stated explicitly.

A covering evenly covers every base point. It is regular covering map when deck transformations act transitively on fibres. For a concrete nonregular example, restrict $z\mapsto z^3-3z$ toIt is a surjective three-sheeted unbranched cover with monodromy $S_3$, so its deck group is not transitive.

A deck transformation of a covering $p:\tilde{X}\to X$ is a homeomorphism $d:\tilde{X}\to \tilde{X}$ with $p\circ d=p$. A connected covering is a regular covering map if its deck transformation group acts transitively on each fiber; for a path-connected base, transitivity on the fiber over the chosen basepoint is equivalent.

Write $H=p_{\ast }{\pi }_1(\tilde{X},\tilde{b})$ and $G={\pi }_1(X,b)$. A based loop represents an element of $H$ exactly when its lift starting at $\tilde{b}$ is closed. Indeed a closed lift itself projects to a representative of an image element; conversely the image of a closed path lifts to that path, and the closed-endpoint property is unchanged by based homotopy.

Suppose the covering is regular, let $h\in H$, and let $g\in G$. Lift a representative of $g$ from $\tilde{b}$ to an endpoint $v$. By regularity some deck transformation sends $\tilde{b}$ to $v$. Applying it to the closed lift of $h$ gives a closed lift of $h$ at $v$, by uniqueness in the path lifting theorem. The concatenation of the lift of $g$, that closed lift of $h$, and the reverse lift of $g$ is therefore a closed lift of $gh{g}^{-1}$. Thus $gH{g}^{-1}\subseteq H$; applying the same argument to $g^{-1}$ gives equality. Hence a regular covering has a normal image subgroup.

For the explicit cover, label the bottom vertex $B=\tilde{b}$, the top $T$, the left $L$, the right $C=c$, and the rear/front equatorial vertices $U,V$, respectively. Reading the arrows gives the complete monodromy table:

Thus the $x$ edges form the cycle $B,V,L,T,C,U,B$, and the $y$ edges form the cycle $B,C,V,T,U,L,B$. In particular every vertex has one incoming and one outgoing edge of each label, as a covering of the two-circle wedge requires. All word paths below are traversed from left to right.

Choose the spanning tree consisting of the first five $x$ edges from $B$ to $U$. The tree paths to $B,V,L,T,C,U$ have labels $1,x,x^2,x^3,x^4,x^5$. For every directed non-tree edge $v\stackrel{z}{\to }w$, the path from $B$ along the tree to $v$, then along the edge, then back along the tree from $w$, has label $t_{v}z{t}_w^{-1}$. Collapsing a spanning tree turns a connected graph into a wedge of circles, so these paths form a free basis of its fundamental group. There are seven non-tree edges, since $12-6+1=7$. Their projected labels give the explicit generating setThe covering map induces an injection on fundamental groups, so these seven generators are also a free basis for the image subgroup. The first comes from the remaining $x$ edge $U\to B$; the other six come from the outgoing $y$ edge at each vertex in the table.

Now the word $h=y{x}^2$ lifts at $B$ around the triangleso $h\in H$. Starting at $C$, however, its lift iswhich is not closed. A deck transformation sending $B$ to $C$ would take the first closed path to a closed lift of the same word at $C$, contradicting uniqueness of lifts and the displayed nonclosed path. No such deck transformation exists. This is also the closed-lift obstruction expressed by the deck transformation group as a monodromy centralizer.

For a connected covering of a connected, locally path-connected, semilocally simply connected space, the classification of connected covering spaces says that the image subgroup is normal if and only if the covering is regular. The two-circle wedge and this connected finite graph satisfy those hypotheses. Since regularity fails, $H$ is not normal in the base group ${\pi }_1(X,b)\cong F(x,y)$. All normality comparisons here are in that base group.

Finally takeThe lift of $gh{g}^{-1}=y^2{x}^2{y}^{-1}$ from $B$ followsIt ends at $V\ne B$, so $gh{g}^{-1}\notin H$. This is the required explicit witness, and independently proves the failure of normality.