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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.

The normal closure $⟨⟨B⟩{⟩}_G$ is the smallest normal subgroup of $G$ containing $B$, equivalently the intersection of all normal subgroups containing $B$. Any such normal subgroup contains every $g{b}^{\pm 1}{g}^{-1}$ and hence every finite product of these elements. Thus $N\subseteq ⟨⟨B⟩{⟩}_G$.

Include the empty product so that $1\in N$, in particular when $B=\varnothing$. Products of elements of $N$ are obtained by concatenating their factor lists. Inversion reverses the list and replaces ${ϵ}_i$ by $-{ϵ}_i$, so $N$ is closed under inverses. For any $k\in G$,Therefore $N$ is a normal subgroup and it contains $B$. Minimality of the normal closure gives the reverse inclusion, provingThe notation $n\in \mathbb{N}$ in a finite-product description is understood to allow the empty product; otherwise the empty generating subset would require a separate identity convention.

In a group presentation, $X$ is a set of formal generators and $R$ is a subset of the free group $F(X)$, consisting of relators which are required to become the identity. The presented group isThe universal property of a group presentation says that a function $X\to H$ extends to a group homomorphism from the presented group precisely when every relator evaluates to $1$ in $H$: first extend freely to $F(X)$, then factor through the quotient because its kernel contains the normal closure of $R$.

A rotation $\alpha$ by $2\pi /n$ and a reflection $\beta$ of a regular $n$-gon satisfyThe last identity holds because reflection reverses the sense of rotation, and it implies $\alpha \beta \alpha \beta =1$. Hence all defining relators vanish, giving a homomorphism $\theta :D_{2n}\to M_n$ with $\theta (\sigma )=\alpha$ and $\theta (\tau )=\beta$.

To prove it is an isomorphism without presupposing the order of the presented group, use $\sigma \tau =\tau {\sigma }^{-1}$. Move every occurrence of $\tau$ to the left, cancel pairs by ${\tau }^2=1$, and reduce exponents modulo $n$. Every element has one of the formsThus $|D_{2n}|\le 2n$. For the usual regular polygon with $n\ge 3$, the $n$ rotations ${\alpha }^j$ are distinct and the $n$ reflections $\beta {\alpha }^j$ are distinct; no rotation is a reflection, since their planar orientation behavior differs. The supplied description says these exhaust $M_n$, so $|M_n|=2n$, and every one lies in the image of $\theta$. Surjectivity and the order bound force $|D_{2n}|=2n$ and trivial kernel. Thus $\theta$ is an isomorphism.

For the requested Cayley graph, use right multiplication by the generators. The $\sigma$ edges are the two directed 3-cyclesThe $\tau$ edges are the undirected pairsThese pairs use ${\sigma }^j\tau =\tau {\sigma }^{-j}$. Thus the underlying graph is a triangular prism, with the two rotation cycles oriented oppositely when reflection pairs are drawn as radial edges.

A homotopy from $f_0:Y\to X$ to $f_1:Y\to X$ is a continuous map $H:Y\times [0,1]\to X$ with $H(y,0)=f_0(y)$ and $H(y,1)=f_1(y)$ for every $y$. A covering map $p:\tilde{X}\to X$ is a surjective continuous map for which every $x\in X$ has an open neighborhood $U$ whose inverse image is a disjoint union of open sets, each mapped homeomorphically onto $U$. A lift of $f:Y\to X$ is a continuous map $\tilde{f}:Y\to \tilde{X}$ satisfying $p\circ \tilde{f}=f$.

The path lifting theorem says that a path $\gamma :[0,1]\to X$ and a point ${\tilde{x}}_0\in p^{-1}(\gamma (0))$ determine a unique path $\tilde{\gamma }:[0,1]\to \tilde{X}$ with $p\tilde{\gamma }=\gamma$ and $\tilde{\gamma }(0)={\tilde{x}}_0$.

On $S^1\setminus \{1\}$ there is a continuous branch ${\mathrm{Arg}}_{(0,2\pi )}$ of the complex argument, taking values in $(0,2\pi )$. If $f$ avoids $1$, set $\theta (z)={\mathrm{Arg}}_{(0,2\pi )}f(z)$. Then the explicit homotopystarts at $f$ and ends at the constant map $-1$. It is continuous because the chosen argument branch is continuous on the whole image. More generally, a map omitting any $w\in S^1$ is null-homotopic: apply the formula to $\overline{w}f$ and multiply the resulting homotopy by $w$.

For the based injective map, use the standard covering map $E(t)=e^{2\pi it}$. Lift the path $t\mapsto g(E(t))$ on $[0,1]$ to a real-valued path $G$ with $G(0)=0$. Since $g(1)=1$, its endpoint $d=G(1)$ is an integer. On $[0,1)$, $G$ is injective: equality $G(s)=G(t)$ implies $g(E(s))=g(E(t))$, then $E(s)=E(t)$ by injectivity, and hence $s=t$ in this half-open interval. A continuous injective real-valued function on an interval is strictly monotone.

If $G$ is increasing, $d>0$. It cannot be at least $2$, because the intermediate value theorem would produce $t\in (0,1)$ with $G(t)=1$, giving $g(E(t))=g(1)$ while $E(t)\ne 1$. Thus $d=1$. If $G$ is decreasing, the same reasoning gives $d=-1$. Now defineAt $t=0$ and $t=1$ this has the same value $1$ for every $s$, since $d$ is an integer. Therefore the formula descends continuously from $[0,1]\times [0,1]$ to $S^1\times [0,1]$. At $s=0$ it is $g$; at $s=1$ it is $z\mapsto z^d$. Hence $g$ is homotopic to the identity or to complex conjugation, with degree $1$ or $-1$. This is the key property of an injective self-map of a circle.

Finally, let $h$ be any continuous injective circle self-map. Normalize it to $g(z)=\overline{h(1)}h(z)$, which is injective and fixes $1$. If $h$ were not surjective, it would omit some circle point, so the explicit missing-point construction would make $h$ null-homotopic. Multiplying this homotopy by $\overline{h(1)}$ would make $g$ null-homotopic too. But $g$ is homotopic to the identity or conjugation, neither of which is null-homotopic by the permitted fact. This contradiction proves every continuous injective circle self-map is surjective. The lifting argument did not assume surjectivity, so the reasoning is not circular.