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For a connected smooth manifold, every component of the orientation double cover maps onto the base by the path lifting theorem. There can therefore be at most two components. A global orientation splits the cover into two copies of the base. Conversely, if there are two components, each meets every fibre once, so its inverse projection gives a smooth global choice of orientation.

For a covering map with nonempty connected base, the degree of a covering map is the common cardinality of its fibres. Each evenly covered neighbourhood makes this number locally constant, so connectedness makes it constant. The total covering space need not be connected. On a path-connected base the path lifting theorem gives a fibre bijection by path lifting. A disconnected base may have different degrees on different connected components.

The fundamental group ${\pi }_1(X,x_0)$ consists of based loops modulo based homotopy. Multiplication follows one based loop and then the other; the constant based loop represents the identity, and path reversal gives the inverse. Passing to homotopy classes makes this operation associative, as explained by well-defined loop concatenation.

Identify the circle with $S^1=\{z\in \mathbb{C}:|z|=1\}$ and choose basepoint $1$. Its based loopshave winding number $n$: positive $n$ counts counterclockwise turns, negative $n$ counts clockwise turns, and $n=0$ gives the constant based loop. Every based loop on the circle is joined by a based homotopy to exactly one ${\gamma }_n$. ConsequentlyThis is Hatcher's Theorem 1.7 in Algebraic Topology, §1.1, pp. 29–30. For example, two counterclockwise turns followed by one clockwise turn represent $2+(-1)=1$, rather than three turns. The winding graph in the figure above makes the signed arithmetic visible.

A proof outline uses the covering map $q:\mathbb{R}\to S^1$, $q(u)=e^{2\pi iu}$. By the path lifting theorem, a based loop has a unique lift starting at zero and ending at an integer $n$. Interpolating its lift to $s\mapsto ns$ produces a based homotopy to ${\gamma }_n$. The homotopy lifting theorem for a covering map keeps that endpoint unchanged during a based homotopy, proving uniqueness.

In the filled disk, every based loop $\gamma$ at $x_0$ has a based homotopy to the constant based loop:Both endpoints remain $x_0$, and the convex set property of the disk keeps every intermediate point inside. Thus ${\pi }_1(D^2,x_0)$ is the trivial group. Together with the deformation retraction above, this computes three concrete answers: the filled disk has trivial fundamental group, while the circle and the annulus have infinite cyclic groups.

Maps have computable algebraic shadows too. The continuous map $p_n:S^1\to S^1$, $p_n(z)=z^n$, fixes $1$ and sends ${\gamma }_k$ to ${\gamma }_{nk}$. Its induced group homomorphism is therefore $k\mapsto nk$, and its mapping degree is $n$. For $n=2$ the image consists of even integers, so this map cannot be a homotopy equivalence. Composing $p_2$ with $p_{-3}$ gives $p_{-6}$: map composition multiplies degrees, whereas concatenating based loops adds winding numbers. Distinct degrees distinguish these maps up to homotopy; this calculation does not classify arbitrary topological spaces.

Historically, Poincaré's 1895 Analysis Situs introduced the fundamental group; its original §12 explicitly names it. The CNRS transcription of Poincaré's §12, Groupe fondamental provides a direct view of this early formulation.

A covering map over a topological graph lifts its vertices to all their preimages and each open edge to disjoint open edges. Their endpoints are supplied by the path lifting theorem applied to each parametrizing path of a closed edge. The covering sheets over small vertex stars verify that the resulting graph topology agrees with the original topology upstairs. A constant finite degree of a covering map $d$ therefore multiplies both vertex and edge counts by $d$, proving the Euler characteristic under a finite covering formula for finite topological graphs.

A homotopy into the base of a covering space has a unique continuous lift once its initial map is continuously lifted. Local inverse sheets construct the lift along each time interval, and compactness of that interval permits a finite subdivision locally around each point of the homotopy's parameter space. Uniqueness on overlapping intervals follows from path lifting theorem uniqueness. In a homotopy of paths with endpoints fixed, a chosen lifted endpoint remains fixed because it varies continuously in a discrete fiber.

For a covering map $p:X\to Y$ and base point $y$, the path lifting theorem lifts each based loop from each point of the fibre $p^{-1}(y)$. Its endpoint defines a permutation of that fibre; path reversal gives the inverse. The homotopy lifting theorem for a covering map makes the permutation depend only on the based homotopy class. These permutations form the monodromy group, the image of the fundamental group actionThe image is independent of the convention for composing routes; one may reverse the loop classes to obtain a left group action when route multiplication means following the first route and then the second.

For $p(w)=w^2$ on the punctured plane, the fibre over $1$ is $\{1,-1\}$, and a once-around based loop exchanges its two points. The monodromy group has two elements: only the parity of the winding number matters. For the covering map $p(w)=e^w$ from the complex plane to the punctured plane, the fibre over $1$ is $2\pi i\mathbb{Z}$, and a positive turn acts by $2\pi ik\mapsto 2\pi i(k+1)$. Its monodromy group is an infinite cyclic group. These are the geometric versions of logarithm and square-root continuation along different paths: the base point returns, while its lifted endpoint can move to another sheet.

A connected smooth manifold is path connected because it is locally path connected. Each connected component $C$ of the orientation double cover is itself path connected. Choose a point of $C$ over $p$. For any $q\in M$, lift a path from $p$ to $q$ starting at that chosen point, using the path lifting theorem. The lifted path lies in $C$ and ends above $q$. Therefore every component maps onto $M$.

Fix one fibre, containing exactly two points. Every component meets that fibre, and distinct components meet disjoint subsets of it, so there are at most two connected components.

If $M$ is orientable, a global continuous choice of tangent-space orientation $o$ gives two disjoint sheetsThey are open: on a sufficiently small oriented coordinate neighborhood the global choice agrees consistently with one of the two local sheets. Each projects homeomorphically to the connected base, so these are precisely two connected components.

Conversely suppose the cover is disconnected. It has exactly two components $C_{+},C_{-}$. Both project onto the base and each fibre contains only two points, so each component meets every fibre exactly once. The restriction $\pi :C_{+}\to M$ is consequently a bijective local diffeomorphism, hence a global diffeomorphism. Transporting the canonical orientation of $C_{+}$ through this map or, equivalently, using its smooth inverse section, gives a global orientation of $M$. Thus the connectedness criterion for the orientation double cover is

Take a homotopy equivalence $f:X\to Y$ with inverse $g$, and universal covering maps $p_X,p_Y$. Since the covering spaces are simply connected, the path lifting theorem constructs lifts $\tilde{f}$ of $f{p}_X$ and $\tilde{g}$ of $g{p}_Y$, after specifying one value in the appropriate fiber. Explicitly, define the lift at a point by lifting the image of a path from the chosen base point; independence from the path follows from simple connectedness and homotopy lifting.

Lift a homotopy $gf\simeq 1_X$ starting at $\tilde{g}\tilde{f}$. The homotopy lifting theorem for a covering map makes its endpoint a map $d_X:\tilde{X}\to \tilde{X}$ with $p_X{d}_X=p_X$. Such a lift of the identity is a deck transformation and is a homeomorphism: lift the identity again with the base point image reversed, and uniqueness of path lifts makes the two compositions identities. Likewise $\tilde{f}\tilde{g}\simeq d_Y$ for a deck transformation of $\tilde{Y}$.

Thus both compositions are homotopy equivalences, even though the lifted homotopies need not end at the particular identity lift. Part (i) then provesKeeping the possible deck transformations is essential when the original homotopies do not fix base points.

The path lifting theorem says that a path $\gamma :[0,1]\to X$ has a unique lift through a covering map once its initial point in the fiber is chosen. The homotopy lifting lemma says that a homotopy $H:Z\times [0,1]\to X$, with a continuous lift specified at time zero, has a unique continuous lifted homotopy. In particular a homotopy of paths with fixed endpoints preserves the endpoint of a chosen lift, since that endpoint varies continuously in a discrete fiber.

Choose $x_0\in X$ and $y_i\in p_i^{-1}(x_0)$. For $y\in Y_1$, choose a path $\gamma$ in $Y_1$ from $y_1$ to $y$, and let $F(y)$ be the endpoint of the lift of $p_1\gamma$ to $Y_2$ starting at $y_2$. Two choices of $\gamma$ are homotopic relative endpoints because $Y_1$ is simply connected. Project that homotopy and apply homotopy lifting: the lifted endpoint remains fixed. Thus $F$ is well-defined and $p_{2}F=p_1$.

To prove continuity without invoking covering-space classification, choose a path-connected open neighborhood $U$ of $p_1(y)$ evenly covered by both maps. Local path connectedness of $X$ permits this choice. On the sheet of $p_1^{-1}(U)$ containing $y$, append paths within that sheet to the chosen path to $y$. Path-lift uniqueness givesthere, for the sheet $V_2$ containing $F(y)$. Hence $F$ is a local homeomorphism and is continuous. Construct $G:Y_2\to Y_1$ in the same way with the basepoints reversed. Lifting the very same projected paths proves $GF$ and $FG$ fix every point. Therefore

A topological graph is a CW complex with only zero-dimensional and one-dimensional cells: its vertices form the zero-skeleton, and each edge is a copy of an open interval attached by its two endpoints to vertices, possibly the same vertex. It has the weak topology of a CW complex. A subgraph is a CW subcomplex, obtained by choosing vertices and edges and retaining the endpoints of every chosen edge. Loops and multiple edges are allowed.

A spanning tree is a connected spanning subgraph containing no cycle in a graph. To construct one in any connected topological graph $\Gamma$, consider all acyclic spanning subgraphs, ordered by inclusion; the subgraph with every vertex and no edges is one. The union of a chain is still acyclic, since any cycle uses only finitely many edges and would already occur in a member of the chain. The Zorn lemma therefore gives a maximal such subgraph $\Delta$.

If $\Delta$ were disconnected, a finite edge path in $\Gamma$ between vertices of two of its components would contain an edge joining different components. Adding that edge creates no cycle, contradicting maximality. Thus $\Delta$ is a spanning tree. Fix a root vertex. Each point of $\Delta$ has a unique finite path to the root; give every edge length one and move the point a fraction $s$ of the way along that path at time $s\in [0,1]$. This defines a contraction of a topological space of $\Delta$. On each closed edge times $[0,1]$, the construction takes place in a finite tree and is continuous. Since the interval is a locally finite CW complex, the product with $\Delta$ has the corresponding weak topology of a CW complex, so these restrictions establish continuity globally. ConsequentlyThis argument includes infinite graphs and does not assume local finiteness of $\Gamma$.

If $p:\tilde{\Gamma }\to \Gamma$ is a covering map, take $p^{-1}({\Gamma }^0)$ as the vertices upstairs. Each component of the inverse image of an open edge is an open interval mapping homeomorphically to that edge, because an interval is simply connected. Its parametrization lifts along the closed edge to attach its endpoints to lifted vertices, by path lifting theorem. Small star neighborhoods at vertices are lifted homeomorphically by the covering map, and give exactly the neighborhoods of the resulting topological graph. Thus the cell structure has the topology of the original covering space. The total space may be disconnected, but every component is a topological graph.

The Nielsen–Schreier formula isFor a proof, realize $F_n$ as the fundamental group of the wedge sum of $n$ circles. Construct its connected covering graph with vertices the right cosets $Hg$, and an oriented edge labeled $x_i$ from $Hg$ to $Hg{x}_i$ for each free generator $x_i$. There is exactly one incoming and one outgoing edge of each label at every vertex, which establishes the local covering map property. Reading a word from $H$ ends at its coset, so the closed paths at $H$ give exactly the words in $H$. The induced map of fundamental groups is injective by the homotopy lifting property of a covering map, and its image is $H$.

For index $d$, this covering graph has $d$ vertices and $dn$ edges. A spanning tree in it has $d-1$ edges. The permitted rank formula for a finite connected graph now givesIn particular this proves both freeness and the index formula.

If $n\ge 2$ and $d>1$, then $1+d(n-1)>n$, so a proper finite-index subgroup cannot be isomorphic to $F_n$. For $n=1$, $F_1\cong \mathbb{Z}$, and $d\mathbb{Z}$ is a proper index-$d$ subgroup isomorphic to $\mathbb{Z}$ for every $d\ge 2$. Hence

A local isometry here is a local diffeomorphism whose pullback metric is the source metric. Use the usual convention that a covering map is surjective. Assume the target is connected and the source is nonempty; this is the connected-manifold setting of the intended assertion. For a disconnected target the precise result is a covering of the union of target components met by the map. For example, inclusion $\mathbb{R}\hookrightarrow \mathbb{R}\bigsqcup \mathbb{R}$ is a local isometry from a complete manifold but is not a surjective covering of the whole target.

The correct Hopf-Rinow theorem says that on a connected finite-dimensional Riemannian manifold, completeness for the induced distance is equivalent to geodesic completeness, and to compactness of every closed bounded subset. These conditions imply that every two points are joined by a minimizing geodesic. The last property by itself does not imply completeness: an open Euclidean ball has minimizing segments between every pair but is incomplete.

A local isometry preserves the Levi-Civita connection, by uniqueness of the torsion-free metric connection on each local isometry chart. Hence it preserves affinely parametrized geodesics, andwhenever the source geodesic is defined. This is geodesic preservation by a local isometry.

Suppose $M$ is complete. Given $y\in N$, join $f(x)$ to $y$ by a minimizing target geodesic using target completeness. The source geodesic whose initial velocity is its inverse image under $d{f}_x$ exists for its full parameter interval by source completeness; its image is the target geodesic by uniqueness. Thus $f$ is onto.

Choose a normal ball $U={\exp }_y(B_r(0))$ with ${\exp }_y$ a diffeomorphism on $B_r(0)\subset T_{y}N$. For every $x\in f^{-1}(y)$ put $U_x={\exp }_x(B_r(0))$. Source completeness defines these exponential maps throughout the tangent balls. The displayed exponential identity makes $f:U_x\to U$ a diffeomorphism: injectivity follows from injectivity of the target exponential map, and differentiating the identity shows the source exponential has invertible derivative on the ball, so its image is open.

These sheets are disjoint. If $z$ lay in two, lift the reversed radial target geodesic from $f(z)$ to $y$, starting at $z$. Its initial velocity is uniquely fixed by $d{f}_z$; uniqueness of the source geodesic forces the two sheet centres to be equal. They exhaust $f^{-1}(U)$: for any point over $U$, lift that reversed radial geodesic using source completeness, obtaining a centre over $y$, and then reverse it. Thus $U$ is evenly covered. This proves the complete local isometry is a covering assertion.

Conversely suppose $f$ is a covering. Given any initial tangent vector in $M$, the corresponding target geodesic extends over all $\mathbb{R}$ by target completeness. The path lifting theorem for coverings lifts it over every compact time interval, with uniqueness gluing the lifts to a curve on $\mathbb{R}$. In local covering charts the lift is a geodesic, because the charts are local isometries. It extends the source initial-value geodesic. Therefore $M$ is geodesically complete, and Hopf-Rinow yields metric completeness on each source component. In the stated connected-target setting,

A function element is a pair $(f,D)$ consisting of a domain $D\subseteq G$ and a holomorphic function $f:D\to \mathbb{C}$ belonging to the given complete analytic function. Its germ of a holomorphic function at $z\in D$ is the equivalence class $[f{]}_z$, where two elements are equivalent when they agree on a neighbourhood of $z$.

For every function element $(f,D)$ and sufficiently small open $U\subseteq D$, setThese sets form a basis for the space of germs of holomorphic functions. The projectionrestricts to a homeomorphism $U_f\to U$. The inverse projections $\pi {|}_{U_f}$ are compatible complex charts, making $\mathcal{G}$ a Riemann surface and $\pi$ a covering map and local biholomorphism. In such a chart, the evaluation map on a space of germs has coordinate expressionIt is therefore holomorphic on every chart, hence analytic on each component of $\mathcal{G}$.

Now let $f:R\to S$ be a nonconstant analytic map of compact connected Riemann surfaces, and let $B$ be its finite branch-value set. Each point of $B$ is a branch value of a holomorphic map. At every point of $R\setminus f^{-1}(B)$ the local degree is one, so the holomorphic inverse-function theorem supplies a neighbourhood on which $f$ is biholomorphic. For $y\in S\setminus B$, the fibre is finite. Choose disjoint inverse neighbourhoods around all its points and shrink their common image; compactness, or equivalently properness of $f$, ensures that no additional preimages enter. Thusis an ordinary unramified covering map.

Fix $P\in S\setminus B$. Given a loop $\gamma$ based at $P$, lift it from every $Q\in f^{-1}(P)$ and send $Q$ to the endpoint of its lift. The path lifting theorem makes this a permutation; concatenation of loops composes the permutations and reversal gives inverses. Its image is the monodromy group of a covering. The punctured surface $R\setminus f^{-1}(B)$ is path connected. Given two points of the fibre, join them there by a path; its projection is a loop whose lift joins those points. Hence the monodromy action is transitive.

Finally considerIt has degree four. Its finite critical points satisfywith branch values $2$ and $-2$; the double poles at zero and infinity give the branch value infinity. For a regular value, the fibre isThe two commuting involutions $z\mapsto -z$ and $z\mapsto z^{-1}$ preserve $F$ and act transitively on this fibre. They generate all four deck transformations, so the covering is normal and its monodromy group isWith the displayed ordering of the fibre, its elements areThis is the monodromy group of z squared plus z to the minus two.

Let $p:\hat{X}\to X$ be the covering map. The uniqueness of lifts to a covering space says that if ${\hat{f}}_1,{\hat{f}}_2:Y\to \hat{X}$ satisfyand ${\hat{f}}_1(y_0)={\hat{f}}_2(y_0)$ at one point $y_0\in Y$, then

For any $y\in Y$, choose a path $\gamma$ from $y_0$ to $y$, using path-connectedness of $Y$. Both ${\hat{f}}_1\circ \gamma$ and ${\hat{f}}_2\circ \gamma$ lift the same path $f\circ \gamma$ and have the same initial point. Uniqueness in the path lifting theorem makes the lifted paths equal, so their endpoints agree. Since $y$ was arbitrary, the two lifts agree everywhere.

The map $p$ is the quotient of ${\mathbb{R}}^2$ by integer translations, andGiven a based loop $\gamma$ at $x_0$, the path lifting theorem gives a unique lift $\tilde{\gamma }$ beginning at $0$. Its endpoint lies in ${\mathbb{Z}}^2$. The homotopy lifting property shows that this endpoint depends only on $[\gamma ]$, so defineLifting a concatenation after translating the second lift by the endpoint of the first shows that $\Phi$ is a group homomorphism. It is surjective because, for every $m\in {\mathbb{Z}}^2$, the path $r\mapsto p(rm)$ is a loop whose lift ends at $m$. If a lift ends at zero, it is a loop in the simply connected space ${\mathbb{R}}^2$ and contracts there; projecting the contraction proves that its original loop is null-homotopic. Thus $\Phi$ is injective, and the lift-endpoint description of the fundamental group of the torus gives

In local coordinates centred at $p$ and $f(p)$, a nonconstant rational map has the formThe point $p$ is a ramification point of a holomorphic map when its ramification index of a holomorphic map satisfies $e_p>1$. A branch value of a holomorphic map, also called a branch point in the target, is a value $f(p)$ of some ramification point.

Let $B$ be the finite set of branch values and put $R=f^{-1}(B)$. For every $w\notin B$, all points of $f^{-1}(w)$ have local degree one, so the holomorphic inverse function theorem supplies disjoint neighbourhoods on which $f$ is biholomorphic. Compactness of the fibre lets their target neighbourhoods be intersected to one evenly covered neighbourhood of $w$. Henceis an unramified covering map. Here the displayed restriction requires $R=f^{-1}(B)$; if “ramification point” is reserved only for points with $e_p>1$, then the source deletion must be written $f^{-1}(B)$.

The monodromy theorem says that analytic continuations along endpoint-fixed homotopic paths have the same terminal germ. Equivalently here, the path lifting theorem lifts a loop $\gamma$ based at $w\notin B$ from each point of $f^{-1}(w)$; taking the endpoint of each lift permutes that fibre. The permutation depends only on $[\gamma ]\in {\pi }_1({\mathbb{C}}_{\infty }\setminus B,w)$, giving the monodromy group of a covering.

For the stated function, make the Möbius change of coordinateThenThe critical points are $t=0,\pm 1/\sqrt{2},\infty$, with branch valuesA loop around $0$ interchanges the two roots born from $t=0$, so its branch cycle is a transposition. A loop around $-1/4$ simultaneously interchanges the two pairs that collide there, so its branch cycle is a product of two disjoint transpositions. With a suitable labelling these areThey generate a transitive group of order eight; their product is a four-cycle. Therefore the full monodromy group isthe dihedral group in its action on the four vertices of a square.

For $t<1/2$, the number $f(t)$ is negative real, so the proposed identity would forceContinuity on the connected interval $[0,1/2)$ makes $\varphi$ equal to one fixed odd multiple of $\pi$ there. For $t>1/2$, the number $f(t)$ is positive real, so $\varphi$ must instead equal one fixed even multiple of $\pi$. These two constants cannot agree, and their one-sided limits at $1/2$ therefore contradict continuity. The fact that $f(1/2)=0$ removes the phase condition at that one point but does not repair the discontinuity.

Now let $g:[0,1]\to \mathbb{C}\setminus \{0\}$ and put $h=g/|g|$. This is a continuous path in the unit circle. The exponential map $\theta \mapsto e^{i\theta }$ is a covering map, so the path lifting theorem supplies a continuous lift $\theta :[0,1]\to \mathbb{R}$ after one value of $\theta (0)$ is chosen. HenceOne can obtain the same lift directly from the allowed special case: by uniform continuity, subdivide $[0,1]$ so that on each subinterval $h(t)/h(t_j)$ lies in the right half-plane, choose its continuous local argument there, and add a multiple of $2\pi$ to match the preceding endpoint.

If $\theta$ and $\tilde{\theta }$ are two such phases, thenThis integer-valued function is continuous and hence constant. Thereforedoes not depend on the chosen lift.

For $u(t)=g(t^2)$, the phase $\theta (t^2)$ givesFor $v(t)=g(t{)}^2$, the phase $2\theta (t)$ gives

Finally, takeThey have the same two endpoints, but $r(g_1)=0$ and $r(g_2)=2\pi$. This endpoint discrepancy records the winding number of the closed path.

Choose a path $\gamma$ in $X$ from $x_0$ to $x_1$. For every $\hat{x}\in p^{-1}(x_0)$, the path lifting theorem gives a unique lift of $\gamma$ starting at $\hat{x}$. Send $\hat{x}$ to the endpoint of this lift. Lifting the reversed path gives the inverse map, so this is the fibre bijection by path liftingIn particular, all fibres have the same cardinality.

A connected covering is a normal covering map when its deck transformations act transitively on a fibre. Equivalently, for a choice of ${\hat{x}}_0$ above $x_0$,The relevant lifting criterion for a covering space says that a based map $f:(Y,y_0)\to (X,x_0)$ lifts through $p$ exactly whenFor a universal covering, $\hat{X}$ is simply connected, so the displayed covering subgroup is trivial and hence normal. Thus a universal covering map is normal.

Now consider connected finite covers of the closed orientable surface ${\Sigma }_g$. By the classification of connected covering spaces, degree-$n$ connected covers correspond to index-$n$ subgroups of the fundamental group of a closed orientable surface, and normal covers correspond to normal subgroups.

The cases in which normality is forced are:

It remains to show that these are the only forced cases. Let $g\ge 2$ and $n\ge 3$. WriteIn the symmetric group $S_n$, putDefine a homomorphism byand send all remaining generators to the identity. The relation is respected becauseThe cycle $\sigma$ and transposition $\tau$ generate $S_n$, so the homomorphism is surjective.

LetThe natural action of $S_n$ is transitive, so $H$ has index $n$. Its image is the point stabilizer $S_{n-1}$, which is not normal in $S_n$ for $n\ge 3$; hence $H$ is not normal. The connected covering corresponding to $H$ is therefore an explicit degree-$n$ nonnormal cover. This is the nonnormal finite cover of a higher-genus orientable surface.

Consequently, among connected covers that exist, normality is forced exactly whenwith the qualification that $g=0$ permits only $n=1$, as summarized by the forced normality of finite connected covers of orientable surfaces.

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.

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.

For a covering map over a nonempty connected space, the degree of a covering map isthe cardinality of any fibre of a function. To show independence of $x$, choose an evenly covered neighbourhood $U$. Each covering sheet contains exactly one point over every $y\in U$, so $|p^{-1}(y)|$ equals the number of covering sheets over $U$. The fibre of a function cardinality is therefore locally constant. For a fixed cardinal number $\kappa$, the points with that cardinality form an open subset, and their complement is also open. Connectedness forces the value to be constant everywhere.

For a path-connected base there is also an explicit fibre bijection by path lifting: lift a path from $x$ to $y$ starting at each point over $x$ and take its endpoint. The path lifting theorem gives existence and uniqueness; path reversal gives the inverse bijection. The bijection may depend on the path, but the resulting equality of cardinalities does not.

A connected base is needed for one global degree. Without it, use a degree on each connected component on which the cardinality is constant. For example, a covering map of a discrete two-point topological space can have one point over the first point and two over the second. Thus the unqualified printed definition needs the usual connected-base convention, or the explicit assumption that every fibre of a function has the same cardinality.