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On a locally path-connected space, on each path component choose a value of a complex logarithm of $f$ at a base point. Continue it along paths using local logarithms. Independence of the path is exactly vanishing of the winding number of a continuous closed path for all loops. This constructs a continuous function $g$ with $f=e^g$. The implication from a logarithm to zero winding requires no local connectivity. For applications to arbitrary planar compact sets, apply the criterion on polygonal neighborhoods rather than assuming the compact set is locally path-connected.

A nonempty compact dynamical system $(X,T)$ is a minimal dynamical system if it has no proper nonempty closed forward-invariant subset. Equivalently, every forward orbit is dense: the orbit closure of any point is a nonempty closed forward-invariant set, and conversely a closed invariant set contains the orbit closure of each of its points.

Consider the collection of nonempty closed forward-invariant subsets of $X$, ordered by reverse inclusion. Every chain has an upper bound, its intersection: the chain has the finite-intersection property, compactness makes the intersection nonempty, and continuity is not even needed for its forward invariance. The Zorn lemma supplies a minimal member $Y$, so $(Y,T{|}_Y)$ is a minimal subsystem. Also $T(Y)$ is compact, nonempty and forward invariant, so minimality gives $T(Y)=Y$. Here the printed word “nontrivial” must mean nonempty; a singleton with its identity map shows that a subsystem with more than one point cannot always be required.

If $y\in Y$, every tail of its forward orbit is dense in $Y$, since it is the orbit of $T^{N}y\in Y$. Therefore every neighborhood of $y$ contains $T^{n}y$ for arbitrarily large $n$. Choosing these times successively for balls of radii $1/j$ gives $n_j\to \infty$ with $T^{n_j}y\to y$. This proves the Birkhoff recurrence theorem: every continuous map of a nonempty compact metric space has a recurrent point. In fact every point of the chosen minimal subsystem is recurrent.

Write the continuous circle extension as the skew productA sufficient condition, and in this setting also a necessary one, is the circle skew-product minimality criterion: for every integer $m\ne 0$ there is no continuous function $f:X\to {\mathbb{S}}^1$ such thatThe cocycle $\rho$ is assumed continuous; a merely measurable cocycle need not define this topological extension.

To prove sufficiency, choose a minimal subsystem $Z$ of the extension. Its projection is a nonempty closed forward-invariant subset of $X$, hence all of $X$. For $a\in \mathbb{R}/\mathbb{Z}$, vertical translation sends $Z$ to another minimal subsystem, since it commutes with $\tilde{T}$. Two such translates either are disjoint or coincide: a nonempty intersection is closed and forward invariant, so minimality forces equality. Define their common stabilizerThis is a closed subgroup of the circle group. Its closedness follows by taking limits of translations of points in the compact set $Z$, using both a translation and its inverse. Each fiber $Z_x$ is nonempty. If $y,z\in Z_x$, the sets $Z$ and $Z+(z-y)$ intersect at $(x,z)$, so $z-y\in H$. Conversely translating any fiber point by any member of $H$ stays in the fiber. Thus every fiber is exactly one coset of $H$.

If $Z$ is proper, $H$ is proper and therefore finite, say $H=\{0,1/m,\dots ,(m-1)/m\}$ for some positive integer $m$. The character $y\mapsto e^{2\pi imy}$ is constant on each fiber coset, giving a well-defined function $f(x)$ from $Z_x$. This function is continuous: its graph is the continuous image of $Z$ in $X\times {\mathbb{S}}^1$, and projection of that compact graph onto the Hausdorff space $X$ is a continuous bijection and hence a homeomorphism. The relation $\tilde{T}(Z)=Z$ gives $f(Tx)=e^{2\pi im\rho (x)}f(x)$, contrary to the criterion. Hence $Z=\tilde{X}$ and the extension is minimal.

Conversely, a solution of the displayed equation makes $F(x,y)=e^{2\pi imy}/f(x)$ a nonconstant continuous invariant function on the extension. Its nonempty level sets are proper closed invariant subsets, contradicting minimality. This also proves necessity of the circle skew-product minimality criterion.

Apply it with $\alpha =\sqrt{2}$, $X=\mathbb{R}/\mathbb{Z}$, $Tx=x+\alpha$, and $\rho (x)=2x+\alpha$. The base is a minimal dynamical system: the closure of the cyclic subgroup generated by an irrational angle is the whole circle group, and its positive forward orbit has the same closure. To see the subgroup assertion, a proper closed subgroup of the circle is finite; if it contained an irrational angle it would be infinite. The elementary classification follows because a closed subgroup with arbitrarily small nonzero elements is dense, while otherwise its least positive representative generates a finite subgroup.

There can be no forbidden $f$ for any $m\ne 0$. Indeed, the winding number of a continuous closed path of $x\mapsto f(x+\alpha )$ equals that of $x\mapsto f(x)$, by translation and homotopy invariance of winding number. Consequently their quotient has winding number zero. But the proposed quotient $e^{2\pi im(2x+\alpha )}$ has winding number $2m\ne 0$. Thus the skew product is minimal. Starting at $(0,0)$, induction givesbecause $n^2\alpha +2n\alpha +\alpha =(n+1{)}^2\alpha$. Its orbit is dense in the two-dimensional torus, so its second coordinate is dense in the circle. Removing the initial point does not change density, since every orbit tail in a minimal dynamical system is dense. Therefore

For a continuous loop in $\mathbb{C}\setminus \{0\}$, choose a continuous argument lift $\theta :[0,1]\to \mathbb{R}$ with $\gamma (t)/|\gamma (t)|=e^{i\theta (t)}$. Such a lift is constructed by subdividing the interval into pieces whose images lie in an argument chart and matching the arguments at their endpoints. The winding number of a continuous closed path isChanging the initial lift adds one constant multiple of $2\pi$ and does not change this integer.

Normalize the loop byChoose a continuous lift $\vartheta :[0,1]\to \mathbb{R}$ satisfyingBecause $\gamma (0)=\gamma (1)$, the difference $\vartheta (1)-\vartheta (0)$ is an integer independent of the chosen lift. The winding number of a continuous closed path is

If $\eta$ is a lift of $\delta /|\delta |$, then $\vartheta +\eta$ is a lift ofTherefore

For a continuous closed path $\gamma :[0,1]\to \mathbb{C}\setminus \{0\}$, choose a continuous argument liftIts winding number of a continuous closed path about zero isFor a piecewise smooth path this equals $(2\pi i{)}^{-1}{\int }_{\gamma }dz/z$.

If $|\gamma (t)|>|\varphi (t)|$, thennever vanishes, since $|H(s,t)|\ge |\gamma (t)|-s|\varphi (t)|>0$. Thus $H$ is a homotopy through closed paths avoiding zero. By homotopy invariance of winding number, or directly by the dominated-perturbation lemma,

More generally, ${\gamma }_0$ and ${\gamma }_1$ are homotopic by paths in $\Gamma$ when there is a continuous functionwith $H(0,t)={\gamma }_0(t)$, $H(1,t)={\gamma }_1(t)$, and $H(s,0)=H(s,1)$ for every $s$. The winding-number theorem states that such a homotopy implies

For the Fundamental theorem of algebra, let $P(z)=a_n{z}^n+\cdots +a_0$ with $n\ge 1$. For sufficiently large $R$,for every $t$. The dominated-perturbation lemma shows that $t\mapsto P(R{e}^{2\pi it})$ has the same winding number as $t\mapsto a_n{R}^n{e}^{2\pi int}$, namely $n$. If $P$ had no zero, however,would be a homotopy in $\Gamma$ from that loop to the constant loop $P(0)$, whose winding number is zero. This contradiction proves that $P$ has a complex root. This is the winding-number proof of the fundamental theorem of algebra.

Finally suppose that a continuous retraction $r:D^2\to S^1$ existed. The boundary loop $\eta (t)=e^{2\pi it}$ has winding number one, while $(1-s)\eta (t)$ contracts it to zero inside the disc. Composing this contraction with $r$ gives a homotopy through loops in $S^1$ from $r\circ \eta =\eta$ to the constant loop $r(0)$. Their winding numbers are respectively one and zero, contradicting homotopy invariance. Hence there is no such retraction, as in the winding-number proof of the no-retraction theorem.