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Algebraic topology translates questions about topological spaces and continuous maps into questions about groups and other structures from algebra. A topological invariant compresses the enormous amount of data in a space into something one can calculate. For example, the fundamental group records which based loops can be continuously deformed into one another. A filled disk has trivial fundamental group, whereas a circle has fundamental group $\mathbb{Z}$: an integer remembers how many times a loop goes around.

This compression also respects maps. By functoriality of the fundamental group, a continuous map preserving the chosen basepoints induces a group homomorphism; successive maps induce successive group homomorphisms. Thus geometry supplies arithmetic constraints. A proposed homotopy equivalence must induce an isomorphism, so incompatible fundamental groups rule it out before anyone attempts to construct the deformation.

The beauty is that a whole annulus can retain exactly the same homotopy type as its inner circle. Its thickness disappears under a deformation retraction, while the obstruction to shrinking a loop survives. Equal fundamental groups alone do not establish a homeomorphism or a homotopy equivalence between arbitrary topological spaces.

This turns a visual question into a reusable test. Suppose a continuous map sends the inner circle of an annulus once around another circle. Could it extend over a filled disk spanning that boundary? Such an extension would supply a homotopy shrinking the image loop to a point. Its winding number would have to change from one to zero, which homotopy invariance of winding number forbids. Algebra rules out the extension even when no particular formula for it has been proposed.

Consider the rational functionInside the unit circle are a double zero of a function at $1/4$ and a simple pole at $-1/2$. The simple zero of a function at $2$ is outside. Thus $N=2$, $P=1$, and the winding number is one. The factorization checks the count exactly; the plot illustrates how the boundary records it.

For the concentric circle $|z|=r$, the winding number is zero for $0<r<1/4$, two for $1/4<r<1/2$, one for $1/2<r<2$, and two for $r>2$. At $r=1/4$ or $r=2$, the boundary contains a zero of a function; at $r=1/2$, it contains a pole. The theorem does not apply at those three radii. Moving a boundary changes the count only when it crosses an object being counted; homotopy invariance of winding number describes the unchanged count between crossings.

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

A loop homotopy is a continuous $H:[0,1{]}^2\to \mathbb{C}\setminus \{0\}$ with $H(0,t)={\gamma }_0(t)$, $H(1,t)={\gamma }_1(t)$ and $H(s,0)=H(s,1)$ for every $s$. Compactness gives a positive lower bound $r$ for $|H|$ and uniform continuity. Partition the $s$ interval finely enough that neighboring loops $\eta ,\delta$ satisfy $|\eta (t)-\delta (t)|<r$. Their quotient lies in $|z-1|<1$, and has a single-valued continuous argument with equal endpoint values. Its winding number is zero. Arguments of a product add, so $w(\eta )=w(\delta )+w(\eta /\delta )=w(\delta )$. Applying this at every step provesThis proves homotopy invariance of winding number directly, including continuous curves without differentiability.

Normalize the nonvanishing loop to $\gamma (t)/|\gamma (t)|$. A continuous argument lift is a continuous function $\theta :[0,1]\to \mathbb{R}$ such that $\gamma (t)/|\gamma (t)|=e^{i\theta (t)}$. Such a lift exists on an interval; any two differ by a constant integer multiple of $2\pi$. The winding number isIts value is independent of the lift. This definition applies to continuous loops without assuming a differentiable contour.

The homotopy invariance of winding number states: if $H:[0,1{]}^2\to \mathbb{C}\setminus \{0\}$ is continuous and $H(s,0)=H(s,1)$ for every $s$, then $w(H(s,\cdot );0)$ is independent of $s$. Neither the loops nor their moving basepoints may pass through zero. One may justify invariance by uniformly subdividing the homotopy into nearby loops whose normalized values have a continuous argument difference of magnitude less than $\pi$; that argument difference has equal endpoint values, so their winding numbers agree.

If the boundary loop meets zero, the assertion already holds. Otherwise suppose, for contradiction, that $f$ has no zero anywhere in the disc. Then $K(s,t)=f((1-s)e^{2\pi it})$ is a homotopy through nonvanishing closed loops from the boundary loop to the constant loop $f(0)$. Constant loops have winding number zero, whereas part (i) gives winding number one. This contradicts homotopy invariance of winding number. Hence $\boxed{f(z)=0\text{ for some }z\in B}$.

Put $z(t)=e^{2\pi it}$ and use the straight homotopy $H(s,t)=z(t)+s(\gamma (t)-z(t))$. For $0\le s<1$, the triangle inequality givesAt $s=1$ it is nonzero by the hypothesis. Thus this is a homotopy of closed loops avoiding the origin, even though the given distance bound is non-strict. The homotopy invariance of winding number gives

The appropriate homotopy invariance of winding number is: if $H:[0,1{]}^2\to \mathbb{C}\setminus \{0\}$ is continuous and $H(s,0)=H(s,1)$ for each $s$, then all loops $t\mapsto H(s,t)$ have the same winding number. The basepoint may move; every intermediate curve must close and avoid zero.

Compactness gives $m={\min }_{s,t}|H(s,t)|>0$. Uniform continuity gives $\delta >0$ such that $|s-s^{\prime }|<\delta$ implies $|H(s,t)-H(s^{\prime },t)|<m$ for every $t$. Part (d) then makes the winding numbers equal for nearby $s,s^{\prime }$. Subdivide $[0,1]$ into steps smaller than $\delta$ and chain these equalities. ThereforeThis proves the theorem directly for continuous loops, without requiring an integral formula or smooth homotopy.

DefineThe strict inequality $|s\sigma (t)|<|\gamma (t)|$ ensures that $H(s,t)\ne 0$ for every $(s,t)$, while $H(s,0)=H(s,1)$. Thus $H$ is a homotopy through closed loops in $\mathbb{C}\setminus \{0\}$ from $\gamma$ to $\gamma +\sigma$. By homotopy invariance of winding number,

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.

Homotopy invariance of winding number states that a homotopy $H(s,t)$ through closed paths in ${\mathbb{C}}^{\ast }$ preserves winding number. Uniform continuity divides the $s$ interval so successive paths differ pointwise by less than the minimum modulus of the earlier path; the supplied perturbation lemma makes their winding numbers equal. Chaining the subdivisions proves the result.

If a continuous retraction from a closed disc to its boundary existed, applying it to a contraction of the boundary circle inside the disc would give a homotopy in $\mathbb{C}\setminus \{0\}$ from a loop of winding number one to a constant loop of winding number zero. Homotopy invariance of winding number rules this out.