Extension over the Disk Forces Zero Degree

Why an Extension Forces a Restriction on Degree

A map from the circle to itself can wind around the target circle several times, and its degree records that winding. Extending the map over the whole disk imposes a strong restriction: the boundary map can be contracted to a constant by moving each boundary point inward along a radius. The resulting homotopy stays in the target circle because it is obtained by evaluating the extension, not by interpolating between points of the target.

We use $D^2=\{z\in\mathbb{C}:|z|\leq 1\}$ with the subspace topology from $\mathbb{C}$, and $S^1=\{z\in\mathbb{C}:|z|=1\}$ with its subspace topology. Thus $S^1$ is the boundary of $D^2$. Let $I=[0,1]$ have its usual topology. A map $f:S^1\to S^1$ extends over the disk if there is a continuous map $F:D^2\to S^1$ such that $F|_{S^1}=f$.

Definition: Let $p:\mathbb{R}\to S^1$ be $p(u)=e^{2\pi i u}$, and let $\gamma:I\to S^1$ be $\gamma(t)=e^{2\pi i t}$. For a continuous map $f:S^1\to S^1$, the path $f\circ\gamma$ has a lift $\widetilde{\gamma}_f:I\to\mathbb{R}$ with $p\circ\widetilde{\gamma}_f=f\circ\gamma$. Its degree is the integer $$ \deg(f)=\widetilde{\gamma}_f(1)-\widetilde{\gamma}_f(0). $$ The difference is an integer because $f\circ\gamma$ is a loop, so its lifted endpoints have the same image under $p$. The lift exists by the path-lifting property of the exponential covering map. Changing the choice of initial lift adds the same integer to both endpoints and does not change the difference.

We use the homotopy invariance of degree: if two maps $S^1\to S^1$ are homotopic, their degrees are equal. This is the earlier result on homotopy invariance of winding number, expressed in degree notation. A constant map has degree zero: if its value is $c=e^{2\pi i a}$, the constant function with value $a$ is a lift, and its endpoint difference is $a-a=0$.

The Radial Homotopy

Theorem: If a continuous map $f:S^1\to S^1$ extends continuously to $F:D^2\to S^1$, then $f$ is homotopic to a constant map and $\deg(f)=0$.

Proof. Define $H:S^1\times I\to S^1$ by $$ H(z,s)=F((1-s)z). $$ For $z\in S^1$ and $s\in I$, $|(1-s)z|=1-s\leq 1$, so $(1-s)z\in D^2$ and the formula is defined. The map $(z,s)\mapsto(1-s)z$ is continuous from $S^1\times I$ to $D^2$, and $F$ is continuous, so their composition $H$ is continuous. At $s=0$, $H(z,0)=F(z)=f(z)$; at $s=1$, $H(z,1)=F(0)$, independent of $z$. Thus $H$ is a homotopy from $f$ to the constant map with value $F(0)$. By homotopy invariance of degree and the degree-zero calculation for constant maps, $\deg(f)=0$. $\square$

The point of the formula is that it contracts the input toward the center of the disk, where the extension is already defined. It does not require the image of a straight-line interpolation between two points of $S^1$ to remain on $S^1$. The map $H$ has values in $S^1$ because every value is an output of $F:D^2\to S^1$.

Worked Extensions and Their Degrees

Worked Example: An Extension Depending on One Coordinate

Define $F:D^2\to S^1$ by $F(x+iy)=e^{ix}$. This is continuous because $x+iy\mapsto x$ and $x\mapsto e^{ix}$ are continuous, and its values have modulus $1$. Its boundary map is $f(e^{2\pi i t})=F(e^{2\pi i t})=e^{i\cos(2\pi t)}$.

Since $p(u)=e^{2\pi i u}$, a lift of this loop is $\widetilde{\gamma}_f(t)=\cos(2\pi t)/(2\pi)$: applying $p$ gives $e^{2\pi i\cos(2\pi t)/(2\pi)}=e^{i\cos(2\pi t)}$. At the endpoints, $\widetilde{\gamma}_f(0)=1/(2\pi)$ and $\widetilde{\gamma}_f(1)=1/(2\pi)$, because $\cos(0)=\cos(2\pi)=1$. Therefore $\deg(f)=1/(2\pi)-1/(2\pi)=0$.

The radial homotopy is $H(e^{2\pi i t},s)=e^{i(1-s)\cos(2\pi t)}$. At $s=0$ this is the boundary map just computed; at $s=1$ it is $e^0=1$ for every $t$. This verifies the theorem directly for this extension.

Worked Example: An Extension Using Both Coordinates

Define $F:D^2\to S^1$ by $F(x+iy)=e^{i(x+y)}$. The function $x+y$ is continuous, so $F$ is continuous, and every value again has modulus $1$. On the boundary, $f(e^{2\pi i t})=e^{i(\cos(2\pi t)+\sin(2\pi t))}$.

A lift is $\widetilde{\gamma}_f(t)=(\cos(2\pi t)+\sin(2\pi t))/(2\pi)$, since applying $p$ multiplies this expression by $2\pi i$ in the exponent. At $t=0$, the numerator is $\cos 0+\sin 0=1+0=1$. At $t=1$, it is $\cos(2\pi)+\sin(2\pi)=1+0=1$. Thus both endpoint values of the lift are $1/(2\pi)$, and $\deg(f)=0$.

In this case the radial homotopy is $H(e^{2\pi i t},s)=e^{i(1-s)(\cos(2\pi t)+\sin(2\pi t))}$. For $s=0$ it is $f$, and for $s=1$ it is the constant $1$. The equal lift endpoints and the explicit homotopy give independent checks of the zero-degree conclusion.

Nonzero Degree Prevents an Extension

Worked Example: A Power Map Cannot Extend

For an integer $n$, consider the power map $f_n:S^1\to S^1$, $f_n(z)=z^n$. Its boundary loop is $f_n(e^{2\pi i t})=e^{2\pi i nt}$, so $t\mapsto nt$ is a lift. Its endpoint difference is $n-0=n$, hence $\deg(f_n)=n$.

For example, when $n=3$, the lift has values $0$ at $t=0$ and $3$ at $t=1$, so $\deg(f_3)=3$. If $f_3$ extended over $D^2$, the theorem would give $\deg(f_3)=0$, contradicting $3\ne0$. Thus $z\mapsto z^3$ does not extend over the disk. More generally, every $f_n$ with $n\ne0$ fails to extend for the same reason. When $n=0$, $f_0$ is the constant map $1$, and the constant extension $F(z)=1$ does exist.

Check: the lift computation gives degree $n$, while any extension would force degree $0$. These statements are compatible for $n=0$, and incompatible for every nonzero integer $n$.

Corollary: If $f:S^1\to S^1$ has nonzero degree, then $f$ has no continuous extension $F:D^2\to S^1$.

Proof. Suppose, to the contrary, that a map $f$ of nonzero degree has a continuous extension over $D^2$. The theorem then gives $\deg(f)=0$, contradicting the assumption $\deg(f)\ne0$. Hence no such extension exists. $\square$

A Common Pitfall: Interpolating in the Wrong Space

A tempting attempt to contract a circle map to a constant $c\in S^1$ is the formula $K(z,s)=(1-s)f(z)+sc$. This interpolation takes place in $\mathbb{C}$, not necessarily in $S^1$, so it need not be a homotopy into the circle.

For a concrete failure, use the extension $F(x+iy)=e^{ix}$ above, its boundary map $f$, and $c=1=F(0)$. At $z=-1$, the boundary coordinates are $x=-1,y=0$, so $f(-1)=e^{-i}$. At $s=1/2$, the proposed interpolation gives $$ K(-1,1/2)=\frac{1+e^{-i}}{2} =e^{-i/2}\cos(1/2). $$ Its modulus is $\cos(1/2)$, which is strictly less than $1$, so this value is not in $S^1$. The formula therefore fails to define a map into the target circle. By contrast, the radial homotopy gives $H(-1,1/2)=F(-1/2)=e^{-i/2}$, whose modulus is $1$. The correction is to use the extension on the disk and move the input radially, rather than interpolate the output values.

This argument proves the necessary condition that a map extending over the disk has degree zero. It does not need a converse: for the purpose of ruling out an extension, a nonzero degree alone is enough. The zero-degree computations in the first two examples are consistent with the theorem, while their explicit extensions verify the stronger fact in those particular cases.

Check Your Understanding

Use the lift definition of degree and the radial formula $H(z,s)=F((1-s)z)$ where appropriate.

For $F(x+iy)=e^{2\pi i y}$ on $D^2$, compute the degree of its boundary map by writing a lift and evaluating its endpoints.
Explain why the formula $H(z,s)=F((1-s)z)$ is defined for every $z\in S^1$ and $s\in I$.
Show that the power map $z\mapsto z^{-2}$ cannot extend continuously over $D^2$.
Why does homotopy invariance imply that every map homotopic to a constant has degree zero?
Give a reason the straight-line interpolation between two points of $S^1$ may leave $S^1$.