Why a map which is close to the identity on the boundary of a disk has a zero?

Question

Let $\mathbb{D}^2$ be the closed unit disk in $2D$. Let $f:\mathbb{D}^2 \to \mathbb{R}^2$ be a smooth map. Suppose that

$$\|f|_{\partial \mathbb{D}^2}-\operatorname{Id}_{\partial \mathbb{D}^2}\|_{\infty,\partial \mathbb{D}^2 }<1. \tag{1}$$

($|| \cdot ||_{\infty}$ is the maximum norm).

Then $f$ vanishes at some point in $\mathbb{D}^2$.

I saw the following argument given somewhere:

The assumption $(1)$ implies that $ f|_{\text{Int}(\mathbb{D}^2)} $ has degree $1$ and so $f(x,y)$ vanishes in some point $(x_0,y_0)\in \text{Int}(\mathbb{D}^2)$.

I don't understand why the degree equals one? And how does this imply the existence of zero?

I would be happy for an explanation of this approach, or any other way of proving this claim.

(I thought the degree of a map is defined only if the domain is a closed manifold, or, in the case of non empty boundary, the map sends boundary to boundary. Both options do not apply here).

Answer 1

1) This is called the "Dog in a Leash" lemma :-) It states that if $\gamma_1, \gamma_2$ are two loops not containing $0$ with $|\gamma_1 - \gamma_2| < |\gamma_2|$ then $\gamma_1, \gamma_2$ have the same winding number around $0$. If you want a proof see page 41 of Fulton's algebraic topology book.

2) Let $f : D^2 \to \Bbb R^2$ as before. If $f$ is nonzero we can assume that $f$ has norm $1$. But then the map $x \mapsto f(x)$ would be a retraction of $D^2$ onto $S^1$, which is impossible.

Edit (see comments) In addition we prove the following lemma, illustrating the well known theorem "the devil is in the details " :-)

Lemma : let $U = D^2 \backslash B(0,r_0)$ and $f : U \to D^2$. Assume $f_{|S^1}$ is of degree $1$. There is an homotopy $H : I \times U \to U$ such that $H_0 = f, {H_1}_{|S^1} = id_{S^1}$ and ${H_t}_{|B} = f_{|B}$ where $B$ is the circle of radius $r_0$.

Proof : Let $G : S^1 \times I \to S^1$ a continuous map with $G_0 = f, G_1 = id_{S^1}$.

I claim that there is a smooth function $\varepsilon(r,e^{i\theta},s) : [r_0,1] \times S^1 \times [0,1]$ such that

• For all $ e^{i\theta}\in S^1$, and for all $s \in [0,1]$, we have $\varepsilon(r_0,e^{i\theta},s) = G_s(e^{i\theta})^{-1}f(e^{i\theta})$
• For all $e^{i\theta} \in S^1$ and $r \in [r_0,1]$, we have $\varepsilon(r,e^{i\theta},0) = 1$
• For all $e^{i\theta} \in S^1$ we have $\varepsilon(1,e^{i\theta}, 1) = 1$

Now you can check that $$ H_s(re^{i\theta}) = f(e^{i\theta})^{-1}f(re^{i\theta})G_s(e^{i\theta})\varepsilon(r, e^{i\theta}, s)$$

is the desired homotopy.

The idea behind this mysterious formula is that looking at $\widetilde{H}_s(re^{i\theta}) := f(e^{i\theta})^{-1}f(re^{i\theta})G_s(e^{i\theta})$ is almost what we want (i.e $\widetilde{H}_0=f$ and ${\widetilde{H}_1}_{|S^1} = id_{S^1}$) but we don't have $\widetilde{H}_s(r_0e^{i\theta}) = f_{|B}$, so we need this extra factor $\varepsilon$ fixing this problem, and the fact that $\varepsilon = 1$ at the right values ensure that we preserved the good properties of $\widetilde{H}$ .

I should write how to construct $\varepsilon$ but this would be tedious and not really interesting. The trick is to define a function $\varepsilon'$ on a "nice" subset of $[r_0,1] \times S^1 \times [0,1]$, which alreay satisfy our requirement. Then, by writing the circle as the union of two intervals, and using that we picked a "nice" subset, this boils down to show that a function defined on a face of a cube can be continuously extended to the whole cube which is clear.