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

algebraic-topologyfixed-point-theoremsgeneral-topology

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).

Best Answer

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.

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