---- > [!theorem] Theorem. ([[Brouwer fixed point theorem for the disc]]) > Any [[continuous]] map from the closed disc $D \subset \mathbb{R}^{2}$ to itself has a [[fixed point]]. > [!generalization] > - [[Brouwer's fixed point theorem]] (uses [[singular homology]] instead of the [[fundamental group]], hence works in higher dimensions) ^generalization > [!proof]- Proof. ([[Brouwer fixed point theorem for the disc]]) > By way of contradiction, suppose there exists a [[continuous]] map $f:D\to D$ such that $f(z) \neq z$ for any $z \in D$. > > We claim that this implies the maps $a,b: \mathbb{S}^{1} \to \mathbb{R}^{2} - \{ 0 \}$ defined by $a(z)=z$ and $b(z)=z-f(z) \neq0$ are freely [[homotopy|homotopic]] (in $\mathbb{R}^{2}-\{ 0 \}$), and that the [[straight-line homotopy]] $H(z,t)=(1-t)z + t(z-f(z))=z-tf(z)$ > witnesses this; we just need to check that $z \neq tf(z)$ for all $t \in [0,1]$ and $z \in \mathbb{S}^{1}$ with $z \neq f(z)$. If $z=tf(z)$ for some pair $(z,t)$, then $1=|z|=|t| \ |f(z)|;$since $|t| \leq 1$ and $|f(z)| \leq 1$, this enforces $t=1$. However, $H(z,1)=z-f(z)\neq 0$ by assumption. Therefore, $H(z,t)$ is never zero. > > We first show that for any nonvanishing vector field $(x,v(x))$ on $D^{2}$, there is a point of $\mathbb{S}^{1}$ where it points directly inward. Take the map $v:D^{2} \to \mathbb{C}-\{ 0 \}$; let $w$ be its restriction to $\mathbb{S}^{1}$. Lemma 55.3 in Munkres tells us that $w$ is nulhomotopic because it extends to a map of $D^{2}$ into $\mathbb{C} - \{ 0 \}$. It is also true that $w$ is homotopic to the inclusion map from part (a), the homotopy being $F(x,t)=t x+(1-t)w(x).$ > We must show that $F(x,t)\neq (0,0)$. This obviously holds for $t=0$ and $t=1$. If $F(x,t) =(0,0)$ for some $t \in (0,1)$ then $tx+(1-t)w(x)=0$ implying $w(x)$ is a negative scalar multiple of $x$, i.e., points directly inward. > > The proof above goes through to show that there is a point where the vector field points directly outward if we use instead the vector field $(x,-v(x))$. > > Recall that we have supposed $f(z)\neq z$ for all $z \in D$. Then $v(z)=f(z)-z$ corresponds to a nonvanishing vector field $(x,v(x))$ on $D$. But the vector field cannot point directly outward at any point $z \in \mathbb{S}^{1}$, as that would imply $f(z)-z=az$ for some $a>0$, such that $f(z)=(1+a)z \notin D$. This is the desired contradiction. ---- #### . #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM outgoing([[]]) FLATTEN file.tags GROUP BY file.tags as Tag