Section 55: Retractions and Fixed Points

A retraction of $X$ onto $A\subset X$ is a continuous $r:X\rightarrow A$ such that $r(a)=a$ for $a\in A$ . If there is such a map, then $A$ is called a retract of $X$ .

If $A$ is a retract of $X$ , then $j_{∗}$ , where $j:A\rightarrow X$ is the inclusion map, is injective (because $r_{∗}\circ j_{∗}=i_{\pi_{1}(A)}$ ), therefore, the cardinality of $\pi_{1}(X)$ should not be less than that of $\pi_{1}(A)$ .
There is no retraction of the unit disc $B^{2}$ onto $S^{1}$ .
Generalized version:
- There is no retraction of $B^{n+1}$ onto $S^{n}$ (can be proved in general using more advanced techniques of the algebraic topology).

If $h:S^{1}\rightarrow X$ is continuous, then $h$ is nulhomotopic iff $h$ extends to a continuous function on $B^{2}$ iff $h_{∗}$ is the trivial homomorphism (the one that maps every element of one group to the identity element of the other).

If $h$ is nulhomotopic, a homotopy between $h$ and a constant map induces a continuous extension $k$ of $h$ on $B^{2}$ by identifying all points $S^{1}\times\{1\}$ (via a quotient map).

Given an extension $k$ , $h=k\circ j$ and $h_{∗}=k_{∗}\circ j_{∗}$ where $j_{∗}$ is trivial (as $\pi_{1}(B^{2})=0$ ).

Finally, if $h_{∗}$ is trivial, and $l$ is a single loop in $S^{1}$ , then the path homotopy between $h\circ l$ and a constant path in $X$ induces a homotopy between $h$ and a constant map in $X$ by identifying all points $(0,t)$ and $(1,t)$ via the quotient map $l\times i_{[0,1]}$ .

The inclusion map $j:S^{1}\rightarrow\mathbb{R}^{2}−0$ and the identity map $i:S^{1}\rightarrow S^{1}$ are not nulhomotopic (in both cases the domain is a retract of the co-domain, so that the group homomorphism $j_{∗}$ and $i_{∗}$ , respectively, is injective, and, hence, non-trivial).
Generalized version:
- If $h:S^{n}\rightarrow X$ is continuous, then it is nulhomotopic iff it has a continuous extension on $B^{n+1}$ .
- The inclusion map $j:S^{n}\rightarrow\mathbb{R}^{n+1}−0$ and the identity map $i:S^{n}\rightarrow S^{n}$ are not nulhomotopic.

A (continuous) vector field on $X\subset\mathbb{R}^{2}$ is an ordered pair $(x,v(x))\in X\times\mathbb{R}^{2}$ for $x\in X$ such that $v:X\rightarrow\mathbb{R}^{2}$ is continuous.

If a (continuous) vector field on $B^{2}$ is non-vanishing, then there is a point of $S^{1}$ where the vector field points directly inward, and a point of $S^{1}$ where the vector field points directly outward ( $v|_{S^{1}}:S^{1}\rightarrow\mathbb{R}^{2}−0$ is extendable, hence, nulhomotopic, and if there is no inward vector, it is also homotopic to $j:S^{1}\rightarrow\mathbb{R}^{2}−0$ , but the latter is not nulhomotopic).
Generalized version:
- A nonvanishing vector field on $B^{n+1}$ points directly outward at some point of $S^{n}$ , and directly inward at some point of $S^{n}$ .

(Brouwer fixed-point theorem for the disc.) If $f:B^{2}\rightarrow B^{2}$ is continuous, then there is a fixed point $x\in B^{2}$ : $f(x)=x$ .

Follows immediately from the previous result: $v(x)=f(x)−x$ cannot point outward on $S^{1}$ , so it is not non-vanishing.

If $f:S^{1}\rightarrow S^{1}$ is nulhomotopic, then it has a fixed point, and also maps some $x$ to $−x$ .
If $A$ is a retract of $B^{2}$ and $f:A\rightarrow A$ is continuous, then $f$ has a fixed point.
Generalized version:
- Every continuous map $f:B^{n+1}\rightarrow B^{n+1}$ has a fixed point.
- If $f:S^{n}\rightarrow S^{n}$ is nulhomotopic, then it has a fixed point, and also maps some $x$ to $−x$ .
- If $A$ is a retract of $B^{n+1}$ and $f:A\rightarrow A$ is continuous, then $f$ has a fixed point.

Some examples of using the fixed point theorem

(Frobenius) A $3\times3$ matrix $A$ of positive real numbers has a positive real eigenvalue (there is a vector $x$ with non-negative coordinates of unit length such that $Ax/||Ax||=x$ ).

$A$ can be assumed to be nonsingular with non-negative entries.
Generalized version:
- Every $n+1$ by $n+1$ matrix with positive real entries (or a nonsingular matrix with non-negative entries) has a positive real eigenvalue.

( $T=\{(x,y)|x,y\ge0\mbox{ and }x+y\le1\}$ has topological dimension of at least 2, related to Section 50.) For some $\epsilon>0$ for every open cover of $T$ by sets of diameter $<\epsilon$ , there is a point covered by at least three open sets.

Subpages

Section 55: Retractions and Fixed Points

Section 55: Retractions and Fixed Points

Some examples of using the fixed point theorem