Let \(A\) be a subspace of \(X\) and let \(j\colon A\hookrightarrow X\) be the corresponding inclusion map. A retraction of \(X\) onto \(A\) is a continuous map \(r\colon X\rightarrow A\) satisfying \(r\vert_A=\id\text{,}\) or equivalently, \(r\circ j=\id\vert_A\text{.}\) The subspace \(A\) is called a retract of \(X\) in this case.
Example2.5.2.\(S^1\) is retract of \(\R^2-\{\boldzero\}\).
Show that \(S^1\) is a retract of \(\R^2-\{\boldzero\}\text{.}\)
Solution.
Definition2.5.3.\(n\)-Ball.
Fix a positive integer \(n\text{.}\) We denote by \(B^n\subseteq \R^n\) the closed unit ball centered at \(\boldzero\text{:}\) i.e.,
There is no retraction of \(B^2\) onto \(S^1\text{.}\)
Proof.
Let \(j\colon S^1\hookrightarrow B^2\) be the inclusion map. Assume by contradiction that there is a retraction \(r\colon B^2\rightarrow S^1\text{.}\) Let \(Q=(1,0)\in S^1\text{.}\) Since \(r\vert_{S^1}=\id\vert_{S^1}\text{,}\) we have \(r(Q)=j(Q)=Q\text{.}\) Let \(r_*\colon \pi_1(B^2, Q)\rightarrow \pi_1(S^1, Q)\) and \(j_*\colon \pi_1(S^1, Q)\rightarrow \pi_1(B^2, Q)\) be the corresponding homomorphisms of fundamental groups. Since \(r\circ j=\id_{S_1}\text{,}\) we have \(r_*\circ j_*=\id\) by Theorem 2.2.6. This implies \(r_*\) is a surjection: a contradiction since \(\pi_1(B^2,Q)=\{e\}\) (\(B^2\) is convex) and \(\pi_1(S^1, Q)=\Z\text{.}\)
Theorem2.5.5.Nullhomotopies from \(S^1\).
Let \(h\colon S^1\rightarrow X \) be a continuous map. The following statements are equivalent.
\(h\) is nullhomotpic.
\(h\) extends to a continuous map \(k\colon B^2\rightarrow X\text{.}\)
For all \(Q\in S^1\) the map \(h_*\colon \pi_1(S_1,Q)\rightarrow \pi_1(X,h(Q))\) is the trivial homomorphism.
Proof: \((1)\implies (2)\).
Assume \(h\colon S^1\rightarrow X\) is nullhomotopic to the constant function \(e_{x_0}\colon S^1\rightarrow X\text{,}\) and let \(H\colon S^1\times I\rightarrow X\) be a homotopy from \(h\) to \(e_{x_0}\text{.}\) Our basic approach is to (a) show that \(B^2\) is a quotient of \(S^1\times I\text{,}\) and \((b)\) use the Theorem 1.11.10 to produce the desired extension \(k\colon B^2\rightarrow X\text{.}\)
To realize \(B^2\) as a quotient of \(S^1\times I\) we wish to map each copy \(S^1\times \{t\}\) of \(S^1\) to the circle \(C_t\colon x^2+y^2=(1-t)^2\) for each \(t\text{.}\) The map \(q\colon S^1\times t\rightarrow B^2\) defined as \(q((\boldx, t))=(1-t)\boldx\) accomplishes this for us. We see by the usual arguments that \(q\) is continuous, and our conceptual description of \(q\) shows that it is surjective. Since \(S^2\times I\) is compact and \(B^2\) is Hausdorff, the map \(q\) is closed, and hence a quotient map. Lastly, observe that \(q\vert_{S^1\times \{t\}}\colon S^1\times \{t\}\rightarrow C_t\) is bijective for all \(t\ne 1\text{,}\) and \(q\vert_{S^1\times \{1\}}\colon S^1\times \{1\}\rightarrow \{\boldzero\}\) collapses all points of \(S^1\times \{1\}\) to \(\boldzero\text{.}\) It follows that
\begin{equation}
q(\boldx, t)=q(\boldy,t') \iff t=t'=1 \text{ or } t=t'\ne 1 \text{ and } \boldx=\boldy\text{.}\tag{2.5.1}
\end{equation}
Now return to the homotopy \(H\text{.}\) Since for \(t=1\) it satisfies \(H(\boldx,1)=H(\boldy, 1)=x_0\text{,}\) we see from (2.5.1) that it is well-defined on the fibers of \(q\text{.}\) By Theorem 1.11.10 there is a continuous map \(k\colon B^2\rightarrow X\) satisfying \(k(q(\boldx, t))=H(\boldx, t)\text{.}\) Observe that \(q(\boldx, 0)=\boldx\) for all \(\boldx\in S^1\text{,}\) and hence we have \(k(\boldx)=H(q(\boldx, 0))=k(\boldx)\) for all \(\boldx\in S^1\text{.}\) This proves \(k\) is an extension of \(h\text{,}\) as desired.
Corollary2.5.6.Some non-nullhomotopies.
The following maps from \(S^1\) are not nullhomotpic:
the inclusion map \(j\colon S^1\hookrightarrow \R^2-\{\boldzero\}\)
the identity map \(\id\colon S^1\rightarrow S^1\text{.}\)
Theorem2.5.7.Nonvanishing vector fields on \(B^2\).
Let \(\boldF\colon B^2\rightarrow \R^2\) be a continuous function: i.e., \(\boldF\) is a continuous vector field on \(B^2\text{.}\) If \(\boldF(\boldx)\ne \boldzero\) for all \(\boldx\in B^2\) (i.e., \(\boldF\) is nonvanishing), then there exist nonzero elements \(\boldx_1, \boldx_2\in B^2\) such that \(\boldF(\boldx_1)=a\boldx_1\) for some positive \(a\in \R\) and \(\boldF(\boldx_2)=-b\boldx_2\) for some positive \(b\in \R\text{.}\) In other words, there is a point in \(B^2\) where the vector field points directly outward, and a point where the vector field points directly inward.
Proof.
Suppose first by contradiction that there is no point where \(\boldF\) points inward. By assumption we have \(\boldF\colon B^2\rightarrow \R^2-\{\boldzero\}\text{,}\) which restricts to continuous function \(F\vert_{S^1}\colon S^1\rightarrow \R^2-\{\boldzero\}\text{.}\) By Theorem 2.5.5\(\boldF\vert_{S_1}\) is nullhomotpic. We will show that \(\boldF\vert_{S^1}\) is also homotopic to the inclusion \(j\colon S^1\rightarrow \R^2-\{\boldzero\}\text{:}\) a contradiction since \(j\) is not nullhomotpic by Corollary 2.5.6.
By abuse of notation, we denote \(\boldF\vert_{S^1}=\boldF\text{.}\) To see \(\boldF\simeq j\text{,}\) consider the function \(H(\boldx, t)=(1-t)\boldx+t\boldF(\boldx)\text{.}\) It is clear that \(H\) defines a continuous function to \(\R^2\text{,}\) and we have \(H(\boldx, 0)=\boldx\) and \(H(\boldx, 1)=\boldF\text{.}\) Thus it appears we have our homotopy. However, we must show that \(H\colon S^1\times I\rightarrow \R^2-\{\boldzero\}\text{:}\) that is, we must show \(H(\boldx, t)\ne \boldzero\) for all \(\boldx\in S^1\) and \(t\in I\text{.}\) But if \(\boldx\in S^1\) were an element satisfying \((1-t)\boldx+t\boldF(\boldx)=\boldzero\) for some \(t\in \R\text{,}\) then we would have \(t\ne 0\) (since \(\boldx\ne\boldzero\)) and \(\boldF(\boldx)=-\frac{1-t}{t}\boldx=-b\boldx\text{,}\) where \(b=\frac{1-t}{t}> 0\text{.}\) This is impossible by our original assumption, since then \(\boldF\) would point directly inward at \(\boldx\text{.}\)
We’ve shown that {\em any} nonvanishing vector field on \(B^2\) must point directly inward at some point. Thus for any nonvanishing vector field \(\boldF\text{,}\) the nonvanishing vector field \(\boldG=-\boldF\) has a point \(\boldx\) in \(B^2\) where it points directly inward. The vector field \(\boldF\) then points directly outward at \(\boldx\text{,}\) as desired.
Theorem2.5.8.Brouwer fixed-point theorem (disc).
If \(f\colon B^2\rightarrow B^2\) is continuous, then \(f\) has a fixed point: i.e., we have \(f(\boldx)=\boldx\) for some point \(\boldx\in B^2\text{.}\)
Proof.
Assume by contradiction that \(f(\boldx)\ne \boldx\) for all \(\boldx\in B^2\text{.}\) The map \(\boldF(\boldx)=f(\boldx)-\boldx\) is a then a nonvanishing vector field on \(B^2\text{.}\) By Theorem 2.5.7 there is a point \(\boldx\in B^2\) such that \(\boldF(\boldx)=f(\boldx)-\boldx=a\boldx\) for some positive \(a\in \R\text{.}\) But then \(f(\boldx)=(a+1)\boldx\notin B^2\text{:}\) a contradiction! Thus \(f\) must have a fixed point.
Corollary2.5.9.Eigenvalues of positive matrices.
If \(A\) is a \(3\times 3\) matrix with positive entries, then \(A\) has a positive real eigenvalue.
