Applications of the Fundamental Group

Since $a\in A$, both $i:(A,a)\to (X,a)$ and $r:(X,a)\to (A,a)$ are pointed, and $r\circ i={\mathrm{id}}_A$.

Functoriality gives $r_{\ast }\circ i_{\ast }=(r\circ i{)}_{\ast }=\mathrm{id}$, so $i_{\ast }$ has a left inverse and is injective.

If $A$ is a deformation retract, the homotopy from ${\mathrm{id}}_X$ to $i\circ r$ fixes $a$, so [L1] also gives $i_{\ast }\circ r_{\ast }=(i\circ r{)}_{\ast }=\mathrm{id}$; hence $i_{\ast }$ and $r_{\ast }$ are mutually inverse isomorphisms.

The disk $D^2$ is nonempty and convex: if $x,y\in D^2$ and $0\le t\le 1$, then $\parallel (1-t)x+ty{\parallel }_2\le (1-t)\parallel x{\parallel }_2+t\parallel y{\parallel }_2\le 1$. Hence ${\pi }_1(D^2,e_0)$ has one element.

The unit circle has ${\pi }_1(S^1,e_0)$ isomorphic to the nontrivial group $\mathbb{Z}$.

Suppose a retraction $r:D^2\to S^1$ existed.

By [L1], $i_{\ast }:{\pi }_1(S^1,e_0)\to {\pi }_1(D^2,e_0)$ would be injective, but steps 1.1 and 1.2 make this a homomorphism from a nontrivial group to a one-element group, which cannot be injective. Thus no such retraction exists.

For $x\in D^2$, put $a=f(x)$, $v=x-a$, $A=⟨v,v⟩$, $B=⟨a,v⟩$, and $C=⟨a,a⟩$. Fixed-point-freeness gives $v\mathrm{\ne }0$ and $A>0$. Define $\lambda (x)=\frac{-B+\sqrt{B^2+A(1-C)}}{A},r(x)=a+\lambda (x)v\mathrm{.}$

The equation $\parallel a+tv{\parallel }_2^2=1$ is $A{t}^2+2Bt+C-1=0$, whose discriminant is $4(B^2+A(1-C))\ge 0$ because $C=\parallel f(x){\parallel }_2^2\le 1$; thus $\lambda (x)$ is its larger root. Since the upward-opening quadratic is nonpositive at both $t=0$ and $t=1$, its larger root satisfies $\lambda (x)\ge 1$ and is the unique intersection parameter of the ray $a+tv$ with $S^1$ for $t\ge 1$.

The maps $a,v,A,B,C$ are continuous; the radicand is nonnegative, its square root is continuous, and the denominator $A$ never vanishes. Hence $\lambda$ is continuous, and componentwise continuity makes $r(x)=a+\lambda (x)v$ continuous.

Step 2.1 gives $\parallel r(x){\parallel }_2=1$, so $r$ maps $D^2$ into $S^1$. If $x\in S^1$, then $t=1$ is a root and is the larger root because $1\ge 0$ lies in the interval on which the quadratic is nonpositive; hence $\lambda (x)=1$ and $r(x)=a+v=x$. Therefore $r$ is a continuous retraction of $D^2$ onto $S^1$.

Suppose that $f$ has no fixed point, so $f(x)\mathrm{\ne }x$ for every $x\in D^2$.

By [L1], the map $f$ then determines a continuous retraction $D^2\to S^1$, contradicting [L2]. Therefore $f$ has a fixed point.

For $\tau \in [0,1]$ and $u\in \mathbb{R}\mathrm{/}\mathbb{Z}$, put $q(\tau ,u)=p(\tau Rh(u))\mathrm{/}p(\tau R)$. Root-freeness makes numerator and denominator nonzero, so $A(\tau ,u):=h^{-1}\left(\frac{q(\tau ,u)}{\mathrm{\mid }q(\tau ,u)\mathrm{\mid }}\right)$ is defined; moreover $q(\tau ,[0])=1$, hence $A(\tau ,[0])=[0]$. Define ${\alpha }_R(u)=A(1,u)$.

Writing complex addition and multiplication in real and imaginary coordinates shows from [F1], [L3], and [L4] that $(\tau ,u)\mapsto p(\tau Rh(u))$ and $\tau \mapsto p(\tau R)$ are continuous. Root-freeness and [L1] make division and radial normalization continuous, and [L2] makes $A$ continuous. At $\tau =0$ one has $q(0,u)=p(0)\mathrm{/}p(0)=1$, so $A(0,u)=[0]$ for every $u$, while step 1.1 keeps the basepoint fixed for all $\tau$. Thus $A$ is a based homotopy on the unit interval from the constant loop to ${\alpha }_R$, including the case $R=0$ without division by $R$.

Hence ${\alpha }_R$ is nullhomotopic for every $R\ge 0$, and [L5] gives $\mathrm{deg}({\alpha }_R)=0$.

If $\mathrm{\mid }z\mathrm{\mid }=R$, then $\mid {\sum }_{j<n}{a}_j{z}^j\mid \le {\sum }_{j<n}\mathrm{\mid }{a}_j\mathrm{\mid }{R}^j\le S{R}^{n-1}<R^n=\mathrm{\mid }{z}^n\mathrm{\mid }\mathrm{.}$ The estimate includes $n=1$ and $S=0$, since $R>1$ and $S<R$.

For $s\in [0,1]$ put $p_s(z)=z^n+s{\sum }_{j<n}{a}_j{z}^j$. Step 1.1 remains strict with $sS\le S$, so $p_s$ never vanishes on $\mathrm{\mid }z\mathrm{\mid }=R$, in particular $p_s(R)\mathrm{\ne }0$. The formula $H(s,u)=h^{-1}\left(\frac{p_s(Rh(u))\mathrm{/}{p}_s(R)}{\mathrm{\mid }{p}_s(Rh(u))\mathrm{/}{p}_s(R)\mathrm{\mid }}\right)$ is therefore a continuous based homotopy by [L2] and [L3]. At $s=1$ it is ${\alpha }_R$, while at $s=0$ it is $u\mapsto h^{-1}(h(u{)}^n)={\omega }_n(u)$.

Homotopy invariance and the standard-loop calculation give $\mathrm{deg}({\alpha }_R)=\mathrm{deg}({\omega }_n)=n$.

Suppose $p$ has no root. Since $p$ is nonconstant, it is nonzero and has degree $n\ge 1$ and leading coefficient $c\mathrm{\ne }0$. Dividing every coefficient by $c$ gives a monic polynomial $q=c^{-1}p$ of the same degree and with the same zero set, so $q$ is also root-free.

By [L1], every normalized radius-$R$ loop of $q$ is nullhomotopic, and therefore has degree zero by [L3].

Write $q(z)=z^n+{\sum }_{j<n}{a}_j{z}^j$, put $S={\sum }_{j<n}\mathrm{\mid }{a}_j\mathrm{\mid }$, and take $R=\max \{1,S\}+1$. Then $R>\max \{1,S\}$, so [L2] says that the normalized radius-$R$ loop has degree $n$.

Steps 2.1 and 2.2 assign the same loop both degree $0$ and degree $n$, impossible because $n\ge 1$. Hence the root-free assumption is false and $p$ has a complex root.

If $[u]=[v]$, then $(u+1\mathrm{/}2)-(v+1\mathrm{/}2)=u-v\in \mathbb{Z}$, so $a([u])=[u+1\mathrm{/}2]$ is well defined by [F1]; moreover $a(a([u]))=[u+1]=[u]$, so $a$ is an involution.

Suppose the loop $t\mapsto h([t])$ were endpoint-fixed homotopic to the constant loop at $h([0])$.

Let $y\in \mathbb{R}$ be arbitrary subject to $p(y)=h([0])$. By [L1] and [L2], the loop $t\mapsto h([t])$ has a unique lift $\tilde{h}$ with $\tilde{h}(0)=y$. Antipodality gives $p(\tilde{h}(1\mathrm{/}2))=h([1\mathrm{/}2])=p(y+1\mathrm{/}2)$, so [F1] gives a unique integer $k$ with $\tilde{h}(1\mathrm{/}2)=y+k+1\mathrm{/}2$.

For $0\le t\le 1\mathrm{/}2$, the paths $t\mapsto \tilde{h}(t+1\mathrm{/}2)$ and $t\mapsto \tilde{h}(t)+k+1\mathrm{/}2$ project to the same path because $h([t+1\mathrm{/}2])=a(h([t]))$, and they agree at $t=0$ by step 2.1. Lift uniqueness gives $\tilde{h}(t+1\mathrm{/}2)=\tilde{h}(t)+k+1\mathrm{/}2$, so at $t=1\mathrm{/}2$ one obtains $\tilde{h}(1)=y+2k+1$.

The constant loop at $h([0])$ has the constant lift beginning at $y$, so [L3] and step 1.2 would force $\tilde{h}(1)=y$. Step 3.1 instead gives $\tilde{h}(1)=y+2k+1\mathrm{\ne }y$ for every integer $k$, including negative $k$. This contradiction shows that the loop is not nullhomotopic and completes the odd-increment claim.

Suppose $f(x)\mathrm{\ne }f(-x)$ for every $x\in S^2$.

The difference $d(x)=f(x)-f(-x)$ is continuous by [L5] and nonzero by step 1.1, so $g(x)=\rho (d(x))$ defines a continuous map $g:S^2\to S^1$. Since $d(-x)=-d(x)$, one has $g(-x)=-g(x)$.

Let $h:\mathbb{R}\mathrm{/}\mathbb{Z}\to S^1$ be the homeomorphism in [F1] and put $b=h^{-1}\circ g\circ e$. The map $e$ is continuous componentwise by [L5]; since $e([t+1\mathrm{/}2])=-e([t])$ and $h([u+1\mathrm{/}2])=-h([u])$, the continuous map $b$ is antipodal. Hence the loop $t\mapsto b([t])$ is not nullhomotopic by [L1].

The loop $t\mapsto e([t])$ in $S^2$ is nullhomotopic because $S^2$ is simply connected. Postcomposing such a nullhomotopy with the continuous map $h^{-1}\circ g$ makes $t\mapsto b([t])$ nullhomotopic in $\mathbb{R}\mathrm{/}\mathbb{Z}$.

Steps 3.1 and 3.2 contradict one another. Therefore the assumption in step 1.1 is false, and some $x\in S^2$ satisfies $f(x)=f(-x)$.

By [L1], choose $x\in S^2$ with $f(x)=f(-x)$.

If $x=-x$, then $2x=0$ and hence $x=0$, contrary to $\parallel x{\parallel }_2=1$. Thus $x$ and $-x$ are distinct points with the same image, so $f$ is not injective.

For $i=1,2$, define ${\delta }_i(x)=d(x,A_i)$ when $A_i\mathrm{\ne }\mathrm{\varnothing }$, and define ${\delta }_i(x)=1$ when $A_i=\mathrm{\varnothing }$. By [L2] and [L3], each ${\delta }_i$ is continuous and its zero set is exactly $A_i$.

Apply [L1] to $\delta =({\delta }_1,{\delta }_2):S^2\to {\mathbb{R}}^2$. There is $x\in S^2$ such that ${\delta }_1(x)={\delta }_1(-x)$ and ${\delta }_2(x)={\delta }_2(-x)$.

If either common value is zero, then $x$ and $-x$ both lie in the corresponding $A_i$. If both common values are positive, neither point lies in $A_1\cup A_2$, so the covering hypothesis puts both in $A_3$. In every case one cover member contains the antipodal pair.

The map $K:[0,1{]}^2\to G$, $K(s,t)=\alpha (s)\beta (t)$, is continuous, and $K(0,0)=K(1,0)=K(0,1)=K(1,1)=e$.

For $u,t\in [0,1]$, put $A_u(t)=\min \{\frac{t}{1-u\mathrm{/}2},1\},B_u(t)=\max \{\frac{t-u\mathrm{/}2}{1-u\mathrm{/}2},0\}\mathrm{.}$ Since $1-u\mathrm{/}2\ge 1\mathrm{/}2$, [L2] and [L3] make these functions continuous on the parameter square. Hence $H_1(t,u)=K(A_u(t),B_u(t))$ is an endpoint-fixed homotopy: at $u=0$ it is $\alpha (t)\beta (t)$, and at $u=1$ it traverses $\alpha$ first and $\beta$ second, so it is $\alpha \ast \beta$ under [F2].

The formula $H_2(t,u)=K(B_u(t),A_u(t))$ is another endpoint-fixed homotopy. At $u=0$ it is again $\alpha (t)\beta (t)$, while at $u=1$ it traverses $\beta$ first and $\alpha$ second, so it is $\beta \ast \alpha$.

If $\alpha$ or $\beta$ is replaced by an endpoint-fixed homotopic loop, multiplying the two homotopies pointwise gives an endpoint-fixed homotopy by [F1] and [L1]. Thus pointwise multiplication is well defined on loop classes, and steps 2.1 and 3.1 identify it with both concatenation orders.

The classes $[\alpha ]$ and $[\beta ]$ have concatenation product $[\alpha \ast \beta ]$, while their pointwise product is represented by $t\mapsto \alpha (t)\beta (t)$; [L1] identifies these two classes.

The same pointwise product is also homotopic to $\beta \ast \alpha$ by [L1]. Therefore $[\alpha ][\beta ]=[\alpha \ast \beta ]=[\beta \ast \alpha ]=[\beta ][\alpha ]$.

Since $[\alpha ]$ and $[\beta ]$ were arbitrary, multiplication in ${\pi }_1(G,e)$ is commutative, so the fundamental group is abelian.

For $n=2$, [L1] and [L2] identify ${\pi }_1(P_2,e_0)$ with ${\pi }_1(S^1,e_0)$, and [L3] identifies the latter with $\mathbb{Z}$.

Let $n\ge 3$ and $y\in S^{n-1}$. Since $n-1\ge 2$, [L4] says that $S^{n-1}$ is simply connected, so ${\pi }_1(S^{n-1},y)$ is trivial; [L1] and [L2] therefore make ${\pi }_1(P_n,y)$ trivial.

For an arbitrary $x\in P_n$, the path ${\gamma }_x(t)=((1-t)+t\mathrm{/}\parallel x{\parallel }_2)x$ runs in $P_n$ from $x$ to $r(x)$. Concatenating an endpoint-fixed homotopy with the fixed paths ${\bar{\gamma }}_x$ and ${\gamma }_x$ preserves it, so ${\mathrm{\Phi }}_x([\alpha ])=[{\bar{\gamma }}_x\ast \alpha \ast {\gamma }_x]$ is well defined. The piecewise formula $K(s,t)={\gamma }_x(2s(1-t))$ for $s\le 1\mathrm{/}2$ and $K(s,t)={\gamma }_x(2(1-s)(1-t))$ for $s\ge 1\mathrm{/}2$ contracts ${\gamma }_x\ast {\bar{\gamma }}_x$ to the constant path at $x$; applying the same formula to ${\bar{\gamma }}_x$ contracts ${\bar{\gamma }}_x\ast {\gamma }_x$ at $r(x)$. Hence the product of ${\mathrm{\Phi }}_x([\alpha ])$ and ${\mathrm{\Phi }}_x([\beta ])$ cancels its middle ${\gamma }_x\ast {\bar{\gamma }}_x$ and equals ${\mathrm{\Phi }}_x([\alpha \ast \beta ])$, while $[\delta ]\mapsto [{\gamma }_x\ast \delta \ast {\bar{\gamma }}_x]$ is a two-sided inverse. Thus ${\mathrm{\Phi }}_x:{\pi }_1(P_n,x)\to {\pi }_1(P_n,r(x))$ is an isomorphism, and step 1.2 makes ${\pi }_1(P_n,x)$ trivial.