Applications of the Fundamental Group — Examples

Restricting the retraction in [L1] to $D^2\setminus \{0\}$ gives a continuous map into $S^1$, and every $x\in S^1$ satisfies $\rho (x)=x$. Thus $\rho$ retracts the punctured disk onto the unit circle.

If a continuous extension $\bar{\rho }:D^2\to S^1$ existed, it would still satisfy $\bar{\rho }(x)=x$ on $S^1$, so it would be a retraction, contrary to [L2].

The failure at the missing point is also visible directly: for every $0<t\le 1$, $\rho (t,0)=(1,0)$ while $\rho (-t,0)=(-1,0)$. The two radial approaches to $0$ therefore have different constant images, so $\rho$ has no limit at $0$.

The coefficient list is $(2,-2,0,1)$, so $p$ is monic of degree $3$ and the lower-coefficient modulus sum is $2+2+0=4<5$.

The hypotheses of [L1] hold with $n=3$ and $R=5$, so the normalized loop has degree $3$. Explicitly, on $\mathrm{\mid }z\mathrm{\mid }=5$ and for $0\le s\le 1$, one has $\mathrm{\mid }s(-2z+2)\mathrm{\mid }\le s(10+2)\le 12<125=\mathrm{\mid }{z}^3\mathrm{\mid }$, so the displayed homotopy never meets zero; its endpoints are $z^3$ and $z^3-2z+2$.

Both coordinate functions of $q$ are continuous, so $q$ is continuous by [L2].

If $v=(x,y,z)\in S^2$ satisfies $q(v)=q(-v)$, then $(x,y)=(-x,-y)$ and hence $x=y=0$. The unit-sphere equation gives $z^2=1$, so $v=(0,0,1)$ or $v=(0,0,-1)$. These are two points forming exactly one antipodal pair, and they do have equal image $(0,0)$, in agreement with [L1].

Let $v_1=\frac{(1,1,1)}{\sqrt{3}},v_2=\frac{(1,-1,-1)}{\sqrt{3}},v_3=\frac{(-1,1,-1)}{\sqrt{3}},v_4=\frac{(-1,-1,1)}{\sqrt{3}},$ and for $1\le i\le 4$ define $A_i=\{x\in S^2:⟨x,v_i⟩\ge ⟨x,v_j⟩\text{ for every }1\le j\le 4\}\mathrm{.}$ The vectors $v_i$ are the vertices of a regular tetrahedron centred at $0$.

By [F2] and [L1], each $A_i$ is a finite intersection of sets defined by a continuous closed inequality $⟨x,v_i-v_j⟩\ge 0$, hence is closed in $S^2$. For every $x\in S^2$, the finite set of four real numbers $⟨x,v_j⟩$ has a maximum, so $x$ belongs to at least one $A_i$. Thus $A_1,A_2,A_3,A_4$ cover $S^2$.

Suppose $x,-x\in A_i$. Then $⟨x,v_i-v_j⟩\ge 0$ and $⟨-x,v_i-v_j⟩\ge 0$ for every $j$, so all these inner products vanish. For $i=1$, the vectors $v_1-v_2$, $v_1-v_3$, and $v_1-v_4$ are scalar multiples of $(0,1,1)$, $(1,0,1)$, and $(1,1,0)$, whose determinant is $2\mathrm{\ne }0$; the other values of $i$ differ only by coordinate sign changes and permutations. Hence the three differences span ${\mathbb{R}}^3$, forcing $x=0$, contrary to $x\in S^2$. No $A_i$ contains an antipodal pair.

Steps 2.1 and 2.2 give a closed four-set cover with no antipodal pair in any member, refuting the proposed extension and showing that the three-set conclusion [F1] cannot be enlarged in this way.

Define $r_n:\mathbb{H}\to C_n$ by $r_n(x)=x$ for $x\in C_n$ and $r_n(x)=0$ for $x\in C_m$ with $m\mathrm{\ne }n$. Distinct circles meet only at $0$: subtracting their equations $x_1^2+x_2^2=2x_1\mathrm{/}m$ and $x_1^2+x_2^2=2x_1\mathrm{/}k$ gives $x_1=x_2=0$. Thus the clauses agree and define a function.

Let $x\in C_m\setminus \{0\}$ and put $d=\parallel x{\parallel }_2>0$. By [L2], choose $N$ with $2\mathrm{/}N<d\mathrm{/}2$; then every $C_k$ with $k\ge N$ lies in $B(0,d\mathrm{/}2)$ and is disjoint from $B(x,d\mathrm{/}2)$. For each of the finitely many $k<N$ with $k\mathrm{\ne }m$, the positive number $\mathrm{\mid }\parallel x-(1\mathrm{/}k,0){\parallel }_2-1\mathrm{/}k\mathrm{\mid }$ and [L3] give a ball about $x$ missing $C_k$; intersect these finitely many balls with $B(x,d\mathrm{/}2)$. The resulting relative neighbourhood in $\mathbb{H}$ meets only $C_m$, so there $r_n$ is the identity when $m=n$ and the constant map $0$ when $m\mathrm{\ne }n$.

Let $V\subseteq C_n$ be open and let $x\in r_n^{-1}[V]$. If $x\mathrm{\ne }0$ and $x\in C_m$, step 2.1 gives a relative open neighbourhood $N$ of $x$ meeting only $C_m$. When $m=n$, write $V=O\cap C_n$ by [F3] and replace $N$ by $N\cap O$; the identity clause of $r_n$ then maps it into $V$. When $m\mathrm{\ne }n$, one has $r_n(x)=0\in V$ and the constant clause maps all of $N$ into $V$. If $x=0$, then $0\in V$; write $V=O\cap C_n$ and choose an ambient open neighbourhood $W$ of $0$ with $W\subseteq O$. Every point of $W\cap C_n$ is fixed and every point of $W\cap C_m$ for $m\mathrm{\ne }n$ maps to $0$, so $W\cap \mathbb{H}\subseteq r_n^{-1}[V]$. Thus every point of $r_n^{-1}[V]$ has a relative open neighbourhood inside it, making that preimage open. The criterion [L1] now shows that $r_n$ is continuous.

The continuous map $r_n$ restricts to the identity on $C_n$, so it is a retraction of $\mathbb{H}$ onto $C_n$.

By [F1], $\mathbb{H}$ satisfies the compactness and path-connectedness hypotheses of the proposed statement.

Let $U$ be any neighbourhood of $0$ in $\mathbb{H}$. By [F4], there is an ambient open set $O\subseteq {\mathbb{R}}^2$ with $0\in O$ and $O\cap \mathbb{H}\subseteq U$. By [F5], choose $\epsilon >0$ with $B(0,\epsilon )\subseteq O$, so $\mathbb{H}\cap B(0,\epsilon )\subseteq U$. Since $C_n\subseteq {\bar{B}}_2(0,2\mathrm{/}n)$, [L4] gives an $n\ge 1$ with $C_n\subseteq U$.

Let $j:C_n\hookrightarrow \mathbb{H}$ be inclusion and $r_n:\mathbb{H}\to C_n$ the retraction of [F2]. Functoriality gives $(r_n{)}_{\ast }\circ j_{\ast }=\mathrm{id}$, so $j_{\ast }$ is injective. The pointed affine homeomorphism $z\mapsto (1\mathrm{/}n,0)-(1\mathrm{/}n)z$ carries the geometric unit circle based at $(1,0)$ onto $C_n$ based at $0$ and carries the once-around loop of [L2] to a loop ${\mathrm{\ell }}_n$ in $C_n$. By [L2] and [L3], $[{\mathrm{\ell }}_n]\mathrm{\ne }1$ in ${\pi }_1(C_n,0)$; injectivity of $j_{\ast }$ therefore makes its image nontrivial in ${\pi }_1(\mathbb{H},0)$. Since $C_n\subseteq U$, the same ${\mathrm{\ell }}_n$ is a loop in $U$.

Since every neighbourhood $U$ of $0$ contains such a loop, no inclusion-induced map ${\pi }_1(U,0)\to {\pi }_1(\mathbb{H},0)$ is trivial. Thus $\mathbb{H}$ is not semilocally simply connected at $0$.

By [L1], the Hawaiian earring has no universal cover. Together with step 1.1, it is a compact path-connected planar counterexample to the proposed universal claim.