The Fundamental Group of the Circle — Examples

The restriction $f=p{\mathrm{\mid }}_{[0,1)}$ is continuous by [L1] and [L3]. It is surjective: for $x\in \mathbb{R}$, [L5] gives $r=x-\lfloor x\rfloor \in [0,1)$ with $p(r)=p(x)$. It is injective: if $r,s\in [0,1)$ and $p(r)=p(s)$, then [L1] gives $r-s\in \mathbb{Z}$ and $\mathrm{\mid }r-s\mathrm{\mid }<1$, so $r=s$. Thus $f$ is a continuous bijection onto $\mathbb{R}\mathrm{/}\mathbb{Z}$.

The set $A=[0,1\mathrm{/}2)$ is relatively open in $[0,1)$, since $A=(-1\mathrm{/}2,1\mathrm{/}2)\cap [0,1)$ by [L4]. Its image satisfies $p^{-1}(p[A])={\bigcup }_{n\in \mathbb{Z}}[n,n+1\mathrm{/}2)$ by [L1]. This union is not open at any integer, in particular at $0$, so [L2] says $p[A]$ is not open in the quotient. Hence the continuous bijection $f$ is not open and is not a homeomorphism by [L6].

The other endpoint convention fails differently: on $[0,1]$ one has $p(0)=p(1)$ by [L1], so the restriction is not injective and cannot be a homeomorphism onto its image. Both intervals have length one, which refutes the proposed replacement of the strict bound in [L7] by length at most one.

The two formulas for $u$ agree at $t=1\mathrm{/}2$, where both equal $1$, and each piece is continuous by [L8], so [L3] makes $u$ continuous. It has $u(0)=0$ and $u(1)=1$, hence [L7] makes $\alpha =p\circ u$ a based loop. Since $u$ starts at zero and projects to $\alpha$, it is the defining lift and [L4] gives $\mathrm{deg}(\alpha )=u(1)=1$.

By [L1], the standard loop ${\omega }_1$ is the projection of $v(t)=t$, and [L5] gives $\mathrm{deg}({\omega }_1)=1=\mathrm{deg}(\alpha )$.

The formula $K(t,s)=(1-s)u(t)+st$ is a continuous homotopy from $u$ to $v$ by [L6]. Since $u(0)=v(0)=0$ and $u(1)=v(1)=1$, it fixes both endpoints for every $s$. Postcomposing with $p$ gives the explicit path homotopy $H(t,s)=p(K(t,s))$ from $\alpha$ to ${\omega }_1$, relative to $t=0,1$.

At $t=1\mathrm{/}2$ both formulas give $1$. Each piece is continuous by [L5], so [L4] makes $u$ continuous. Its values at the boundary are $u(0)=0$, $u(1\mathrm{/}2)=1$, and $u(1)=0$.

By [L1] and [L7], $\alpha =p\circ u$ is a continuous based loop at $[0]$. The function $u$ is a lift of $\alpha$ beginning at zero, so [L2] gives $\mathrm{deg}(\alpha )=u(1)=0$.

Let $[x]\in \mathbb{R}\mathrm{/}\mathbb{Z}$. By [L6], $r=x-\lfloor x\rfloor \in [0,1)$ and $p(r)=p(x)$. For $t=r\mathrm{/}2\in [0,1\mathrm{/}2)$, the first formula gives $u(t)=r$, so $\alpha (t)=[x]$; hence $\alpha$ is surjective. It is nonconstant because $\alpha (0)=[0]$ while $\alpha (1\mathrm{/}4)=[1\mathrm{/}2]\mathrm{\ne }[0]$, the latter inequality following from $1\mathrm{/}2\mathrm{\notin }\mathbb{Z}$ in [L1].

Step 2.1 gives degree zero, so the reverse direction of [L3] makes $\alpha$ nullhomotopic. Step 3.1 shows that nullhomotopy here neither forces constancy nor prevents surjectivity.

Under the homeomorphism of [L1], the loop in [L3] has image $t\mapsto (\cos (2\pi nt),\sin (2\pi nt))={\gamma }_n(t)$.

The quotient loop has degree $n$ by [L2], so the transported degree of its geometric image ${\gamma }_n$ is also $n$. At $n=0$ this is the constant loop, and the same computation covers every negative $n$.

By [L4], ${\omega }_0(0)={\omega }_0(1)=[0]$ and ${\omega }_1(0)=[0]=[1]={\omega }_1(1)$, so both satisfy the two endpoint equalities in [L1]. Their degrees are $0$ and $1$ by [L3].

If ${\omega }_0$ and ${\omega }_1$ were path-homotopic relative to the endpoints, then ${\omega }_1$ would be path-homotopic to the constant loop and hence nullhomotopic. But [L2] and [L3] rule this out because $\mathrm{deg}({\omega }_1)=1\mathrm{\ne }0$. Thus equal endpoints do not imply path homotopy.

The real line is a nonempty convex subset of ${\mathbb{R}}^1$, so [L1] with $n=1$ makes $\mathbb{R}$ simply connected.

The same explicit map $p$ is a quotient map by [L2] and a covering map by [L3].

Its base is not simply connected by [L4], whereas its total space is simply connected by step 1.1. Step 1.2 therefore supplies both announced failures of preservation.

Suppose, for contradiction, that the identity is nullhomotopic. By [A1], there are $c\in \mathbb{R}\mathrm{/}\mathbb{Z}$ and a homotopy $H$ with $H(y,0)=y$ and $H(y,1)=c$. Reverse its time coordinate to obtain $K(y,t)=H(y,1-t)$, so $K(y,0)=c$ and $K(y,1)=y$.

Since $p$ is surjective, choose $a\in \mathbb{R}$ with $p(a)=c$. The constant map $y\mapsto a$ lifts $K(-,0)$, so [L1] and [L2] give a lift $\tilde{K}:(\mathbb{R}\mathrm{/}\mathbb{Z})\times I\to \mathbb{R}$. Define $s(y)=\tilde{K}(y,1)$. Then $s$ is continuous and $p(s(y))=K(y,1)=y$, so $p\circ s$ is the identity: $s$ is a section of $p$.

The path $s\circ {\omega }_1$ is a lift of ${\omega }_1$ because $p\circ s$ is the identity, and it is closed because ${\omega }_1(0)={\omega }_1(1)=[0]$. Put $m=s([0])\in \mathbb{Z}$ by [L5]. The path $t\mapsto m+t$ is another lift of ${\omega }_1$ starting at $m$, since [L4] and [L5] give $p(m+t)=[t]$; it is continuous by [L6]. Uniqueness in [L3] forces $s({\omega }_1(t))=m+t$, whose endpoint is $m+1\mathrm{\ne }m$, contradicting that $s\circ {\omega }_1$ is closed.

The contradiction discharges the assumption of step 1.1. Hence the identity is not nullhomotopic, and the universal claim is false.