The Fundamental Group of the Circle

A real $x$ lies in $p^{-1}(p[U])$ exactly when $p(x)=p(u)$ for some $u\in U$, which by [L1] is equivalent to $x-u=n$ for some $n\in \mathbb{Z}$; hence $p^{-1}(p[U])={\bigcup }_{n\in \mathbb{Z}}(U+n)$. Each translate $U+n$ is open: translating by $n$ and by $-n$ gives mutually inverse continuous maps by [L4]. The union is open, so the quotient-topology criterion in [L1] makes $p[U]$ open. Thus $p$ is open in the sense of [L3], including when $U=\mathrm{\varnothing }$.

Suppose $x,y\in J$ and $p(x)=p(y)$. Then $k:=x-y\in \mathbb{Z}$ by [L1], while $\mathrm{\mid }k\mathrm{\mid }=\mathrm{\mid }x-y\mathrm{\mid }\le \mathrm{\ell }<1$. If $0<k<1$, both $0$ and $k$ satisfy the integer-part inequalities for the real $k$, contrary to uniqueness in [L2]; applying the same argument to $-k$ excludes $-1<k<0$. Hence $k=0$ and $x=y$, so $p{\mathrm{\mid }}_J$ is injective. This also covers a singleton interval; for an empty interval injectivity is vacuous.

The restriction $p{\mathrm{\mid }}_J$ is continuous and is a bijection onto $p[J]$ by step 1.2. To prove its inverse continuous, let $O$ be relatively open in $J$ and $x\in O$. Choose $\delta >0$ with $J\cap (x-\delta ,x+\delta )\subseteq O$, and put $r=\frac{1}{2}\min \{\delta ,1-\mathrm{\ell }\}>0$. Step 1.1 makes $W:=p[(x-r,x+r)]$ open. If $z\in J$ and $p(z)\in W$, choose $y\in (x-r,x+r)$ with $p(z)=p(y)$; then $z-y\in \mathbb{Z}$ by [L1] and $\mathrm{\mid }z-y\mathrm{\mid }\le \mathrm{\mid }z-x\mathrm{\mid }+\mathrm{\mid }x-y\mathrm{\mid }<\mathrm{\ell }+r<1$, so [L2] gives $z=y$. Thus $z\in J\cap (x-\delta ,x+\delta )\subseteq O$, and $W\cap p[J]\subseteq p[O]$. Every point of $p[O]$ therefore has a relative open neighbourhood contained in $p[O]$, so $p[O]$ is open in $p[J]$. The empty case has the unique empty inverse. Hence $p{\mathrm{\mid }}_J$ is a homeomorphism onto its image, and therefore an embedding by [L3].

The interval $J_x=(x-1\mathrm{/}3,x+1\mathrm{/}3)$ has length $2\mathrm{/}3<1$. Its image $U_x=p[J_x]$ is open by [L1] and contains $p(x)=[x]$.

By the saturation formula in [L1], $p^{-1}(U_x)={\bigcup }_{n\in \mathbb{Z}}(J_x+n)$. These open intervals are pairwise disjoint: if $y$ belonged to $J_x+m$ and $J_x+n$, then $\mathrm{\mid }(y-m)-(y-n)\mathrm{\mid }=\mathrm{\mid }n-m\mathrm{\mid }<2\mathrm{/}3$, and an integer of absolute value below one is zero, so $m=n$.

Every translate $J_x+n$ has length $2\mathrm{/}3<1$, so [L1] makes $p{\mathrm{\mid }}_{J_x+n}$ a homeomorphism onto its image; its image is $p[J_x+n]=p[J_x]=U_x$. The map $p$ is already a continuous surjection because it is a quotient projection, and steps 1.1 and 1.2 give an open neighbourhood with a disjoint union of open sheets. Thus every clause of [L2] holds, and $p$ is a covering map.

For $x\in \mathbb{R}$, let $m=\lfloor x\rfloor$ from [L1] and put $r=x-m$. Then $0\le r<1$ and $p(r)=p(x)$ by [L5]. Hence $p{\mathrm{\mid }}_{[0,1]}$ is surjective onto $\mathbb{R}\mathrm{/}\mathbb{Z}$.

The interval $[0,1]$ is closed and bounded, so [L2] makes it compact for the usual metric and [L6] makes it compact as a topological subspace. The quotient projection is continuous by [L5], and its restriction remains continuous. By step 1.1 its image is all of $\mathbb{R}\mathrm{/}\mathbb{Z}$, so [L3] proves that $\mathbb{R}\mathrm{/}\mathbb{Z}$ is compact.

Let $[x],[y]\in \mathbb{R}\mathrm{/}\mathbb{Z}$. The affine map $a(t)=(1-t)x+ty$ is continuous by [L7], and $\gamma =p\circ a$ is continuous by [L5] and [L8]. Its endpoints are $\gamma (0)=[x]$ and $\gamma (1)=[y]$. Thus [L4] gives a path between every pair of classes, so the quotient is path-connected. If the classes agree, the same conclusion also follows from the constant path.

Choose representatives $x,y\in \mathbb{R}$ of $\xi ,\eta$ and put $a=x-\lfloor x\rfloor$, $b=y-\lfloor y\rfloor$. By [L2], $a,b\in [0,1)$; by [L4], $[a]=\xi$ and $[b]=\eta$. Distinctness gives $a\mathrm{\ne }b$, so $d:=\mathrm{\mid }a-b\mathrm{\mid }$ and $e:=1-d$ are both positive.

Put $r=\frac{1}{3}\min \{d,e\}>0$, and let $U=p[(a-r,a+r)]$ and $V=p[(b-r,b+r)]$. Both are open by [L1], and they contain $\xi$ and $\eta$, respectively.

Suppose $U\cap V\mathrm{\ne }\mathrm{\varnothing }$. Then some $u\in (a-r,a+r)$ and $v\in (b-r,b+r)$ have $p(u)=p(v)$, so $k:=u-v\in \mathbb{Z}$ by [L4] and $\mathrm{\mid }k-(a-b)\mathrm{\mid }<2r$. But $a-b\in (-1,1)\setminus \{0\}$, and its distance from every integer is at least $\min \{\mathrm{\mid }a-b\mathrm{\mid },1-\mathrm{\mid }a-b\mathrm{\mid }\}=\min \{d,e\}>2r$: the candidates $0$ and the nearer of $1,-1$ give those two distances, while every other integer is farther away. This is a contradiction. Thus $U$ and $V$ are disjoint open neighbourhoods, and [L3] proves the Hausdorff condition.

Let $\tilde{\alpha }$ and $\tilde{\beta }$ be the unique lifts used in [L1]. Both begin at the common point $0$.

Since the base loops are endpoint-fixed homotopic and the lifts have the same initial point, [L2] gives $\tilde{\alpha }(1)=\tilde{\beta }(1)$.

Reading these terminal values through [L1] yields $\mathrm{deg}(\alpha )=\mathrm{deg}(\beta )$.

If $[\alpha ]=[\beta ]$ in the set of [L2], then $\alpha$ and $\beta$ are path-homotopic relative to their endpoints, so [L1] gives $\mathrm{deg}(\alpha )=\mathrm{deg}(\beta )$.

Therefore the displayed rule is independent of the representative and defines one function on ${\pi }_1(S^1,[0])$. No representative-selection function is used.

The path $t\mapsto nt$ starts at zero and projects to ${\omega }_n$ by [L1]. The uniqueness clause of [L3] therefore identifies it with the lift used to define the degree of ${\omega }_n$.

Its terminal value is $n\cdot 1=n$, so [L2] gives $\mathrm{deg}({\omega }_n)=n$. This calculation is uniform for $n=0$, positive $n$, and negative $n$.

Define $\gamma$ by the displayed two-piece formula. At $t=1\mathrm{/}2$ the left value is $\tilde{\alpha }(1)=m$ and the right value is $m+\tilde{\beta }(0)=m$, so [L4] and [L6] make $\gamma$ continuous. It starts at zero. By [L5], its first half projects to $\alpha (2t)$ and its second half to $\beta (2t-1)$, in the order fixed by [L2], so $p\circ \gamma =\alpha \ast \beta$; its endpoint is $m+n$.

Define $\delta (t)=\tilde{\alpha }(1-t)-m$. It is continuous by [L6], begins at $m-m=0$, and ends at $0-m=-m$. Since $m\in \mathbb{Z}$ by [L1], [L5] gives $p(\delta (t))=p(\tilde{\alpha }(1-t))=\alpha (1-t)=\bar{\alpha }(t)$.

Both $\gamma$ and $\delta$ are lifts with initial point zero, so uniqueness in [L3] identifies them with the defining lifts of $\alpha \ast \beta$ and $\bar{\alpha }$. Their endpoints are therefore $m+n$ and $-m$, respectively.

By [L1], the lift of $\alpha \ast \beta$ from zero ends at $\mathrm{deg}(\alpha )+\mathrm{deg}(\beta )$. Reading that endpoint through [L2] gives $\mathrm{deg}(\alpha \ast \beta )=\mathrm{deg}(\alpha )+\mathrm{deg}(\beta )$.

By the reversal formula in [L1], the lift of $\bar{\alpha }$ from zero ends at $-\mathrm{deg}(\alpha )$, so [L2] gives $\mathrm{deg}(\bar{\alpha })=-\mathrm{deg}(\alpha )$.

The constant path at zero is a lift of the constant loop $c_{[0]}$ and starts at zero; uniqueness in [L3] makes it the defining lift. Its terminal value is zero, so [L2] gives $\mathrm{deg}(c_{[0]})=0$.

By [L5], [L1], and the concatenation law in [L2], $\mathrm{Deg}([\alpha ][\beta ])=\mathrm{Deg}([\alpha \ast \beta ])=\mathrm{deg}(\alpha \ast \beta )=\mathrm{deg}(\alpha )+\mathrm{deg}(\beta )=\mathrm{Deg}([\alpha ])+\mathrm{Deg}([\beta ])$.

The target is the additive group of the integers by [L4], and step 1.1 is exactly the product-preservation condition of [L3]. Hence $\mathrm{Deg}$ is a group homomorphism. Its identity and inverse laws also agree with the zero and negation formulas of [L2].

Let $\tilde{\alpha }$ and $\tilde{\beta }$ be the lifts from [L1]. Both begin at zero, and the degree hypothesis with [L1] gives $\tilde{\alpha }(1)=\tilde{\beta }(1)$.

Since $\mathbb{R}$ is convex, [L2] makes $H(t,s)=(1-s)\tilde{\alpha }(t)+s\tilde{\beta }(t)$ continuous. Step 1.1 gives $H(0,s)=0$ and $H(1,s)=\tilde{\alpha }(1)=\tilde{\beta }(1)$ for every $s$, so this homotopy fixes both endpoints.

Postcomposing with the continuous quotient projection, [L3] and [L4] give an endpoint-fixed homotopy $p\circ H$. The defining lift equations in [L1] identify its endpoints as $p\circ \tilde{\alpha }=\alpha$ and $p\circ \tilde{\beta }=\beta$. Hence the loops are path-homotopic.

If $\alpha$ and $\beta$ are path-homotopic, then [L1] gives $\mathrm{deg}(\alpha )=\mathrm{deg}(\beta )$.

Conversely, if $\mathrm{deg}(\alpha )=\mathrm{deg}(\beta )$, then [L2] gives an endpoint-fixed path homotopy from $\alpha$ to $\beta$.

Steps 1.1 and 1.2 prove the forward and reverse implications, respectively, so the stated biconditional holds.

If $\alpha$ is nullhomotopic, then it is path-homotopic to the constant loop ${\omega }_0$ by [L3]. The forward implication of [L1] and [L2] give $\mathrm{deg}(\alpha )=\mathrm{deg}({\omega }_0)=0$.

Conversely, if $\mathrm{deg}(\alpha )=0$, then [L2] gives $\mathrm{deg}(\alpha )=\mathrm{deg}({\omega }_0)$. The reverse implication of [L1] makes $\alpha$ path-homotopic to ${\omega }_0$, which is the required based nullhomotopy by [L3].

Steps 1.1 and 1.2 establish both directions of the degree-zero criterion.

By [L1], $\mathrm{Deg}$ is a group homomorphism into the additive group of integers supplied by [L5].

If $\mathrm{Deg}([\alpha ])=\mathrm{Deg}([\beta ])$, then $\mathrm{deg}(\alpha )=\mathrm{deg}(\beta )$, so [L2] makes $\alpha$ and $\beta$ path-homotopic. Their classes are equal by [L6], and $\mathrm{Deg}$ is injective.

Let $n\in \mathbb{Z}$. By [L3], $\mathrm{Deg}([{\omega }_n])=\mathrm{deg}({\omega }_n)=n$, so every integer is attained and $\mathrm{Deg}$ is surjective. This includes $n=0$, $n=1$, and negative integers.

Steps 1.1, 1.2, and 1.3 show that $\mathrm{Deg}$ is a bijective group homomorphism, hence an isomorphism by [L4]. Step 1.3 also shows that $n\mapsto [{\omega }_n]$ is its inverse, since injectivity makes this preimage unique.

The quotient is nonempty because it contains $[0]$ by [L5], and it is path-connected by [L1].

By [L3], $\mathrm{deg}({\omega }_1)=1\mathrm{\ne }0$. The criterion [L2] therefore shows that ${\omega }_1$ is not nullhomotopic, so its loop class differs from the constant-loop class.

Thus the fundamental group at $[0]$ does not have exactly one element. Although step 1.1 supplies the other two clauses of [L4], this failure at one basepoint violates the definition, so $\mathbb{R}\mathrm{/}\mathbb{Z}$ is not simply connected.