The Fundamental Group of the Circle, Part 6
At last we come to the sixth and final post in our proof that the fundamental group of the circle is $\mathbb{Z}$. In the first five posts, we showed that the map $\Phi:\mathbb{Z}\to\pi_1(S^1)$ given by $n\mapsto[\omega_n]$ where $\omega_n:[0,1]\to S^1$ is the loop $s\mapsto (\cos{2\pi ns}, \sin{2\pi ns})$ is a group isomorphism. Our outline has been the following:
Part 1: Set-up/observations
Part 2: Show $\Phi$ is well defined
Part 3: Show $\Phi$ is a group homomorphism
Part 4: Show $\Phi$ is surjective
Part 5: Show $\Phi$ is injective
Part 6: Prove the two lemmas used in parts 4 and 5
While proving $\Phi$ is one to one and onto, we used two lemmas whose proofs we have deferred until now. For reference, we restate them here:
Lemma 1: For each path $f:I\to S^1$, with $f(0)=x_0$ and for each $\tilde{x_0}\in p^{-1}(x_0)$, there is a unique lift $\tilde{f}:I\to\mathbb{R}$ such that $\tilde{f}(0)=x_0$.
Lemma 2: For each homotopy $f_t:I\to S^1$ of paths starting at $f_t(0)=x_0$ and for each $\tilde{x}_0\in p^{-1}(x_0)$, there is a unique lift $\tilde{f}_t:I\to\mathbb{R}$ such that $\tilde{f}_t(0)=\tilde{x}_0$.
It turns out that both lemmas are special cases of a more general result, namely the following proposition:
Proposition: Given any space $Y$, a map $F:Y\times I\to S^1$, and a map $\tilde{F}:Y\times\{0\}\to\mathbb{R}$ which lifts $F\big|_{Y\times\{0\}}$, there exists a unique map $G:Y\times I\to \mathbb{R}$ which lifts $F$ and is an extension of $\tilde{F}$.
Notice that Lemma 1 and Lemma 2 are the special cases when $Y=\{y_0\}$ is a single point and $Y=I$ is the unit interval (pictured above), respectively.
Proving the Proposition*
We begin with an
Observation: There is an open cover $\{U_\alpha\}$ of $S^1$ such that for each $\alpha$, the set $p^{-1}(U_\alpha)$ is a disjoint union of open sets each of which is mapped homeomorphically to $U_\alpha$ by $p$. (For an example, see the footnote ** below)
Let $\{U_\alpha\}$ be an open cover of $S^1$ of the kind mentioned in the remark. Suppose $F:Y\times I\to S^1$ is any map and let $\tilde{F}:Y\times\{0\}\to\mathbb{R}$ be any map which lifts $F\big|_{Y\times\{0\}}$. Let $(y_0,t)\in Y\times I$ be any point and observe that it has a neighborhodd $N_t\times (a_t,b_t)\subset Y\times I$ such that $F(N_t\times (a_t,b_t))\subset U_\alpha \subset S^1$ for some $\alpha$. (This last bit holds simply because $\{U_\alpha\}$ is a cover for $S^1$.)
By compactness of $\{y_0\}\times I$ (since it's homeomorphic to $I$), finitely many such products $N_t\times (a_t,b_t)$ cover $y_0\times I$. Thus we can choose a single neighborhood $N$ of $y_0$ and a partition $0=t_0< t_1 < \cdots < t_m=I$ of $I$ so that for each $i$, $F(N\times [t_i,t_{i+1}])$ is containd in $U_\alpha$ for some $\alpha$, which we'll now denote by $U_i$.
Our present goal is to construct a lift $G:N\times I\to\mathbb{R}$ where $N\times I$ is the "rainbow tube." Assume by induction on $i$ that $G$ has already been constructed on $N\times [0,t_i]$. (Notice! This is true for $i=0$ since $N\times [0,0]$ is a subset of $Y\times\{0\}$ and by assumption we already have the map $\tilde{F}:Y\times\{0\}\to\mathbb{R}$.) Now since $F(N\times[t_i,t_{i+1}])\subset U_i$ we know by our opening Observation that $p^{-1}(U_i)$ is a disjoint union of open sets in $\mathbb{R}$, each of which is mapped homeomorphically onto $U_i$. Hence there exists a $\tilde{U}_i\subset p^{-1}(U_i)$ that contains the point $G(y_0,t_i)$ (this point exists by the induction hypothesis) since $(y_0,t_i)$ sits inside $N\times [0,t_i]$, and the latter maps into $U_i$ by $F$.
Now it's possible that $G(N\times\{t_i\}))$ (the lift of the top slice of $N\times [0,t_i]$) isn't fully contained within $\tilde{U}_i$ but we can just assume it is.*** Finally we may define $G$ on $N\times [t_i,t_{i+1}]$ by $$\left( p\big|_{\tilde{u}_i} \right)^{-1}\circ F\big|_{[t_i,t_{t+1}]}.$$ (This is exactly what you think it should be! Imagine for a moment that $N\times[t_i,t_{i+1}]$ is the green rectangle in our "rainbow tube" above. We can map it to $\mathbb{R}$ by first mapping it across to $S^1$ via $F$ and then lifting it up to $\tilde{U}_i$ via $p^{-1}$!)
This is well-defined since $p\big|_{\tilde{u}_i}$ is a homeomorphism, so we can repeat this inductive step finitely many times to obtain the desired lift $G:N\times I\to\mathbb{R}$ for some neighborhood $N$ of $y_0$. And since $y_0\in Y$ was arbitrary, we have our lift $G$ on all of $Y\times I$.
QED!
Footnotes:
*This proof is directly from Hathcer's Algebraic Topology chapter 1.1, but I've included the pictures to help make sense of things.
**For example, take $\{U_\alpha\}=\{U_1,U_2\}$ where $U_1=S^1\smallsetminus \{(1,0)\}$ and $U_2=S^1\smallsetminus \{(-1,0)\}$. Then $p^{-1}(U_1)=\{(n,n+1)\}_{n\in\mathbb{Z}}$ is a disjoint collection and $p(n,n+1)\cong U_1$ for all $n$. Similarly $p^{-1}(U_2)=\{(n-1/2,n+1/2)\}_{n\in\mathbb{Z}}$ is disjoint and $p(n-1/2,n+1/2)\cong U_2$ for all $n$.
*** Since we can replace $N\times {t_i}$ with the intersection $(N\times {t_i})\cap \tilde{G}\big|_{N\times\{t_i\}}^{-1}(\tilde{U}_i)$.