Elaborating why the fundamental group of the circle is infinite cyclic.

Question

The book of "Introduction to knot theory" of Richard H. Crowell, gives the following proof:

where (5.7) is:

"Equivalent loops have equal winding numbers."

And (5.5) is:

$$r_{a.b} = r_{a} + r_{b}.$$

where $r_{a}$ is a uniquely determined integer, which we call the winding number of $a$. Geometrically, $r_{a}$ is the algebraic number of times the loop $a$ wraps around the circle.

And (5.6) is:

Loops with equal winding numbers are equivalent.

My questions are:

1-why (5.6) proves that the homomorphism is, in fact, an isomorphism ?

2-why there is no retraction from a space with trivial fundamental group to a subspace of an infinite cyclic fundamental group?

3-Intuitively what does it mean for a subspace to have infinite cyclic fundamental group?

4-Why the definition of multiplication in the fundamental group and (5.5) show that this association is a homomorphism into the additive group of integers?

Could anyone help me in answering these questions please?

Answer 1

I haven't read Crowell's book, but I have read the proof in Massey's Algebraic Topology, which appears to be similar in flavor. Let's start with an outline of the proof:

• Given a differentiable function $f:[0,1] \to S^1$ with $f(0)=f(1)=1\in S^1$, we can calculate the winding number $\frac{1}{2\pi}\int_f d\theta$, where $d\theta$ measures changes in angle, measured in radians. (This is where covering space theory might come in: there is no corresponding global parameterization $\theta$ of $S^1$, since if you go around the circle once it will have to have a discontinuity, a fact the fundamental group will measure. The $\mathbb{R}$ cover of $S^1$ is a sort of "extended angle" approach, where the "extended angles" $2\pi k+s\in\mathbb{R}$ for $k\in\mathbb{Z}$ all represent the same point of $S^1$. There is a way to take a path like $f$ and "lift" it to a path $\widetilde{f}:[0,1]\to \mathbb{R}$ through the "extended angles," and it turns out that $\frac{1}{2\pi}(\widetilde{f}(1)-\widetilde{f}(0))$ is the winding number!)
• Observe that winding numbers are $\mathbb{Z}$-valued. We actually do not need differentiability of the path, using the following idea. The circle can be written as the union of two open arcs $U,V\subset S^1$ whose intersection is a disjoint union of two smaller open arcs containing $1$ and $-1\in S^1$, respectively. (I am thinking of $S^1$ as being the complex unit circle.) The preimages $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint unions of arcs. By compactness of $[0,1]$, there is a finite collection of (relatively) open arcs $J_1,\dots, J_n\subseteq [0,1]$ with $\bigcup_iJ_i=[0,1]$ such that for all $i$ either $f(J_i)\subseteq U$ or $f(J_i)\subseteq V$. This means there is a sequence $0=t_0<t_1<\dots<t_n=1$ such that $f([t_i,t_{i+1}])\subset U$ or $f([t_i,t_{i+1}])\subset V$ (or both) for all $i$. We can arrange for $f(t_i)\in\{1,-1\}$ for all $i$ by homotoping $f$ to have small spurs, which is possible since each component of $U\cap V$ is homeomorphic to $\mathbb{R}$, so straight-line homotopies exist. Since $U$ and $V$ are both homeomorphic to $\mathbb{R}$, we can homotope $f|_{[t_i,t_{i+1}]}$ so that it takes a shortest path for each $i$ from $f(t_i)$ to $f(t_{i+1})$. If $f$ is stationary on any interval, we can homotope $f$ to remove that interval. Thus, on each interval $f$ either makes a half turn clockwise or counterclockwise. If $f$ ever changes directions, then there is some pair of intervals $[t_{i-1},t_i]$ and $[t_i,t_{i+1}]$ where $f$ goes in opposite directions, and you can homotope $f$ so it remains stationary on $[t_{i-1},t_{i+1}]$. Thus, we may eventually homotope $f$ so that it is monotonically clockwise or counterclockwise. After one more homotopy, we can assume $t_i=i/n$ for each $i$. This implies $f$, in this form, can be parameterized by $f(t)=e^{2\pi i kt}$ for some $k\in\mathbb{Z}$, and we can call $k$ the winding number of the original $f$.
• This version of winding number is well-defined since homotopy is an equivalence relation. If $f\sim g$, $f\sim e^{2\pi ikt}$, and $g\sim e^{2\pi i\ell t}$, then $e^{2\pi ikt}\sim e^{2\pi i\ell t}$. The generalized Stokes theorem implies line integrals over homotopic loops are equal, so the integral-based winding number definition implies $k=\ell$. This also implies there is a well-defined function $\phi:\pi_1(S^1)\to \mathbb{Z}$ defined by taking the winding number of a representative loop from a given homotopy class.
• The $\phi$ function is a homomorphism. We already showed each homotopy class is represented by a function $e^{2\pi ikt}$ for some $k$. The loop composition of $e^{2\pi ikt}$ and $e^{2\pi i\ell t}$ is a piecewise differentiable loop with winding number $k+\ell$, so $\phi$ is a homomorphism to $\mathbb{Z}$ as an additive group.
• Every winding number is attainable since $e^{2\pi ikt}$ has winding number $k$. Hence, $\phi$ is surjective.
• Injectivity was that every loop is homotopic to some $e^{2\pi ikt}$, which has zero winding number only if $k=0$, which would imply the original loop was nullhomotopic.
• Therefore $\phi:\pi_1(S^1)\to \mathbb{Z}$ is an isomorphism.

Your questions

1. 5.6 is that the homomorphism is an injection. Since it is also a surjection, the homomorphism is an isomorphism.

2. For a subspace $A\subset X$ with inclusion map $i:A\to X$, a retraction $r:X\to A$ is a map such that $r\circ i=\operatorname{id}_A$. That is, where $r(a)=a$ for all $a\in A$. Since $\pi_1$ is a functor, the induced maps satisfy $r_*\circ i_*=\operatorname{id}_{\pi_1(A)}$, which means $r_*$ is surjective. If $\pi_1(X)=1$, then by surjectivity $\pi_1(A)=1$. This is why $\pi_1(A)$ cannot be nontrivial if $\pi_1(X)$ is trivial.

3. An infinite cyclic fundamental group means there is one thing you can go around. If you try to define a function $X\to \mathbb{R}$ locally, there is no guarantee the local definitions will come together into a global definition; for example, the complex logarithm $\log:\mathbb{C}-\{0\}\to \mathbb{C}+2\pi i\mathbb{Z}$ is only defined up to addition by integer multiples of $2\pi i$, which is sort of a consequence of $\pi_1(\mathbb{C}-\{0\})\cong \mathbb{Z}$. If you've heard of branch cuts, that is where you cut $\mathbb{C}-\{0\}$ along a ray to trivialize its fundamental group.

4. For groups, all you need to check of a function to see it is a homomorphism is that $r_{a\cdot b}=r_a+r_b$ (that the composition in one group is carried to the composition of the next; if $\psi:G\to A$ is a function from a multiplicative group to an additive group, then if $\psi(gh)=\psi(g)+\psi(h)$ for all $g,h\in G$, the function is a homomorphism). Here, $r:\pi_1(S^1)\to \mathbb{Z}$ is the function that 5.5 is describing to be a homomorphism.