Fundamental Group of the Circle

While on holiday, I started reading a couple of on-line texts on algebraic topology on my Kindle (Hatcher 2002), (May 1999). Kindle’s aren’t great for reading maths but they save having to carry around heavy books of which one might only read a fraction. Along with the usual holiday activities (ok not that usual as we did cycle into Burma), I made it as far as calculating the fundamental group of the circle, \pi_1(S^1, 1).

The idea is that two continuous functions are homotopic if they can be deformed continuously into each other. All functions from now on are considered continuous unless otherwise specified.

More formally if f, g : X \rightarrow Y are homotopic then there exists a function F : X \times [0,1] \rightarrow Y such that F(x,0) = f(x) and F(x,1) = g(x).

A path is a map f : \mathbb{I} \rightarrow X where \mathbb{I} is the unit interval [0,1]. A homotopy of paths is a function F : \mathbb{I} \times \mathbb{I} \rightarrow X such that F(0,t) = x_0 and F(1,t) = x_1. We write f_t(x) \triangleq F(x,t). The two paths f_0(x) and f_1(x) are homotopic and we write f_0 \simeq f_1. Homotopy of paths is an equivalence relation, a fact which we do not prove here. The equivalence class of a path f is denoted [f].

Where the end point of one path is the start point of another path, we can join these together to form new paths. This joining operation respects homotopy classes as the diagram below illustrates.

A loop is a path with f(0) = f(1) = x_0.

We can join loops based at the same point together to form new loops giving a group operation on equivalence classes under homotopy. In more detail, suppose that f and g are loops then we can form the new loop f \cdot g which is the obvious loop formed by first traversing f and then traversing g. The inverse of the loop f is the loop f^{-1}(t) = f(1-t) and the identity is the constant loop e(t) = x_0. The way we have defined homotopy ensures that this is all well defined (see either of the on-line books for the details).

The group of loops under these group operations is called the fundamental group or the first homotopy group and is denoted \pi_1(X, x_0).

Theorem \pi(S^1, 1) \cong (\mathbb{Z}, +) where the latter denotes the integers with the group structure being given by addition.

Proof Define

\displaystyle   i : \mathbb{Z} \rightarrow \pi_1(S^1, 1)


\displaystyle   i(n) = \big[\lambda t : \mathbb{I} \rightarrow e^{2\pi int}\big]

that is for each integer we define an equivalence class of loops. In more standard notation (not using \lambda as a way of introducing anonymous functions)

\displaystyle   i(n) = [f_n]


\displaystyle   f_n(t) = e^{2\pi int}

Further we have i(n + m) = [f_{n+m}] and clearly f_{n+m} \simeq f_n \cdot f_m. Thus i(n+m) = [f_n][f_m] =i(n)i(m) making i a homomorphism.

We wish to show that i is an isomorphism. To do this we are going to take a loop f on S^1 and lift it to a path on \mathbb{R} so that the map p : \mathbb{R} \rightarrow S^1 which wraps \mathbb{R} around the circle p(t) = e^{2\pi\imath t} projects this lifted map \tilde{f} back to the original map: p \circ \tilde{f} = f. Since f(1) = 1 we must have that \tilde{f}(1) = n for some integer n (this will be the number of times the loop winds around the origin). Let us call the map that creates this integer from a loop on the circle

\displaystyle   j : \pi_1(S^1,1) \rightarrow \mathbb{Z}

Note that we also have f(0) = 1 so the lifted path could start at any integer. For the sake of definiteness let us start it at 0.

We now construct this map, j, show that it is well defined and is a homomorphism.

We know that S^1 is a smooth manifold. Let us give it an atlas.

\displaystyle   \begin{aligned}  U_1 &= \{ z : z = e^{2\pi\imath t}, \pi / 6 < t < 5\pi / 6\} \\  U_2 &= \{ z : z = e^{2\pi\imath t}, 4\pi / 6< t < 8\pi / 6\} \\  U_3 &= \{ z : z = e^{2\pi\imath t}, 7\pi / 6< t < 11\pi / 6\} \\  U_4 &= \{ z : z = e^{2\pi\imath t}, 10\pi / 6< t < 14\pi / 6\}  \end{aligned}

We define the co-ordinate maps \xi_i : U_i \rightarrow \mathbb{R} as \xi_i(e^{2\pi\imath t}) = t - (1 + 3(i - 1))\pi / 6. The transition maps are then just a translation around the circle. For example

\displaystyle   \xi_2^{-1}\circ\xi_1(e^{2\pi\imath t}) = \xi_2^{-1}(t - \pi / 6) = e^{\pi\imath(2t + 1 / 2)}

And these are clearly smooth.

By continuity of f, every point in \mathbb{I} is contained in some open interval whose map under f is contained in one of the charts. This gives an open cover of \mathbb{I}. By the compactness of \mathbb{I} (it is closed and bounded), we therefore have a finite subcover. From this we can construct t_0 (= 0), t_1, \ldots, t_{n-1}, t_n (= 1) such that f[t_k, t_{k+1}] \subset U_l for some k and l.

Since f[0,t_1] \subset U_{l_0} we could to define \tilde{f}\restriction_{[0,t_1]} = f\restriction_{[0,t_1]}. But how do we define \tilde{f} beyond t_1?

We know that

\displaystyle   p^{-1}(U_l) = \ldots \cup \tilde{U}^{(-1)}_l \cup \tilde{U}^{(0)}_l \cup \tilde{U}^{(1)}_l \cup \ldots

where p\restriction_{\tilde{U}^{(k)}_l} is a homeomorphism from each \tilde{U}^{(k)}_l to U_l.

So we can equivalently define

\displaystyle   \tilde{f}\restriction_{[0,t_1]} = p^{-1}\restriction_{\tilde{U}^{(k_0)}_{l_0}} \circ                                    f\restriction_{[0,t_1]}

where \tilde{U}^{(k_0)}_{l_0} has been chosen to contain 0.

Given our specific atlas we have

\displaystyle   p^{-1}(U_4) = \ldots \cup (-7/6, -5/6) \cup (-1/6, 1/6) \cup (5/6, 7/6) \cup \ldots

Suppose we have the loop f(t) = e^{4\pi\imath t}, then we could define

\displaystyle   \tilde{f}\restriction_{[0,1/16]}(t) = p^{-1}\restriction_{\tilde{U}^{(k_0)}_4} \circ                                        f\restriction_{[0,1/16]}(t)

So that

\displaystyle   \tilde{f}\restriction_{[0,1/16]}(t) = 2t

And now we know how to continue. There must be a U_{l_1} such that f[t_1, t_2] \subset U_{l_1}.

So we can define

\displaystyle   \tilde{f}\restriction_{[t_1, t_2]} = p^{-1}\restriction_{\tilde{U}^{(k_1)}_{l_1}} \circ                                       f\restriction_{[t_1, t_2]}

where \tilde{U}^{(k_1)}_{l_1} has been chosen to contain \tilde{f}(t_1).

Given our specific atlas we have

\displaystyle   p^{-1}(U_1) = \ldots \cup (-11/12, -7/12) \cup (1/12, 5/12) \cup (13/12, 17/12) \cup \ldots

With our specific example we could define

\displaystyle   \tilde{f}\restriction_{[1/16, 3/16]}(t) = p^{-1}\restriction_{\tilde{U}^{(k_1)}_1} \circ                                            f\restriction_{[1/16,3/16]}(t)

So that (again)

\displaystyle   \tilde{f}\restriction_{[1/16, 3/16]}(t) = 2t

Continuing in this fashion, after a finite number of steps we have defined \tilde{f} on the entirety of \mathbb{I}. Note that this construction gives a unique path as at each point t_i the value of \tilde{f} on [t_i,t_{i+1}] is uniquely determined by its value at t_i (and of course by f itself).

We still have a problem that if g is homotopic to f then \tilde{g}(1) might not equal \tilde{f}(1).

Assume we have a homotopy h : f \simeq g then since \mathbb{I} \times \mathbb{I} is compact we can proceed as above and choose a partition of of rectangles

\displaystyle   T_{ij} = \{ (x, y) : s_i \le x \le s_{i+1}, t_j \le y \le t_{j+1} \}

with s_0 = 0, s_n = 1, t_0 = 0 and t_n = 1 such that h(T_{ij}) \subset U_k for some k.

Thus we can define \tilde{h} on [0,s_1] \times [0,t_1] as

\displaystyle   \tilde{h}\restriction_{[0,s_1] \times [0,t_1]} =     p^{-1}\restriction_{\tilde{U}^{(k_0)}_{l_0}} \circ     h\restriction_{[0,s_1] \times [0,t_1]}

where \tilde{U}^{(k_0)}_{l_0} has been chosen to contain 0.

We can continue as before; we know there must be a U_{l_1} such that h([0,s_1] \times [t_1,t_2]) \subset U_{l_1} so we can define

\displaystyle   \tilde{h}\restriction_{[0,s_1] \times [t_1,t_2]} =     p^{-1}\restriction_{\tilde{U}^{(k_1)}_{l_1}} \circ     h\restriction_{[0,s_1] \times [t_1,t_2]}

where \tilde{U}^{(k_1)}_{l_1} has been chosen to contain \tilde{h}(s,t_1) for all s \in[0,s_1] using the previously defined \tilde{h}.

Eventually we will have defined \tilde{h} on the whole of [0,s_1] \times \mathbb{I}. But now we can start the same process for [s_1,s_2] and by choosing \tilde{h}(s1,0) to be the value of the previously defined \tilde{h}(s_1,0) and by the uniquess of lifts of paths we can define \tilde{h} on the whole of [0,s_2] \times \mathbb{I}.

Carrying on in this way we will have defined \tilde{h} on the whole of \mathbb{I} \times \mathbb{I}.

\tilde{h}(0,t) and \tilde{h}(1,t) must be constant paths because h(0,t) = h(1,t) for all t \in [0,1] (constant lifted paths get projected onto these and they are unique).

Since we require that \tilde{h}(0,0) = 0, we must have that \tilde{h}(0,t) = 0.

By the uniquess of lifted paths we must then have

\displaystyle   \begin{aligned}  \tilde{f}(s) &= \tilde{h}(s,0) \\  \tilde{g}(s) &= \tilde{h}(s,1)  \end{aligned}

Since \tilde{h}(1,t) is constant we must have that \tilde{f}(1) = \tilde{h}(1,0) = \tilde{h}(1,1) = \tilde{g}(1). Thus j is well defined on homotopy classes.

Recall that f_{n+m} \simeq f_n \cdot f_m so that j(f_n \cdot f_m) = m + n = j(f_n) + j(f_m) so that j is a homomorphism.

Suppose that i(n) = i(m) then f_n \simeq f_m with a homotopy h of loops. This lifts to a unique homotopy of paths \tilde{h} with \tilde{h}_0 = \tilde{f}_n and \tilde{h}_1 = \tilde{f}_m. Since \tilde{h} is a homotopy of paths, the end point \tilde{h}_t(1) is fixed. Thus n = \tilde{f}_n(1) = \tilde{h}_0(1) = \tilde{h}_1(1) = \tilde{f}_m(1) = m and i is injective.

Suppose that f : \mathbb{I} \rightarrow S^1 is a loop starting at 1 then there is a lift \tilde{f} starting at 0. We also have that \tilde{f} \simeq \tilde{f}_n by the linear homotopy t\tilde{f} + (1-t)\tilde{f}_n. Thus i(n) = [f_n] = [p \circ \tilde{f}_n] = [p\circ \tilde{f}] = [f] and i is surjective.



Hatcher, A. 2002. Algebraic Topology. Tsinghua University.

May, J.P. 1999. A Concise Course in Algebraic Topology. Chicago Lectures in Mathematics. University of Chicago Press.

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