So far we have not shown that any fundamental group is not trivial. We shall now show that this is the case for the circle, based on so called lifting of paths and homotopies to covering spaces. This is a very elegant and important construction.

We shall throughout consider spaces $X$ and $Y$ together with a map $p: Y\to X$. The basic example is when $X = S^1$, $Y= \mathbb{R}$ and $p(t) = e^{2\pi i t}$ - the map associating to a real number $t$ the point on the circle forming an angle $2\pi t$ with the horizontal.

Even covers

Fix $X$, $Y$ and $p$ as above. An open set $U\subset X$ is said to be evenly covered if we can express $p^{-1}(U)$ as $p^{-1}(U) = \coprod_{i\in I} V_i$ so that for each $i\in I$, $V_i$ is open in $Y$ and $p\vert_{V_i}: V_i\to U$ is a homeomorphism.

It is easy to see that each open interval in the circle is evenly covered. We say that $p: Y\to X$ is a covering map if each point in $X$ has an evenly covered neighbourhood. Using the word covering with its two different meanings, we can express this as saying that $X$ has an open cover $ \{U_\alpha\}_{\alpha\in A} $ consisting of evenly covered neighbourhoods. In this case we say $Y$ is a covering space (or simply a cover) of $X$.

Note that if $p$ is a covering map and $x\in X$, then $p^{-1}(x)$ is a discrete space.

Lifting

Let $f: A \to X$ be a map. A lift of $f$ is a map $F: A \to Y$ so that $f=p \circ F$. For instance, given a map to the circle, a lift is a map to the real numbers so that which lifts the point on the circle to an angle corresponding to that point. We see that for paths in the space $X$ (i.e., maps from the interval to $X$) lifts always exist and are determined by their starting points.

Theorem: (path lifting) Let $\alpha: [0,1] \to X$ be a path and let $y\in Y$ be such that $p(y) = \alpha(0)$. Then there is a unique path $\tilde{\alpha} :[0,1] \to Y$ such that $\alpha = p \circ \tilde\alpha$ and $\tilde{\alpha}(0) = y$.

Firstly, we see that we can subdivide the interval $[0,1]$ into $n$ equal parts so that each sub-interval has image contained in an evenly covered neighbourhood. Namely, the inverse images of evenly covered neighbourhoods under $\alpha$ form an open cover of $[0, 1]$, and the Lesbesgue number theorem implies that any sufficiently small sub-interval is contained in an open set in the cover.

Next, we construct the lift (and see uniqueness) for the union of the first $k$ sub-intervals, for $k=1$, …, $n$. At each stage, we have an interval $[x, y]$ contained in an evenly covered neighbourhood $U$ and the lift $\tilde\alpha$ defined on a set that includes $x$ but no other point in the interval. Let $p^{-1}(U) = \coprod_{i\in I} V_i$, and assume $\tilde\alpha(0)\in V_j$. Then we extend $\tilde\alpha$ by defining it on $[x, y]$ to be $\tilde\alpha(s)= (p\vert_{V_j})^{-1}(\alpha(s))$. It is easy to see that this is continuous, and (using connectedness) the only continuous extension of the given lift.

Local lifting

To generalise path lifting, we observe that the proof of the second step extends to the following more general situation.

Lemma Let $A$ be a space, $B\subset A$, $f: A \to U$ a map to an evenly covered neighbourhood of $X$ and $F_0: B\to Y$ a lift of $f\vert_B$. Suppose $B$ and $A$ are both connected. Then there is a unique lift $F: A\to X$ extending $F_0$.

We sketch the proof. Let $p^{-1}(U) = \coprod_{i\in I} V_i$. As $B$ is connected, $F_0(B)$ is contained in a single component, say $V_j$, of $p^{-1}(U)$. We define $F = (p\vert_{V_j})^{-1} \circ f$ to get a lift extending $F_0$. Using the connectedness of $A$, we see that any lift extending $F_0$ must have image in $V_j$, and so coincides with the lift we constructed.

Homotopy lifting

Using the more general lemma from above, we see that we can lift maps from any metric space $A$, uniquely given the lift of a point $O$, provided we can write $A$ as a union of sets $A_k$ so that, if $C_k$ denotes $A_0\cup A_1\cup A_k$, with such a decomposition chosen so that

• $A_0=\{O\}$.
• Each $A_k$ has diameter smaller than any specified $\epsilon>0$.
• For each $k\geq 1$, $A_k$ and $A_k\cap C_{k-1}$ are connected.

We shall use this result for the case of homotopies. We record this here.

Theorem: (Homotopy lifting) Let $H: [0,1] \times[0, 1] \to X$ be a path and let $y\in Y$ be such that $p(y) = H(0, 0)$. Then there is a unique homotopy $\tilde{H} : [0, 1]\times [0,1]\to Y$ such that $H = p \circ \tilde{H}$ and $\tilde{H}(0, 0) = y$.

Fundamental group of the circle

Armed with path lifting and homotopy lifting, we can compute the fundamental group of the circle.

Theorem: $\pi_1(S^1, 1) = \mathbb{Z}$

We give the main steps of the proof.

• Firstly, a map $\varphi: \pi_1(S^1, 1)\to \mathbb{Z}$ is constructed as follows. Given a path $\alpha: [0, 1]\to S^1$ with $\alpha(0)=\alpha(1)=1$, there is a unique lift $\tilde\alpha: [0, 1]\to Y$ with $\tilde\alpha (0) = 0$. Define $\varphi([\alpha]) = \tilde\alpha(1)$, which we see is an integer as $p(\tilde\alpha(1))=1$.
• To see the above lift is well-defined, use the fact that a path-homotopy $H$ lifts to a homotopy. Further, note that the lift $\tilde{H}$ is itself a path-homotopy, as $t\mapsto \tilde{H}(0, t)$ and $t\mapsto \tilde{H}(1, t)$ are continuous functions to the discrete set $p^{-1}(1)$. Hence the end points of lifts of path-homotopic loops are equal.
• Observe that any lift of a path $\alpha$ is obtained from any other by post-composing with a translation, and the product of lifts, if defined, is a lift. Using this we see that $\varphi$ is a homomorphism.
• By construction, we see that $\varphi$ is a surjection.
• Finally, paths in $\mathbb{R}$ with the same endpoints are path-homotopic. Using this we see that $\varphi$ is injective.