Section 1.1 Homotopy of Paths

Definition 1.1.

Let \(X\) be a space. A path from a point \(x_0\) to a point \(x_1\) is a continuous map \(f \colon I \to X\) such that \(f(0) = x_0\) and \(f(1) = x_1\text{.}\) Two paths \(f, g \colon I \to X\) from \(x_0\) to \(x_1\) are path-homotopy equivalent if there is a map \(H \colon I \times I \to X\) such that

\begin{align*} H(s, 0) &= f(s), \quad H(s, 1) = g(s),\\ H(0, t) &= x_0, \quad H(1, t) = x_1. \end{align*}

The map \(H\) represents a continuous deformation of \(f\) to \(g\) fixing endpoints, and is called a path homotopy. We write \(f \simeq_p g\) when such an \(H\) exists.

Exercise 1.2.

Prove that path homotopy is an equivalence relation on paths.

We write \([f]\) for the path-homotopy equivalence class of a path \(f\text{.}\)

Definition 1.3.

If \(f \colon I \to X\) is a path from \(x_0\) to \(x_1\) and \(g \colon I \to X\) is a path from \(x_1\) to \(x_2\text{,}\) their product \(f * g \colon I \to X\) is

\begin{equation*} (f * g)(s) = \begin{cases} f(2s) & 0 \le s \le \tfrac{1}{2}, \\ g(2s - 1) & \tfrac{1}{2} \le s \le 1. \end{cases} \end{equation*}

Exercise 1.4.

Prove that concatenation induces a well-defined operation on homotopy classes: \([f] * [g] := [f * g]\text{.}\)

All three parts of the next proposition follow from a single observation: up to path homotopy, a path does not remember the speed at which it is traversed. We isolate this as a lemma, since the same trick recurs throughout the course.

Definition 1.5.

A reparametrisation is a continuous map \(\varphi \colon I \to I\) with \(\varphi(0) = 0\) and \(\varphi(1) = 1\text{.}\)

Lemma 1.6.

Let \(f \colon I \to X\) be a path and \(\varphi\) a reparametrisation. Then \(f \circ \varphi \simeq_p f\text{.}\)

Proof.

Since \(I\) is convex, the straight-line homotopy

\begin{equation*} \Phi(s,t) = (1-t)\varphi(s) + t s \end{equation*}

is a homotopy \(\varphi \simeq \mathrm{Id}_I\) through reparametrisations: \(\Phi(0,t) = 0\) and \(\Phi(1,t) = 1\) for all \(t\text{.}\) Then \(H = f \circ \Phi\) is continuous, satisfies \(H(s,0) = f(\varphi(s))\) and \(H(s,1) = f(s)\text{,}\) and has \(H(0,t) = f(0)\text{,}\) \(H(1,t) = f(1)\text{.}\) So \(H\) is a path homotopy.

Proposition 1.7.

The operation \(*\) on homotopy classes of paths in \(X\) has the following properties.

Associativity. If \(f\) is a path from \(x_0\) to \(x_1\text{,}\) \(g\) from \(x_1\) to \(x_2\text{,}\) and \(h\) from \(x_2\) to \(x_3\text{,}\) then
\begin{equation*} ([f] * [g]) * [h] = [f] * ([g] * [h]). \end{equation*}
Identities. Given \(y \in X\text{,}\) let \(c_y \colon I \to X\) denote the constant path at \(y\text{.}\) Then
\begin{equation*} [f] * [c_{x_1}] = [f] \quad \text{and} \quad [c_{x_0}] * [f] = [f]. \end{equation*}
Inverses. Let \(\bar{f}\) be the path from \(x_1\) to \(x_0\) defined by \(\bar{f}(s) = f(1-s)\text{.}\) Then
\begin{equation*} [f] * [\bar{f}] = [c_{x_0}] \quad \text{and} \quad [\bar{f}] * [f] = [c_{x_1}]. \end{equation*}

Proof.

For (1), both \((f * g) * h\) and \(f * (g * h)\) traverse \(f\text{,}\) then \(g\text{,}\) then \(h\text{;}\) they differ only in the subintervals of \(I\) allotted to each. Concretely, \(f * (g * h)\) runs \(f\) on \([0, \tfrac12]\text{,}\) \(g\) on \([\tfrac12, \tfrac34]\) and \(h\) on \([\tfrac34, 1]\text{,}\) whereas \((f * g) * h\) runs them on \([0, \tfrac14]\text{,}\) \([\tfrac14, \tfrac12]\) and \([\tfrac12, 1]\text{.}\) Let \(\varphi \colon I \to I\) be the piecewise-linear reparametrisation carrying \([0,\tfrac14]\text{,}\) \([\tfrac14,\tfrac12]\text{,}\) \([\tfrac12,1]\) affinely onto \([0,\tfrac12]\text{,}\) \([\tfrac12,\tfrac34]\text{,}\) \([\tfrac34,1]\) respectively, that is

\begin{equation*} \varphi(s) = \begin{cases} 2s & 0 \le s \le \tfrac14, \\ s + \tfrac14 & \tfrac14 \le s \le \tfrac12, \\ \tfrac{s+1}{2} & \tfrac12 \le s \le 1. \end{cases} \end{equation*}

Then \((f * g) * h = \bigl(f * (g * h)\bigr) \circ \varphi\text{,}\) so the two agree up to path homotopy by Lemma 1.6.

For (2), the path \(f * c_{x_1}\) traverses \(f\) at double speed on \([0,\tfrac12]\) and then stands still. That is, \(f * c_{x_1} = f \circ \varphi\) for the reparametrisation \(\varphi(s) = \min(2s, 1)\text{,}\) and Lemma 1.6 applies. Similarly \(c_{x_0} * f = f \circ \psi\) with \(\psi(s) = \max(2s - 1, 0)\text{.}\)

Part (3) is not a reparametrisation, and we exhibit the homotopy directly. The idea is to run \(f\) only partway along and come straight back, retreating a little further at each stage. Set

\begin{equation*} H(s,t) = \begin{cases} f\bigl(2s(1-t)\bigr) & 0 \le s \le \tfrac12, \\ f\bigl(2(1-s)(1-t)\bigr) & \tfrac12 \le s \le 1. \end{cases} \end{equation*}

The two formulas agree at \(s = \tfrac12\text{,}\) where both give \(f(1-t)\text{,}\) so \(H\) is continuous by the pasting lemma. At \(t = 0\) we recover \(f * \bar{f}\text{,}\) since \(\bar{f}(2s-1) = f(2-2s)\text{;}\) at \(t = 1\) we get the constant path \(c_{x_0}\text{;}\) and \(H(0,t) = H(1,t) = f(0) = x_0\) throughout. Hence \([f] * [\bar{f}] = [c_{x_0}]\text{.}\) Applying this to \(\bar{f}\) in place of \(f\text{,}\) and noting \(\bar{\bar{f}} = f\text{,}\) gives \([\bar{f}] * [f] = [c_{x_1}]\text{.}\)

Remark 1.8.

Note that the equalities in Proposition 1.7 genuinely fail before passing to homotopy classes: \((f * g) * h\) and \(f * (g * h)\) are different maps \(I \to X\text{.}\) Paths in \(X\) do not form a group, or even a monoid; only their homotopy classes do. Keeping track of the coherence data lost by passing to homotopy classes is one of the motivations for higher category theory.

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