Chapter 187Core route

Elementary Topology and Classical Homotopy

Let $e : R \to R^{2}$ be the map $e (x) : = (\cos 2 π x, \sin 2 π x),$ and for each integer $n$ let $ℓ_{n} (s) : = e (n s)$ for $s \in [0, 1]$ . Every $ℓ_{n}$ starts and ends at the point $b : = (1, 0)$ , and every $ℓ_{n}$ takes its values in the set $S^{1} = {x \in R^{2} : ‖ x ‖ = 1}$ . Inside $R^{2}$ each $ℓ_{n}$ can be shrunk to the constant map at $b$ along $(s, t) \mapsto (1 - t) ℓ_{n} (s) + t b$ . Inside $R^{2} ∖ {0}$ that formula passes through $0$ as soon as $n \neq 0$ , and no substitute formula is available. The assertion that no substitute exists is the first genuinely topological statement in this book, and it cannot even be formulated yet: it quantifies over continuous maps $[0, 1] \times [0, 1] \to R^{2} ∖ {0}$ , and continuity for maps out of a square into a punctured plane has not been defined here.

This chapter builds that vocabulary and then performs three calculations with it: the loops $ℓ_{m}$ and $ℓ_{n}$ concatenate, up to deformation, to $ℓ_{m + n}$ ; the punctured plane deforms onto $S^{1}$ ; and the circle, described three different ways, is one space up to homeomorphism while being three different sets. The invariant that finally separates $ℓ_{0}$ from $ℓ_{1}$ is not constructed here. What is constructed here is the exact language in which such an invariant can be defined, together with the constructions — products, subspaces, quotients, wedges, cones, suspensions — that later mechanisms consume.

Open sets and continuous maps

The $ε$ – $δ$ definition of continuity for maps $R^{m} \to R^{n}$ mentions distances, but the deformations we must compare are insensitive to distance: stretching the square does not change which deformations exist. The structure that survives stretching is the collection of open sets, so we take that collection as the primitive datum.

Definition 187.1 — Topological space ¶

A topology on a set $X$ is a collection $τ$ of subsets of $X$ , called the open sets of $X$ , such that

$\emptyset \in τ$ and $X \in τ$ ;
if $U_{i} \in τ$ for every $i$ in a set $I$ , then $⋃_{i \in I} U_{i} \in τ$ ;
if $U, V \in τ$ then $U \cap V \in τ$ .

A topological space is a pair $(X, τ)$ ; we write $X$ for the pair when $τ$ is determined. A subset $C \subseteq X$ is closed when $X ∖ C$ is open. A neighborhood of $x \in X$ is an open set containing $x$ .

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Condition (iii) is stated for two sets and therefore gives finite intersections by induction; it is not stated for arbitrary intersections, and exercise 187.1 exhibits an infinite intersection of open subsets of $R$ that is not open.

Example 187.2 — The spaces used in this chapter ¶

Metric topology. Let $(M, d)$ be a metric space and let $B (x, ε) : = {y \in M : d (x, y) < ε}$ . Call $U \subseteq M$ open when for every $x \in U$ there is $ε > 0$ with $B (x, ε) \subseteq U$ . Conditions (i) and (ii) hold because the witness $ε$ for a point of a union may be taken from any member containing it; for (iii) take the smaller of the two witnesses. With $d (x, y) = ‖ x - y ‖$ this gives the standard topology on $R^{n}$ , and it is the topology meant whenever $R^{n}$ is named below.
Subspace topology. For $A \subseteq X$ put $τ_{A} : = {U \cap A : U open in X}$ . The three conditions follow from those for $X$ because intersection with $A$ commutes with unions and with binary intersections. The circle $S^{1}$ , the interval $[0, 1]$ , and the punctured plane $R^{2} ∖ {0}$ always carry this topology, inherited from $R^{2}$ or $R$ .
Discrete and indiscrete topologies. On any set, $τ = P (X)$ and $τ = {\emptyset, X}$ are topologies.
The Sierpiński space. On ${0, 1}$ the collection ${\emptyset, {1}, {0, 1}}$ is a topology. It is neither discrete nor indiscrete, and it is the smallest space in which two points are topologically distinguishable while one of them cannot be separated from the other.

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Definition 187.3 — Continuity, homeomorphism ¶

A map $f : X \to Y$ of topological spaces is continuous when $f^{- 1} (V)$ is open in $X$ for every open $V \subseteq Y$ . It is a homeomorphism when it is a continuous bijection whose inverse is continuous; $X$ and $Y$ are homeomorphic, written $X \approx Y$ , when such a map exists.

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Lemma 187.4 — The metric case ¶

Let $(M, d)$ and $(N, d^{'})$ be metric spaces with their metric topologies. A map $f : M \to N$ is continuous in the sense of definition 187.3 if and only if for every $x \in M$ and every $ε > 0$ there is $δ > 0$ , depending on $x$ and $ε$ , with $d (x, y) < δ ⟹ d^{'} (f (x), f (y)) < ε$ .

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Proof of Lemma 187.4 — The metric case

Proof. Assume preimage continuity, and fix $x$ and $ε > 0$ . The ball $B (f (x), ε)$ is open in $N$ , so $f^{- 1} (B (f (x), ε))$ is open in $M$ and contains $x$ ; by the definition of the metric topology there is $δ > 0$ with $B (x, δ) \subseteq f^{- 1} (B (f (x), ε))$ , which is the displayed implication.

Conversely assume the $ε$ – $δ$ condition and let $V \subseteq N$ be open. Let $x \in f^{- 1} (V)$ . Since $V$ is open there is $ε > 0$ with $B (f (x), ε) \subseteq V$ , and the hypothesis gives $δ = δ (x, ε) > 0$ with $f (B (x, δ)) \subseteq B (f (x), ε) \subseteq V$ . Hence $B (x, δ) \subseteq f^{- 1} (V)$ , and as $x$ was an arbitrary point of $f^{- 1} (V)$ , that set is open. ◻

So the new definition agrees with the old one where both apply, and the maps $e$ , $ℓ_{n}$ , and $x \mapsto x / ‖ x ‖$ of the chapter opening are continuous for the reason they were continuous in analysis.

Lemma 187.5 — Composites and restrictions ¶

Let $f : X \to Y$ and $g : Y \to Z$ be continuous.

$g \circ f$ is continuous, and ${id}_{X}$ is continuous.
If $A \subseteq X$ carries the subspace topology, the inclusion $ι : A \to X$ and the restriction $f \circ ι : A \to Y$ are continuous.
If $f (X) \subseteq B \subseteq Y$ and $B$ carries the subspace topology, then the corestriction $f |^{B} : X \to B$ is continuous.

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Proof of Lemma 187.5 — Composites and restrictions

Proof. For (i), $(g \circ f)^{- 1} (W) = f^{- 1} (g^{- 1} (W))$ , and $g^{- 1} (W)$ is open by continuity of $g$ , so its $f$ -preimage is open by continuity of $f$ ; and ${id}_{X}^{- 1} (U) = U$ . For (ii), $ι^{- 1} (U) = U \cap A$ , which is open in $A$ by the definition of $τ_{A}$ ; the restriction is then a composite. For (iii), an open set of $B$ is $V \cap B$ with $V$ open in $Y$ , and $(f |^{B})^{- 1} (V \cap B) = f^{- 1} (V)$ because $f$ lands in $B$ . ◻

Definition 187.6 — Product topology ¶

For spaces $X$ and $Y$ , call $W \subseteq X \times Y$ open when for every $(x, y) \in W$ there are open $U ∋ x$ and $V ∋ y$ with $U \times V \subseteq W$ . The three conditions of definition 187.1 hold, taking for a binary intersection the pairwise intersections of the two witnessing rectangles. The projections $π_{1}, π_{2}$ are continuous, since $π_{1}^{- 1} (U) = U \times Y$ .

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Lemma 187.7 — Maps into a product ¶

A map $h = (h_{1}, h_{2}) : Z \to X \times Y$ is continuous if and only if $h_{1}$ and $h_{2}$ are continuous.

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Proof of Lemma 187.7 — Maps into a product

Proof. If $h$ is continuous then $h_{1} = π_{1} \circ h$ and $h_{2} = π_{2} \circ h$ are composites of continuous maps. Conversely let $W \subseteq X \times Y$ be open and $z \in h^{- 1} (W)$ . Choose $U ∋ h_{1} (z)$ and $V ∋ h_{2} (z)$ open with $U \times V \subseteq W$ . Then $h_{1}^{- 1} (U) \cap h_{2}^{- 1} (V)$ is an open set containing $z$ and contained in $h^{- 1} (U \times V) \subseteq h^{- 1} (W)$ . ◻

The standard topology of $R^{2}$ and the product topology of $R \times R$ coincide: an open ball contains an open rectangle about each of its points and conversely (exercise 187.2). We use the two descriptions interchangeably from here on, and in particular a homotopy defined on $[0, 1] \times [0, 1]$ may be tested for continuity by lemma 187.4.

Example 187.8 — A continuous bijection that is not a homeomorphism ¶

Let $f : [0, 1) \to S^{1}$ be $f (t) = e (t)$ . It is continuous by lemma 187.4, lemma 187.5(iii), and it is a bijection because $e (t) = e (t^{'})$ with $t, t^{'} \in [0, 1)$ forces $t - t^{'} \in Z$ and hence $t = t^{'}$ . Its inverse is not continuous. Take $U : = [0, \frac{1}{2})$ , which is open in $[0, 1)$ because $U = (- \frac{1}{2}, \frac{1}{2}) \cap [0, 1)$ . Its image $f (U)$ is not open in $S^{1}$ : a neighborhood of $b = f (0)$ in $S^{1}$ contains $B (b, ε) \cap S^{1}$ for some $ε > 0$ , and that set contains $e (t)$ for all $t$ in some interval $(1 - η, 1)$ , whereas no such point lies in $f (U)$ . Since $(f^{- 1})^{- 1} (U) = f (U)$ , the inverse is not continuous.

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The example is the reason definition 187.3 asks for continuity of the inverse separately, and it is worth keeping in view: the two spaces $[0, 1)$ and $S^{1}$ are in bijection by a continuous map, and they will nevertheless be distinguished by every homotopy invariant constructed later.

Definition 187.9 — Connectedness ¶

A space $X$ is connected when the only subsets of $X$ that are both open and closed are $\emptyset$ and $X$ .

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Lemma 187.10 — Intervals are connected ¶

Every interval $I \subseteq R$ — in particular $[0, 1]$ — is connected, and a continuous image of a connected space is connected.

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Proof of Lemma 187.10 — Intervals are connected

Proof. Let $A \subseteq I$ be open, closed, nonempty, and different from $I$ ; choose $a \in A$ and $c \in I ∖ A$ , and assume $a < c$ (otherwise exchange the roles of $A$ and $I ∖ A$ , which is also open and closed). Put $s : = sup {x \in [a, c] : [a, x] \subseteq A}$ , which exists because the set contains $a$ and is bounded by $c$ . If $s \in A$ , then $A$ open gives $ε > 0$ with $(s - ε, s + ε) \cap I \subseteq A$ , and since $s < c$ — because $c \notin A$ — points slightly above $s$ lie in $A$ , contradicting the definition of $s$ as a supremum. If $s \notin A$ , then $I ∖ A$ is open, so some $(s - ε, s + ε) \cap I$ misses $A$ , contradicting the fact that $[a, x] \subseteq A$ for $x$ arbitrarily close to $s$ from below. Both cases are impossible, so no such $A$ exists.

For the second claim let $f : X \to Y$ be continuous and surjective with $X$ connected, and let $B \subseteq Y$ be open and closed. Then $f^{- 1} (B)$ is open and closed, hence $\emptyset$ or $X$ , hence $B = \emptyset$ or $B = Y$ by surjectivity. ◻

Exercise 187.1 ¶

★☆☆ Exhibit open subsets $U_{n} \subseteq R$ for $n \geq 1$ with $⋂_{n \geq 1} U_{n} = {0}$ , and conclude that clause (iii) of definition 187.1 cannot be strengthened to arbitrary intersections.

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Exercise 187.2 ¶

★☆☆ Prove that a subset of $R^{2}$ is open for the standard metric topology if and only if it is open for the product topology of $R \times R$ . (Compare an open ball of radius $ε$ with the square of side $ε / \sqrt{2}$ about the same point.)

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Exercise 187.3 ¶

★★☆ Let $Σ$ be the Sierpiński space of example 187.2(d). Prove that for every space $X$ the map sending a continuous $f : X \to Σ$ to $f^{- 1} ({1})$ is a bijection from the set of continuous maps $X \to Σ$ to the set of open subsets of $X$ . State which clause of definition 187.1 each direction uses.

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Compactness, and one gluing lemma

Two facts about $[0, 1]$ are used repeatedly below: a map defined by different formulas on $[0, \frac{1}{2}]$ and $[\frac{1}{2}, 1]$ is continuous when the formulas agree at $\frac{1}{2}$ , and a continuous bijection out of a quotient of $[0, 1]$ has a continuous inverse. The first is elementary; the second is false without a hypothesis — example 187.8 is a counterexample — and the hypothesis that repairs it is compactness.

Lemma 187.11 — Gluing along closed pieces ¶

Let $X = A \cup B$ with $A$ and $B$ closed in $X$ , and let $f : A \to Y$ and $g : B \to Y$ be continuous with $f (x) = g (x)$ for all $x \in A \cap B$ . Then the map $h : X \to Y$ with $h |_{A} = f$ and $h |_{B} = g$ is well defined and continuous.

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Proof of Lemma 187.11 — Gluing along closed pieces

Proof. Well-definedness is the agreement hypothesis. Let $C \subseteq Y$ be closed. Then $h^{- 1} (C) = f^{- 1} (C) \cup g^{- 1} (C)$ . Now $f^{- 1} (C)$ is closed in $A$ by continuity of $f$ , and a closed subset of a closed subspace is closed in $X$ : if $f^{- 1} (C) = D \cap A$ with $D$ closed in $X$ , then $D \cap A$ is an intersection of two closed subsets of $X$ . The same argument applies to $g^{- 1} (C)$ , and a union of two closed sets is closed. So $h^{- 1} (C)$ is closed for every closed $C$ , which is equivalent to continuity by taking complements. ◻

Definition 187.12 — Compactness, Hausdorff spaces ¶

A space $X$ is compact when every collection of open sets whose union is $X$ has a finite subcollection whose union is $X$ . A space $X$ is Hausdorff when any two distinct points $x \neq y$ have disjoint neighborhoods. Every metric space is Hausdorff: take the two balls of radius $d (x, y) / 2$ .

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Lemma 187.13 — Heine–Borel for the unit interval ¶

$[0, 1]$ is compact.

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Proof of Lemma 187.13 — Heine–Borel for the unit interval

Proof. Let $U$ be a collection of open subsets of $[0, 1]$ with union $[0, 1]$ , and let $S : = {x \in [0, 1] : [0, x] is covered by finitely many members of U} .$ Then $0 \in S$ , because some $U \in U$ contains $0$ . Let $s : = sup S \in [0, 1]$ and choose $U_{0} \in U$ with $s \in U_{0}$ . Since $U_{0}$ is open there is $ε > 0$ with $(s - ε, s + ε) \cap [0, 1] \subseteq U_{0}$ . By the definition of the supremum there is $x \in S$ with $x > s - ε$ , so $[0, x]$ has a finite subcover $F$ ; then $F \cup {U_{0}}$ is a finite subcover of $[0, y]$ for every $y < s + ε$ with $y \in [0, 1]$ . Hence $s \in S$ , and if $s < 1$ then some $y > s$ lies in $S$ , contradicting $s = sup S$ . Therefore $s = 1$ and $1 \in S$ . ◻

Lemma 187.14 — Compactness transfers ¶

If $f : X \to Y$ is continuous and $X$ is compact, then $f (X)$ is compact.
A closed subspace of a compact space is compact.
A compact subspace $K$ of a Hausdorff space $Y$ is closed in $Y$ .
If $X$ is compact and $Y$ is compact then $X \times Y$ is compact.

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Proof of Lemma 187.14 — Compactness transfers

Proof. (i) Let $V$ cover $f (X)$ by sets open in $f (X)$ . The preimages $f^{- 1} (V)$ are open and cover $X$ ; a finite subfamily $f^{- 1} (V_{1}), \dots, f^{- 1} (V_{k})$ covers $X$ , and then $V_{1}, \dots, V_{k}$ cover $f (X)$ .

(ii) Let $C \subseteq X$ be closed and let $U$ cover $C$ by sets open in $C$ ; write each as $U_{i} \cap C$ with $U_{i}$ open in $X$ . Then the $U_{i}$ together with $X ∖ C$ cover $X$ ; extract a finite subcover and discard $X ∖ C$ .

(iii) Let $y \in Y ∖ K$ . For each $x \in K$ choose disjoint neighborhoods $U_{x} ∋ x$ and $V_{x} ∋ y$ , using the Hausdorff property. The sets $U_{x} \cap K$ cover $K$ , so finitely many $U_{x_{1}}, \dots, U_{x_{k}}$ cover $K$ , and $V : = ⋂_{j \leq k} V_{x_{j}}$ is a neighborhood of $y$ disjoint from $K$ . Hence $Y ∖ K$ is open.

(iv) Let $W$ be an open cover of $X \times Y$ . Fix $x \in X$ . For each $y \in Y$ choose $W \in W$ containing $(x, y)$ and, by definition 187.6, open sets $U_{y} ∋ x$ , $V_{y} ∋ y$ with $U_{y} \times V_{y} \subseteq W$ . The $V_{y}$ cover $Y$ , so finitely many $V_{y_{1}}, \dots, V_{y_{m}}$ do; put $U^{x} : = ⋂_{j \leq m} U_{y_{j}}$ , an open neighborhood of $x$ such that $U^{x} \times Y$ is covered by $m$ members of $W$ . The $U^{x}$ cover $X$ , so finitely many $U^{x_{1}}, \dots, U^{x_{k}}$ do, and the corresponding $k \cdot m$ members of $W$ cover $X \times Y$ . ◻

Proposition 187.15 — Continuous bijections out of compact spaces ¶

Let $f : X \to Y$ be a continuous bijection with $X$ compact and $Y$ Hausdorff. Then $f$ is a homeomorphism.

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Proof of Proposition 187.15 — Continuous bijections out of compact spaces

Proof. It suffices to prove that $f$ maps closed sets to closed sets, since for a bijection $f (C) = (f^{- 1})^{- 1} (C)$ and continuity of $f^{- 1}$ is exactly the statement that preimages of closed sets under $f^{- 1}$ are closed. Let $C \subseteq X$ be closed. By lemma 187.14(ii) $C$ is compact, by (i) $f (C)$ is compact, and by (iii) $f (C)$ is closed in $Y$ . ◻

Example 187.8 is consistent with this proposition: $[0, 1)$ is not compact, since the open cover by the sets $[0, 1 - \frac{1}{n})$ has no finite subcover.

Paths, homotopies, and the first calculation

Definition 187.16 — Paths and loops ¶

A path in a space $X$ is a continuous map $p : [0, 1] \to X$ ; its endpoints are $p (0)$ and $p (1)$ . A loop at $x_{0} \in X$ is a path with $p (0) = p (1) = x_{0}$ . The constant path at $x$ is $c_{x} (s) : = x$ . For paths $p, q$ with $p (1) = q (0)$ , the concatenation is $(p \cdot q) (s) : = {\begin{cases} p (2 s), & 0 \leq s \leq \frac{1}{2}, \\ q (2 s - 1), & \frac{1}{2} \leq s \leq 1, \end{cases}$ and the reversal of $p$ is $\overset{―}{p} (s) : = p (1 - s)$ .

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Concatenation is well defined and continuous: the two closed pieces $[0, \frac{1}{2}]$ and $[\frac{1}{2}, 1]$ cover $[0, 1]$ , the formulas agree at $s = \frac{1}{2}$ because $p (1) = q (0)$ , and each formula is a composite of a continuous affine map with a continuous path, so lemma 187.11 applies. Reversal is a composite with $s \mapsto 1 - s$ .

Example 187.17 — The winding loops ¶

For $n \in Z$ the loop $ℓ_{n} = e (n -)$ of the chapter opening is a loop at $b = (1, 0)$ in $S^{1}$ . Concatenating two of them and evaluating gives $(ℓ_{1} \cdot ℓ_{1}) (s) = {\begin{cases} e (2 s), & s \leq \frac{1}{2}, \\ e (2 s - 1), & s \geq \frac{1}{2}, \end{cases} ℓ_{2} (s) = e (2 s) .$ At $s = \frac{3}{4}$ the first is $e (\frac{1}{2}) = (- 1, 0)$ and the second is $e (\frac{3}{2}) = (0, - 1)$ , so $ℓ_{1} \cdot ℓ_{1} \neq ℓ_{2}$ as maps. The two loops traverse the same circle the same number of times at different speeds. The relation that identifies them is defined next.

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Definition 187.18 — Homotopy ¶

Let $f, g : X \to Y$ be continuous. A homotopy from $f$ to $g$ is a continuous map $H : X \times [0, 1] \to Y$ with $H (x, 0) = f (x)$ and $H (x, 1) = g (x)$ for all $x \in X$ ; we write $f ≃ g$ when one exists. If $A \subseteq X$ and $H (a, t) = f (a)$ for all $a \in A$ and $t \in [0, 1]$ , the homotopy is rel $A$ , written $f ≃ g$ rel $A$ . A path homotopy between paths $p, q : [0, 1] \to X$ is a homotopy rel ${0, 1}$ ; explicitly, $H (s, 0) = p (s)$ , $H (s, 1) = q (s)$ , $H (0, t) = p (0)$ , and $H (1, t) = p (1)$ .

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Proposition 187.19 — Path homotopy is an equivalence relation ¶

Fix $x_{0}, x_{1} \in X$ . Path homotopy is an equivalence relation on the set of paths from $x_{0}$ to $x_{1}$ .

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Proof of Proposition 187.19 — Path homotopy is an equivalence relation

Proof. Reflexivity. $H (s, t) : = p (s)$ is continuous as the composite of the projection $π_{1}$ with $p$ , and it is a path homotopy from $p$ to $p$ .

Symmetry. If $H$ is a path homotopy from $p$ to $q$ , then $H^{'} (s, t) : = H (s, 1 - t)$ is continuous as a composite with the continuous map $(s, t) \mapsto (s, 1 - t)$ , and it is a path homotopy from $q$ to $p$ .

Transitivity. Let $H$ be a path homotopy from $p$ to $q$ and $K$ one from $q$ to $r$ . Define $L (s, t) : = {\begin{cases} H (s, 2 t), & t \leq \frac{1}{2}, \\ K (s, 2 t - 1), & t \geq \frac{1}{2} . \end{cases}$ The sets $[0, 1] \times [0, \frac{1}{2}]$ and $[0, 1] \times [\frac{1}{2}, 1]$ are closed in $[0, 1]^{2}$ and cover it; at $t = \frac{1}{2}$ both formulas give $H (s, 1) = q (s) = K (s, 0)$ ; so $L$ is continuous by lemma 187.11. Its boundary values are $L (s, 0) = p (s)$ , $L (s, 1) = r (s)$ , and $L (0, t) = x_{0}$ , $L (1, t) = x_{1}$ , the last two because both $H$ and $K$ are rel ${0, 1}$ . ◻

Lemma 187.20 — Reparametrization ¶

Let $p$ be a path in $X$ and let $φ, ψ : [0, 1] \to [0, 1]$ be continuous with $φ (0) = ψ (0)$ and $φ (1) = ψ (1)$ . Then $p \circ φ ≃ p \circ ψ$ rel ${0, 1}$ .

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Proof of Lemma 187.20 — Reparametrization

Proof. Put $H (s, t) : = p ((1 - t) φ (s) + t ψ (s))$ . The inner map is continuous into $R$ by lemma 187.4, and its values lie in $[0, 1]$ because $[0, 1]$ is convex, so $H$ is continuous by lemma 187.5. At $t = 0$ and $t = 1$ it is $p \circ φ$ and $p \circ ψ$ . At $s = 0$ the inner value is $(1 - t) φ (0) + t ψ (0) = φ (0)$ , so $H (0, t) = p (φ (0))$ is constant in $t$ , and the same calculation applies at $s = 1$ . ◻

The interpolation in lemma 187.20 takes place in the parameter interval, not in $X$ : it is the convexity of $[0, 1]$ that is used, and no convexity of $X$ is assumed. The chapter opening shows why this distinction is not pedantic — straight-line interpolation inside $R^{2} ∖ {0}$ is exactly what fails.

The next theorem is the chapter’s first substantial calculation. Its mechanism: both $ℓ_{m} \cdot ℓ_{n}$ and $ℓ_{m + n}$ are of the form $e \circ u$ for a path $u$ in $R$ from $0$ to $m + n$ , and paths in $R$ with fixed endpoints can be interpolated linearly. The interpolation happens upstairs, in $R$ , and is transported to $S^{1}$ by $e$ ; the reason the transported homotopy is a path homotopy is that $e$ takes the same value at every integer.

Theorem 187.21 — Concatenation adds winding ¶

For all $m, n \in Z$ there is a path homotopy $ℓ_{m} \cdot ℓ_{n} ≃ ℓ_{m + n}$ of loops at $b$ in $S^{1}$ .

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Proof of Theorem 187.21 — Concatenation adds winding

Proof. Define $u : [0, 1] \to R$ by $u (s) : = {\begin{cases} 2 m s, & s \leq \frac{1}{2}, \\ m + n (2 s - 1), & s \geq \frac{1}{2}, \end{cases} and v (s) : = (m + n) s .$ Both formulas of $u$ give $m$ at $s = \frac{1}{2}$ , so $u$ is continuous by lemma 187.11; and $e \circ u = ℓ_{m} \cdot ℓ_{n}$ by definition 187.16, since $e (2 m s) = ℓ_{m} (2 s)$ and $e (m + n (2 s - 1)) = e (m) e (n (2 s - 1)) = ℓ_{n} (2 s - 1)$ , using $e (x + y) = e (x) e (y)$ for the multiplication of $R^{2} = C$ and $e (m) = 1$ . Also $e \circ v = ℓ_{m + n}$ . Now set $H (s, t) : = e ((1 - t) u (s) + t v (s)) .$ It is continuous as a composite of continuous maps. Its four boundary values are $\begin{aligned} H (s, 0) & = e (u (s)) = (ℓ_{m} \cdot ℓ_{n}) (s), & H (s, 1) & = e (v (s)) = ℓ_{m + n} (s), \\ H (0, t) & = e (0) = b, & H (1, t) & = e (m + n) = b, \end{aligned}$ since $(1 - t) u (0) + t v (0) = 0$ and $(1 - t) u (1) + t v (1) = (1 - t) (m + n) + t (m + n) = m + n$ , and $e (m + n) = b$ because $m + n \in Z$ and $e$ is $1$ -periodic. So $H$ is a path homotopy of loops at $b$ . ◻

Remark 187.22 — What the theorem does not say ¶

Theorem 187.21 produces homotopies; it does not distinguish $ℓ_{0}$ from $ℓ_{1}$ . Every step above constructed a map, and no step could have shown that a map fails to exist. The last equality in the proof is the load-bearing one: it uses that $v (1) - u (1) = 0$ , so that the interpolation does not move the endpoint. If one replaces $v$ by $s \mapsto (m + n + \frac{1}{2}) s$ the same formula defines a homotopy of maps, but not rel ${0, 1}$ , and the resulting relation identifies all loops. A separation theorem therefore needs a construction that reads something off a loop and is unchanged by every path homotopy.

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Exercise 187.4 ¶

★★☆ Prove that $ℓ_{n} \cdot ℓ_{- n} ≃ c_{b}$ rel ${0, 1}$ , by exhibiting the path $u$ in $R$ that the proof of theorem 187.21 requires and checking its two endpoint values. Then prove $\overset{―}{ℓ_{n}} = ℓ_{- n}$ as maps.

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Exercise 187.5 ¶

★★☆ Let $p ≃ p^{'}$ rel ${0, 1}$ and $q ≃ q^{'}$ rel ${0, 1}$ with $p (1) = q (0)$ . Prove $p \cdot q ≃ p^{'} \cdot q^{'}$ rel ${0, 1}$ by gluing the two given homotopies along $s = \frac{1}{2}$ , and state where the hypothesis $p (1) = q (0)$ is used.

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Exercise 187.6 ¶

★★☆ Using lemma 187.20, prove $(p \cdot q) \cdot r ≃ p \cdot (q \cdot r)$ rel ${0, 1}$ for composable paths. Display the two functions $φ, ψ : [0, 1] \to [0, 1]$ and the path of which both sides are a reparametrization.

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Homotopy equivalence and deformation retraction

Homotopy compares two maps. The corresponding comparison of two spaces relaxes the two equations of an inverse to homotopies.

Definition 187.23 — Homotopy equivalence ¶

A continuous map $f : X \to Y$ is a homotopy equivalence when there is a continuous $g : Y \to X$ with $g \circ f ≃ {id}_{X}$ and $f \circ g ≃ {id}_{Y}$ ; such a $g$ is a homotopy inverse of $f$ . Spaces admitting such a pair are homotopy equivalent, written $X ≃ Y$ . A space homotopy equivalent to a one-point space is contractible.

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Definition 187.24 — Deformation retraction ¶

Let $A \subseteq X$ carry the subspace topology. A deformation retraction of $X$ onto $A$ is a continuous map $D : X \times [0, 1] \to X$ with $\begin{aligned} D (x, 0) & = x and D (x, 1) \in A & (x \in X), \\ D (a, t) & = a & (a \in A, t \in [0, 1]) . \end{aligned}$

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Proposition 187.25 — Retraction gives equivalence ¶

If $D$ is a deformation retraction of $X$ onto $A$ , then the inclusion $ι : A \to X$ is a homotopy equivalence with homotopy inverse $r (x) : = D (x, 1)$ , corestricted to $A$ .

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Proof of Proposition 187.25 — Retraction gives equivalence

Proof. The map $r$ is continuous by lemma 187.5(iii), because $x \mapsto D (x, 1)$ is continuous and lands in $A$ . For $a \in A$ we have $r (ι (a)) = D (a, 1) = a$ , so $r \circ ι = {id}_{A}$ , which is in particular homotopic to it. In the other order, $D$ itself is a homotopy from ${id}_{X}$ to $ι \circ r$ . ◻

Theorem 187.26 — The punctured plane deforms onto the circle ¶

The map $D (x, t) : = (1 - t) x + t \frac{x}{‖ x ‖}$ is a deformation retraction of $R^{2} ∖ {0}$ onto $S^{1}$ . Consequently $S^{1} ≃ R^{2} ∖ {0}$ .

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Proof of Theorem 187.26 — The punctured plane deforms onto the circle

Proof. Values. For $x \neq 0$ write $λ (x, t) : = (1 - t) + t / ‖ x ‖$ . Then $D (x, t) = λ (x, t) x$ and $λ (x, t) > 0$ , being a convex combination of the positive numbers $1$ and $1 / ‖ x ‖$ . Hence $D (x, t) \neq 0$ , so $D$ lands in $R^{2} ∖ {0}$ .

Continuity. The norm $‖ - ‖$ is continuous and nonzero on $R^{2} ∖ {0}$ , so $x \mapsto x / ‖ x ‖$ is continuous there; $D$ is then built from continuous maps by products and sums, and lands in the subspace by the previous paragraph.

Boundary conditions. $D (x, 0) = x$ . Next, $‖ D (x, 1) ‖ = ‖ x / ‖ x ‖ ‖ = 1$ , so $D (x, 1) \in S^{1}$ . Finally, if $a \in S^{1}$ then $‖ a ‖ = 1$ and $D (a, t) = (1 - t) a + t a = a$ .

The consequence is proposition 187.25. ◻

Example 187.27 — Three relations, three answers ¶

Let $P$ be the one-point space.

$R^{n} ≃ P$ : the maps $R^{n} \to P$ and $0 \mapsto R^{n}$ compose to ${id}_{P}$ in one order, and in the other to the constant map $0$ , which is homotopic to ${id}_{R^{n}}$ by $H (x, t) = (1 - t) x$ . So $R^{n}$ is contractible.
$R^{n} ≉ P$ for $n \geq 1$ : a homeomorphism is in particular a bijection, and $R^{n}$ has more than one point. Hence homotopy equivalence is strictly weaker than homeomorphism.
$S^{1} ≃ R^{2} ∖ {0}$ by theorem 187.26, while the two spaces are not equal as sets: the second contains $(2, 0)$ . The three relations — equality of sets, homeomorphism, homotopy equivalence — are therefore distinct, and each later construction states which one it preserves.

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Remark 187.28 — A distinction not yet available ¶

Example 187.27 separates equality from homeomorphism, and homeomorphism from homotopy equivalence, in one direction only. To show that $S^{1} ≄ P$ — that the circle is not contractible — one needs a quantity attached to a space that is unchanged by homotopy equivalence and differs for $S^{1}$ and $P$ . No such quantity has been constructed in this chapter, and none of the constructions above could produce one: each of them built a map.

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Exercise 187.7 ¶

★★☆ Prove that $≃$ of definition 187.23 is reflexive, symmetric, and transitive on spaces. For transitivity, state the two homotopies you glue and the map you insert between them; the composite $H (x, t)$ of a homotopy with a fixed continuous map is again a homotopy.

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Exercise 187.8 ¶

★☆☆ Prove that the annulus $A : = {x \in R^{2} : 1 \leq ‖ x ‖ \leq 2}$ deformation retracts onto $S^{1}$ , and write the analogue of $λ (x, t)$ from the proof of theorem 187.26 that shows the homotopy stays inside $A$ .

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Exercise 187.9 ¶

★★☆ Let $X : = {0} \cup {1 / n : n \geq 1} \subseteq R$ with the subspace topology. Prove that $X$ is not discrete, that every map from $X$ to a space $Y$ that is continuous at $0$ is determined there by the sequence of values at $1 / n$ , and that ${0}$ is not open in $X$ . Conclude that $X$ is not homeomorphic to the discrete space on the same set.

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Pointed spaces and loops

Every construction of section 187.3 fixed a point of $S^{1}$ and kept it fixed. Carrying that choice in the data removes it from the hypotheses of later statements.

Definition 187.29 — Pointed spaces ¶

A pointed space is a pair $(X, x_{0})$ with $x_{0} \in X$ , called the base point. A pointed map $f : (X, x_{0}) \to (Y, y_{0})$ is a continuous $f : X \to Y$ with $f (x_{0}) = y_{0}$ . A pointed homotopy is a homotopy $H$ with $H (x_{0}, t) = y_{0}$ for all $t$ , that is, a homotopy rel ${x_{0}}$ .

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Definition 187.30 — Loop space ¶

The loop space $Ω (X, x_{0})$ is the set of loops at $x_{0}$ , topologized by the compact-open topology: the open sets are the unions of finite intersections of the sets $⟨ K, U ⟩ : = {p a loop at x_{0} : p (K) \subseteq U},$ for $K \subseteq [0, 1]$ compact and $U \subseteq X$ open. Its base point is the constant loop $c_{x_{0}}$ . In this chapter only two features of $Ω (X, x_{0})$ are used: its underlying set, and the path homotopy relation on that set from proposition 187.19. Iterating gives $Ω^{k + 1} (X, x_{0}) : = Ω (Ω^{k} (X, x_{0}), c)$ , with $c$ the constant loop at the previous stage.

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Example 187.31 — Loops on the circle, and one warning ¶

The loops $ℓ_{n}$ are points of $Ω (S^{1}, b)$ , and theorem 187.21 says that concatenation sends the pair $(ℓ_{m}, ℓ_{n})$ into the path-homotopy class of $ℓ_{m + n}$ . It does not say that $Ω (S^{1}, b)$ has a group structure: concatenation is not associative on the nose, by the calculation of exercise 187.6, and $ℓ_{m} \cdot ℓ_{- m}$ is not the constant loop but only homotopic to it (exercise 187.4). A group appears only after passing to homotopy classes, and that construction, with its base-point bookkeeping, is the subject of the next chapter.

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Exercise 187.10 ¶

★☆☆ Prove that pointed maps compose and that a pointed homotopy between pointed maps $(X, x_{0}) \to (Y, y_{0})$ is exactly a homotopy $H$ such that each $H (-, t)$ is a pointed map.

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Exercise 187.11 ¶

★★☆ In $Ω (S^{1}, b)$ , describe explicitly a subbasic open set $⟨ K, U ⟩$ containing $ℓ_{0}$ but not $ℓ_{1}$ , with $K = [0, 1]$ . Then show that no subbasic set with $K = {0}$ separates any two loops at $b$ .

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Quotients, wedges, cones, and suspensions

The spaces the later chapters must build — spheres in every dimension, wedges, mapping cones — are all obtained by gluing simple pieces. Gluing is a quotient, so the quotient topology comes first.

Definition 187.32 — Quotient topology ¶

Let $X$ be a space and $\sim$ an equivalence relation on its underlying set, with quotient set $X / \sim$ and projection $q : X \to X / \sim$ . Declare $V \subseteq X / \sim$ open exactly when $q^{- 1} (V)$ is open in $X$ . This is a topology, because preimage commutes with unions and intersections, and $q$ is continuous by construction.

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Lemma 187.33 — Universal property of a quotient ¶

Let $q : X \to X / \sim$ be as above and let $f : X \to Y$ be continuous with $f (x) = f (x^{'})$ whenever $x \sim x^{'}$ . Then the induced map $\bar{f} : X / \sim \to Y$ with $\bar{f} \circ q = f$ is well defined and continuous, and it is the unique such map.

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Proof of Lemma 187.33 — Universal property of a quotient

Proof. Well-definedness and uniqueness are the set-level statements, since $q$ is surjective. For continuity let $V \subseteq Y$ be open; then $q^{- 1} ({\bar{f}}^{- 1} (V)) = f^{- 1} (V)$ is open in $X$ , which by definition 187.32 says exactly that ${\bar{f}}^{- 1} (V)$ is open in $X / \sim$ . ◻

Definition 187.34 — Disjoint union and pushout ¶

The disjoint union $X ⊔ Y$ carries the topology whose open sets are the sets $U ⊔ V$ with $U$ open in $X$ and $V$ open in $Y$ . Given continuous maps $f : A \to X$ and $g : A \to Y$ , the pushout $X \cup_{A} Y$ is the quotient of $X ⊔ Y$ by the smallest equivalence relation with $f (a) \sim g (a)$ for every $a \in A$ .

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Lemma 187.35 — Mapping out of a pushout ¶

Let $u : X \to Z$ and $v : Y \to Z$ be continuous with $u \circ f = v \circ g$ . There is a unique continuous $w : X \cup_{A} Y \to Z$ whose composites with the two canonical maps $X \to X \cup_{A} Y$ and $Y \to X \cup_{A} Y$ are $u$ and $v$ .

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Proof of Lemma 187.35 — Mapping out of a pushout

Proof. The map $[u, v] : X ⊔ Y \to Z$ is continuous, since $[u, v]^{- 1} (W) = u^{- 1} (W) ⊔ v^{- 1} (W)$ . The hypothesis $u \circ f = v \circ g$ makes it constant on the generating pairs of the equivalence relation, hence on each of its classes, so lemma 187.33 applies. Uniqueness holds because the two canonical maps are jointly surjective. ◻

Definition 187.36 — Wedge, cone, suspension, mapping cone ¶

Let $(X, x_{0})$ and $(Y, y_{0})$ be pointed spaces.

The wedge $X \lor Y$ is the quotient of $X ⊔ Y$ that identifies $x_{0}$ with $y_{0}$ ; its base point is the class of $x_{0}$ .
The cone $C X$ is the quotient of $X \times [0, 1]$ that identifies all of $X \times {1}$ to a single point.
The suspension $Σ X$ is the quotient of $X \times [0, 1]$ that identifies all of $X \times {0}$ to one point $N$ and all of $X \times {1}$ to another point $S$ .
For a continuous $f : X \to Y$ , the mapping cone $C_{f}$ is the pushout of $C X \leftarrow X \overset{f}{\to} Y$ , where $X \to C X$ is $x \mapsto [(x, 0)]$ .

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Example 187.37 — The circle as a quotient ¶

Let $\sim$ identify $0$ with $1$ in $[0, 1]$ and change nothing else. The map $e |_{[0, 1]} : [0, 1] \to S^{1}$ is continuous, surjective, and satisfies $e (0) = e (1)$ , so lemma 187.33 gives a continuous $\bar{e} : [0, 1] / \sim \to S^{1}$ . It is injective: $e (s) = e (s^{'})$ with $s, s^{'} \in [0, 1]$ forces $s - s^{'} \in Z$ , hence $s = s^{'}$ or ${s, s^{'}} = {0, 1}$ , and those two points are identified. The domain is compact, being a continuous image of the compact $[0, 1]$ by lemma 187.14(i), and $S^{1}$ is Hausdorff as a metric space, so $\bar{e}$ is a homeomorphism by proposition 187.15. Thus $[0, 1] / (0 \sim 1) \approx S^{1},$ and the two sets are certainly not equal: one consists of equivalence classes of real numbers, the other of pairs of real numbers.

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Theorem 187.38 — Suspending the circle ¶

$Σ S^{1} \approx S^{2}$ , where $S^{n} : = {x \in R^{n + 1} : ‖ x ‖ = 1}$ .

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Proof of Theorem 187.38 — Suspending the circle

Proof. Regard $S^{1} \subseteq R^{2}$ and define $F : S^{1} \times [0, 1] \to S^{2} \subseteq R^{3}, F (z, t) : = (\sin (π t) z, \cos (π t)) .$ Values. $‖ F (z, t) ‖^{2} = \sin^{2} (π t) ‖ z ‖^{2} + \cos^{2} (π t) = 1$ , so $F$ lands in $S^{2}$ .

Continuity. Each of the three coordinates is a product of continuous real-valued functions of $(z, t)$ .

Surjectivity. Given $(y, c) \in S^{2}$ with $y \in R^{2}$ and $c \in [- 1, 1]$ , choose $t : = \frac{1}{π} \arccos c \in [0, 1]$ ; then $\sin (π t) = \sqrt{1 - c^{2}} = ‖ y ‖$ . If $‖ y ‖ \neq 0$ take $z : = y / ‖ y ‖$ ; otherwise $c = \pm 1$ , $t \in {0, 1}$ , and any $z$ works.

Identifications. $F (z, t) = F (z^{'}, t^{'})$ forces $\cos π t = \cos π t^{'}$ , hence $t = t^{'}$ as both lie in $[0, 1]$ ; then either $\sin π t \neq 0$ and $z = z^{'}$ , or $t \in {0, 1}$ and both sides equal $(0, 0, \pm 1)$ . So $F$ identifies exactly the two ends $S^{1} \times {0}$ and $S^{1} \times {1}$ , which is the relation defining $Σ S^{1}$ .

By lemma 187.33 the induced map $\bar{F} : Σ S^{1} \to S^{2}$ is continuous, and by the previous paragraph it is a bijection. Its domain is compact, being a continuous image of the compact space $S^{1} \times [0, 1]$ (lemma 187.14(i),(iv), using that $S^{1}$ is a closed bounded subset of the compact square $[- 1, 1]^{2}$ ), and $S^{2}$ is Hausdorff. Proposition 187.15 finishes the proof. ◻

The same formula with $z$ ranging over $S^{n - 1} \subseteq R^{n}$ and the target $S^{n} \subseteq R^{n + 1}$ proves $Σ S^{n - 1} \approx S^{n}$ for every $n \geq 1$ : each step of the proof used only $‖ z ‖ = 1$ and compactness of $S^{n - 1} \times [0, 1]$ , never the dimension. This is the inductive description of the spheres that the later synthetic constructions imitate.

Example 187.39 — Cones are contractible ¶

For any $X$ , the cone $C X$ is contractible. Define $G : (X \times [0, 1]) \times [0, 1] \to C X$ by $G ((x, s), t) : = [(x, (1 - t) s + t)]$ . It is continuous as a composite of a continuous map into $X \times [0, 1]$ with the projection $q$ . Since $G ((x, 1), t) = [(x, 1)]$ is the cone point for every $t$ , the map $G$ is constant on the classes of the defining relation in its first argument, so lemma 187.33 — applied for each fixed $t$ , and then to the map $C X \times [0, 1] \to C X$ obtained from $q \times {id}_{[0, 1]}$ , which is a quotient map because $[0, 1]$ is compact — gives a homotopy from ${id}_{C X}$ to the constant map at the cone point.

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Remark 187.40 — Where compactness entered ¶

Example 187.37, Theorem 187.38 both produced a continuous bijection and then invoked proposition 187.15. Without compactness the conclusion fails: example 187.8 is a continuous bijection $[0, 1) \to S^{1}$ with the same image and no continuous inverse. A quotient construction therefore does not automatically produce the expected space; it produces it when the source is compact and the target is Hausdorff.

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Exercise 187.12 ¶

★★☆ Prove that $S^{1} \lor S^{1}$ is homeomorphic to the subspace of $R^{2}$ consisting of the two circles of radius $1$ centered at $(\pm 1, 0)$ . Say where compactness and the Hausdorff property are used.

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Exercise 187.13 ¶

★★☆ Let $f : X \to Y$ be constant at $y_{0}$ . Prove that $C_{f} \approx (Σ X) \lor Y$ when $X$ is compact, $Y$ is Hausdorff, and both are pointed. Identify the equivalence classes on both sides before writing a map.

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Exercise 187.14 ¶

★★☆ Let $\sim$ identify $x$ and $y$ in $R$ whenever $x - y \in Q$ . Prove that the quotient topology on $R / \sim$ is indiscrete, and conclude that a quotient of a Hausdorff space need not be Hausdorff. Which hypothesis of proposition 187.15 does this example remove?

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Geometric simplices

One more family of spaces is needed later, and only its geometry is needed here.

Definition 187.41 — Standard simplex ¶

The standard $n$ -simplex is $Δ^{n} : = {(x_{0}, \dots, x_{n}) \in R^{n + 1} : x_{i} \geq 0, \sum_{i = 0}^{n} x_{i} = 1},$ with the subspace topology. For $0 \leq i \leq n$ , its $i$ -th face is the subspace ${x \in Δ^{n} : x_{i} = 0}$ , and the map $d^{i} : Δ^{n - 1} \to Δ^{n}$ inserting $0$ in position $i$ is a homeomorphism onto that face.

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Example 187.42 — The first three simplices ¶

$Δ^{0}$ is a point. $Δ^{1}$ is homeomorphic to $[0, 1]$ by $(x_{0}, x_{1}) \mapsto x_{1}$ , whose inverse $t \mapsto (1 - t, t)$ is continuous; so a path may be described equivalently as a continuous map $Δ^{1} \to X$ . $Δ^{2}$ is a triangle with its interior, and its three faces are the images of the three maps $d^{0}, d^{1}, d^{2}$ . A continuous map $Δ^{2} \to X$ restricted to the three faces gives three paths, and the compatibility of their endpoints is exactly the statement that the images of $d^{i}$ meet as the vertices of the triangle require.

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No combinatorics of simplices is developed here: the face maps are recorded because example 187.42 is the geometric picture that later combinatorial definitions must reproduce, not because any argument of this chapter uses them.

A translation ledger

The type theory of chapter 30 has objects that behave formally like several notions above. The correspondence is stated here as a ledger, not as a theorem: each row names a classical notion, the object of the type theory already constructed in this book that obeys the same laws, and the exact respect in which the correspondence is currently incomplete.

Classical notion	Object already available	Status
Space $X$	A type $A$	No topology is present on $A$ ; only the path algebra is matched.
Point $x \in X$	A term $a : A$	Exact.
Path $p$ from $x$ to $y$	An identification $p : a =_{A} b$ (definition 30.1)	Exact for the algebra below; $[0, 1]$ has no internal counterpart.
Constant path, reversal, concatenation	$refl$ , $p^{- 1}$ , $p \cdot q$ (theorem 30.20)	The classical operations satisfy the same laws only up to homotopy, and the internal ones only up to higher identifications; the match is exact.
Path homotopy	An identification between identifications	Exact.
Continuous map $f$ and its action on paths	A function $f$ and ${ap}_{f}$ (proposition 30.21)	Exact; internally every function acts on paths, which is the internal counterpart of continuity rather than a theorem about it.
Homotopy $f ≃ g$	A pointwise family of identifications $\prod_{x} (f (x) = g (x))$	The classical notion is one map out of $X \times [0, 1]$ ; the internal notion is a family.
Homotopy equivalence	—	Not yet defined internally; a definition and its comparison with quasi-inverses are still owed.
Deformation retraction	—	No internal counterpart is available.
Quotient, wedge, cone, suspension	—	Not available: the type theory of chapter 28 generates types by constructors for points only.

Three entries in the ledger are blank, and they are blank for one reason: the classical constructions of section 187.6 attach new points and new paths to a space, while every type former available so far attaches only new points. That is the construction problem the synthetic development must solve, and the classical statements proved above are what its solutions will be measured against.

Suggested first pass.

None of these problems is a prerequisite for a later chapter. Begin with exercise 187.15, then exercise 187.17; the implementation project exercise 187.19 may be attempted at any time.

Exercise 187.15 ¶

★★☆ Extend lemma 187.11 to a finite closed cover $X = A_{1} \cup \dots \cup A_{k}$ , and then show by example that it fails for the infinite closed cover of $R$ by the singletons ${x}$ : exhibit a family of continuous maps on the pieces agreeing on overlaps whose union is not continuous.

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Exercise 187.16 ¶

★★★ Let $C : = ({0} \cup {1 / n : n \geq 1}) \times [0, 1] \cup [0, 1] \times {0} \subseteq R^{2}$ , the comb space. Prove that $C$ is contractible, and that the deformation cannot be taken rel the point $(0, 1)$ : there is no homotopy $D$ with $D (x, 0) = x$ , $D (x, 1) = (0, 1)$ and $D ((0, 1), t) = (0, 1)$ for all $t$ . (For the second part, examine the first coordinate of $D ((0, 1), t)$ for $t$ near $0$ and use continuity at $(0, 1)$ together with the fact that a path in $C$ starting at $(0, 1)$ and leaving the segment ${0} \times [0, 1]$ must pass through $(0, 0)$ .) Conclude that “contractible” and “contractible rel a chosen point” are different statements.

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Elementary Topology and Classical Homotopy

Exercise 187.17 ¶

★★☆ Let $T : = ([0, 1] \times [0, 1]) / \sim$ identify $(s, 0) \sim (s, 1)$ and $(0, t) \sim (1, t)$ . Prove $T \approx S^{1} \times S^{1}$ by constructing the map $(s, t) \mapsto (e (s), e (t))$ , checking exactly which pairs it identifies, and naming the two hypotheses of proposition 187.15 that make the induced bijection a homeomorphism.

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Exercise 187.18 ¶

★★☆ Prove that $Σ (X \lor Y) \approx Σ X \lor Σ Y$ for compact pointed $X, Y$ with Hausdorff suspensions. Begin by describing the equivalence classes of both sides, and use lemma 187.35 in both directions before appealing to proposition 187.15.

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Exercise 187.19 ¶

★★★ Practical project.circle-loop-normalizer Implement in Agda or Kappa the following edge-path model of loops on the circle and its normalizer. Fix $k \geq 1$ and let a word be a finite list over ${+, -}$ . A word $w$ denotes the loop $γ_{w}$ in $S^{1}$ obtained by traversing consecutive arcs of length $1 / k$ , forward for $+$ and backward for $-$ , starting at $b$ ; only words whose partial sums return to $0$ at the end denote loops, and your representation must make that condition checkable. Implement: concatenation of words; reversal, which reverses the list and swaps the two symbols; the degree $\deg (w) : = (# of + - # of -) / k$ , defined only when $k$ divides the difference; and the normalizer that repeatedly deletes adjacent pairs $+ -$ or $- +$ .

The invariant your implementation must maintain is that deletion of an adjacent cancelling pair changes neither the endpoints of $γ_{w}$ nor its path-homotopy class: the deleted pair is a path of the form $p \cdot \overset{―}{p}$ , and exercise 187.4 shows that such a path is homotopic rel endpoints to a constant. The concrete result is a function taking a word to its normal form together with its degree.

Acceptance test. With $k = 4$ : the words $(+)^{4 m}$ and $(+)^{4 n}$ concatenate to a word whose normal form is $(+)^{4 (m + n)}$ and whose degree is $m + n$ , for the four pairs $(m, n) \in {(1, 1), (2, 3), (1, - 1), (0, 5)}$ — the negative case being represented by $(-)^{4}$ ; the word $+ + - - + + - -$ normalizes to the empty word with degree $0$ ; the word $+ - + - + + + +$ normalizes to $(+)^{4}$ with degree $1$ ; and the normalizer applied to $w$ followed by reversal gives the reversal of the normal form of $w$ , with degree negated, on all of the above inputs. A mutation of the normalizer that deletes a non-adjacent pair must fail the third test.

Your program computes with words, not with continuous loops: it illustrates theorem 187.21 and does not prove it, and it establishes nothing about loops that are not of the form $γ_{w}$ .

Referenced from 3 locations

Elementary Topology and Classical Homotopy

Bibliographic notes

The point-set material of section 187.1, section 187.2 follows Hatcher’s introductory notes [Hat], which take open sets as primitive and reach quotients quickly; the proofs of lemma 187.13, lemma 187.14 are the standard ones given there. Viro, Ivanov, Netsvetaev and Kharlamov [VINK08] organize the same material as a problem sequence, and several exercises above — in particular exercise 187.14, exercise 187.17 — are of the kind they use to force the definitions to be applied rather than recalled. Morris [Mor24] is the slower alternative exposition for a reader who wants more examples before the homotopy material.

The homotopy material of section 187.3, section 187.4 is the opening of May’s concise treatment [May99] and of Hatcher’s introductory chapter; the deformation retraction of theorem 187.26 and the suspension homeomorphism of theorem 187.38 are their standard examples. Theorem 187.21 is stated here with the explicit interpolation in the exponent, which is the calculation that the covering-space machinery of the next chapter turns into a general method.

The ledger of section 187.8 follows the reading of identity types as paths that organizes the HoTT Book [Uni13] and Rijke’s introduction [Rij25]; those texts are the source of the correspondence, not of any theorem proved in this chapter. Nothing above is proved by that analogy, and no classical statement proved here is used as evidence for an internal one.