Computing Topological Groups

We explored algebraic topology with the goal of an alternate approach to distinguishing topological spaces, taking heavy advantage of the continuity in the definition of path-homotopy. We can first generalise the idea to homotopy, where we don’t require the paths to have the same start and end points.

Definition 1. Given topological spaces $K, L$ and continuous maps $f, g : K \to L$ , a homotopy $H$ from $f$ to $g$ , denoted $f \simeq g$ , is a continuous map $H : K \times [0, 1] \to L$ such that

$H( \cdot , 0) = f,\quad H (\cdot , 1) = g.$

In the case $K = [0, 1]$ and that $H(0, \cdot)$ and $H(1, \cdot)$ are constant, we call $H$ a path-homotopy.

Lemma 1. $\simeq$ forms an equivalence relation on $\mathcal C(K, L)$ .

In fact, if $f$ is a homeomorphism with inverse map $g = f^{-1}$ , then the paths $f$ and $f^{-1}$ are homotopic to one another. Thus, homotopy-equivalence generalises homeomorphisms as a form of topological invariance.

Lemma 2. Given any continuous map $p : K \to L$ , define the map $p_* : \pi_1(K, x) \to \pi_1(L, p(x))$ by $p_*([f]) := [p \circ f]$ . Then for any continuous map $q : L \to M$ , $q_* \circ p_* = (q \circ p)_*$ .

Intuitively, since $f(0) = f(1) = x$ , $p_*([f])$ denotes the collection of loops around a given base point $p(x) = (p \circ f)(0) = (p \circ f)(1)$ (that, of course, are path-homotopic to one another). Given continuous maps $p, q : K \to L$ such that $p \simeq q$ , would it follow that $p_* = q_*$ ? Unfortunately not, since we don’t get $p(x) = q(x)$ in general, and thus these two induced maps may not even have the same base points. However if $p \simeq q$ , then there is a homotopy $\alpha$ between them. Let’s spell out its details for further analysis:

$\alpha(\cdot, 0) = p,\quad \alpha( \cdot , 1) = q.$

Roughly speaking we should get some kind of connection between them. In fact,

$p(x) = \alpha(x, 0),\quad q(x) = \alpha(x, 1),$

suggesting that if we viewed $\tilde \alpha := \alpha(x, \cdot )$ as a path from $p(x)$ to $q(x)$ , we can get a connection. The key tool of choice is $\gamma_{\tilde \alpha}$ as discussed in the previous post.

Theorem 1. Given homotopic continuous maps $p, q : K \to L$ with homotopy $\alpha$ , $\gamma_{[\tilde{ \alpha }]} \circ p_* = q_*$ .

Proof. We first spell out the goal. Given any loop $f$ based on $x$ ,

$(\gamma_{[\tilde{ \alpha }]} \circ p_*)([f]) = \gamma_{[\tilde{ \alpha }]} (p_*([f])) = \gamma_{[\tilde{ \alpha }]} ([p \circ f]) = \tilde \alpha \cdot (p \circ f) \cdot \tilde \alpha^{-1}.$

Thus, we need to prove that $\tilde \alpha \cdot (p \circ f) \cdot \tilde \alpha^{-1} \simeq q \circ f$ . Define the path-homotopy $H = \alpha(f(\cdot), \cdot)$ so that

$H(\cdot, 0) = p \circ f,\quad H(\cdot , 1) = q \circ f, \quad H(0, \cdot) = H(1, \cdot) = \tilde \alpha.$

Hence, we can define a loop based on $(0, 0) \in [0,1 ] \times [0, 1]$ via the straight-line edges $H(0, 0) \to H(1 ,0) \to H(1,1) \to H(0, 1)$ , resulting in a path that looks like

$\beta := \tilde \alpha^{-1} \cdot (q \circ f)^{-1} \cdot \tilde \alpha \cdot (p \circ f),$

after proper rescaling of the parameters. To prove that this path and $e_x$ are equivalent, for each $t \in [0, 1]$ , define the path $r_t$ that is a shrinkage of the original unit square into $[0, t] \times [0, t]$ .

Intuitively this shrinking continuously approaches $(0, 0)$ as $t \to 0$ . Then, define the continuous map $K$ by

$K(s, t) = H(r_t (s))$ .

We leave it to verify that

$K( \cdot , 0) = e_{p(x)},\quad K(\cdot , 1) = \beta, \quad K(0, \cdot) = p(x) = K(1, \cdot),$

so that $K$ is the desired path-homotopy, as required.

Much of our discussion still rings hollow: we have not computed a single fundamental group! Well, let’s use our tools to evaluate two fundamental groups, a trivial one and a not-so-trivial one.

Example 1. $\pi_1(\mathbb R, 0) = 0$ , where $0 = \{0\} = \{ [e_0] \}$ denotes the trivial sub-group in $(\mathbb R, +)$ , and $=$ refers to group isomorphism.

Proof. Fix $[f] \in \pi_1(\mathbb R, 0)$ . Firstly, define the homotopy $H$ from $\mathrm{id}_{\mathbb R}$ to $0$ by the straight-line homotopy

$H(\cdot, t) = (1-t) \cdot \mathrm{id}_{\mathbb R} + t \cdot 0 = (1-t) \cdot \mathrm{id}_{\mathbb R}$

for any $t \in [0, 1]$ , since

$H(\cdot , 0) = \mathrm{id}_{\mathbb R},\quad H(\cdot , 1) = 0.$

To prove that $f \simeq e_0$ , define the path-homotopy $K = H(f(\cdot), \cdot)$ from $f$ to $0$ , since

$K(\cdot , 0) = H(f(\cdot), 0) = f,\quad K(\cdot , 1) = 0,\quad K(0, \cdot) = 0 = K(1, \cdot).$

For our not-so-trivial example $\pi_1(S^1, 1)$ , we will need to introduce the idea of path-lifting. Define $S^1 := \{z \in \mathbb C : |z| = 1\}$ , where $\mathbb C$ is equipped with the usual topology, and the usual exponential map $p : \mathbb R \to S^1$ is defined by $s \mapsto e^{2 \pi i s}$ .

Lemma 1. For any connected open set $V \subseteq S^1$ , there exists $r_V \in [0, 1)$ such that

$\displaystyle p^{-1}(V) \subseteq \bigcup_{n \in \mathbb Z} (r_V + n, r_V + n+1)$

and $p|_{(r_V + n, r_V + n+1)}$ is injective for each $n \in \mathbb Z$ .

Proof. Fix any connected open set $V \subseteq S^1$ . Then $p^{-1}(V) \cap [0, 1]$ will take either the form $(a, b) \subseteq (0, 1)$ if $1 \notin V$ or $[0, a) \cup (b, 1]$ if $1 \in V$ for some $0 < a < b < 1$ . In the latter case, defining $r := b$ , $p^{-1}(V) \cap [r, r+1] \subseteq (r, r+1)$ . In general, there exists $r_V \in [0, 1)$ such that

$p^{-1}(V) \cap [r_V, r_V+1] \subseteq (r_V, r_V+1).$

Subsequently,

$\displaystyle p^{-1}(V) \subseteq \bigcup_{n \in \mathbb Z} (r_V + n, r_V + n+1),$

and $p|_{(r_V + n, r_V + n+1)}$ is injective for each $n \in \mathbb Z$ .

Lemma 2. Let $K \subseteq \mathbb R^n$ be compact. For any continuous map $f : K \to S^1$ with $f(0) = 1$ , there exists a unique continuous map $\tilde f : K \to \mathbb R$ such that $\tilde f(0) = 0$ and $f = p \circ \tilde f$ .

Proof. We will use $f$ and $p$ to construct $\tilde f$ . By the argument in Lemma 1, to obtain $\tilde f$ such that $f = p \circ \tilde f$ , we can define, roughly speaking, that $\tilde f = p^{-1} \circ f$ . This definition, of course, is nonsense, since $p : \mathbb R \to S^1$ is not injective: $p(0) = p(1) = 1$ .

Now, since $f$ is continuous, $f(K) \subseteq S^1$ is compact. Thus, we can cover $f(K)$ with a finite open cover $\{V_1,\dots,V_n\}$ of connected open sets. We obtain the finite open cover $\{U_1,\dots,U_n\}$ of $K$ , where $U_i = f^{-1}(V_i)$ . By Lemma 1, for each $i$ , there exists $r_i \in [0, 1)$ such that

$\displaystyle p^{-1}(V_i) \subseteq \bigcup_{k \in \mathbb Z} (r_i + k, r_i + k+1)$

and $p_{i,k} := p|_{(r_i + k, r_i + k+1)}$ is injective for each $k \in \mathbb Z$ . We may assume $0 < r_1 < r_2 < \dots < r_n < 1$ without loss of generality.

Hence, we define $\tilde f|_{U_i} := p_{i, -1}^{-1} \circ f|_{U_i}$ and paste all continuous pieces together to create $\tilde f : K\to \mathbb R$ . Furthermore, since $f(0) = 1 \in p_{n,-1}((r_n-1, r_n))$ , we define

$\tilde f(0) = (p_{1,-1}^{-1} \circ f)(0) = p_{1,-1}^{-1}(f(0)) = p|_{1, -1}^{-1}(1) = 0,$

establishing existence.

The uniqueness follows from the decomposition of $p$ into piece-wise bijective segments, then assuming $\tilde f_1, \tilde f_2$ satisfies

$f = p \circ \tilde f_1 = p \circ \tilde f_2,$

we obtain $\tilde f_1 = p|_{i, -1}^{-1} \circ f = \tilde f_2$ for each $i$ , so that $\tilde f_1 = \tilde f_2$ .

Lemma 3. If $f, g$ are loops with $f(0) = g(0) = 1$ and $f \simeq g$ , then $\tilde f(1) = \tilde g(1)$ .

Proof. Let $H : [0, 1] \times [0, 1] \to S^1$ be a path homotopy from $f$ to $g$ :

$H( \cdot, 0) = f, \quad H( \cdot, 1) = g, \quad H(0, \cdot) = H(1, \cdot) = 1.$

Since $[0, 1] \times [0, 1]$ is compact, we use Lemma 2 to obtain a unique map $\tilde H : [0, 1] \times [0, 1] \to \mathbb R$ such that $H = p \circ \tilde H$ . In particular,

$p(\tilde H(1, \cdot)) = H(1, \cdot) = 1,$

so that $\tilde H(1, \cdot)$ is some fixed integer. Furthermore,

$p(\tilde f(\cdot)) = f = H( \cdot, 0) = p(\tilde H( \cdot, 0)).$

Taking advantage of local injectivity, $\tilde f = \tilde H( \cdot, 0)$ and likewise $\tilde g = \tilde H(\cdot, 1)$ , yielding

$\tilde f(1) = \tilde H(1, 0) = H(1, 1) = \tilde g(1),$

as required.

Example 2. $\pi_1(S^1, 1) = \mathbb Z$ .

Proof. Fix any loop $f : [0, 1] \to S^1$ with $f(0) = f(1) = 1$ . Since $f = p \circ \tilde f$ , we have

$e^{2 \pi \tilde f(1) i} = p(\tilde f(1)) = f(1) = 1,$

which implies that $\tilde f(1) \in \mathbb Z$ . By Lemma 3, $\tilde f(1) = \tilde g(1)$ whenever $f \simeq g$ . Hence, the map

$j : \pi_1(S^1, 1) \to \mathbb Z,\quad [f] \mapsto \tilde f(1)$

is well-defined. We claim that $j$ is a group isomorphism. To that end, observe that for each $n$ , defining $f_n(s) = e^{2 n \pi i s}$ , $\tilde f_n(s) = ns$ so that

$j([f_n]) = \tilde f_n(1) = n.$

This establishes $j$ as surjective. To establish $j$ as injective, suppose $j([f]) = j([e_1])$ . Then $\tilde f(1) = 0$ so that $\tilde f$ is a loop at $0 \in \mathbb R$ .

Since $\pi_1(\mathbb R, 0) = \{0\} = \{[e_0]\}$ , $[\tilde f] = [e_0]$ . Hence,

$[f] = p_*([\tilde f]) = p_*([e_0]) = [p \circ e_0] = [e_1],$

as required. Finally, to establish $j$ as an isomorphism, notice that for any $f \in \pi_1(S^1, 1)$ , there exists $f_n \in \pi_1(S^1, 1)$ such that

$j([f]) = n = j([f_n]).$

Since $j$ is injective, $[f] = [f_n]$ so that $f \simeq f_n$ . Hence, if $g \simeq f_m$ ,

$\begin{aligned} j([g] \cdot [f]) &= j([f_m] \cdot [f_n]) \\ &= j([f_m \cdot f_n]) \\ &= j([f_{m+n}]) \\ &= m + n \\ &= j([g]) + j([f]). \end{aligned}$

Whew, that was a LOT of hard work! There is a whole lot more to explore about homotopy, but we will not venture there any time soon. For now, let’s use our algebraic topology toolkit to prove several famous results in surprisingly elegant ways. For now, the key takeaways are:

$\pi_1(\mathbb R, 0) = \{0\},\quad \pi_1(S^1, 1) = \mathbb Z.$

—Joel Kindiak, 16 Jun 25, 1825H

KindiakMath

Computing Topological Groups

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Computing Topological Groups

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