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A Taste of Algebraic Topology

Previously, we used topological ideas to motivate the higher-level study of differentiable geometry. Here, we will use topological ideas to motivate the higher-level study of algebraic topology, inspired by this concise guide.

Let K be a topological space. A path f in K is a continuous map f : [0, 1] \to K. That is, f \in \mathcal C([0, 1], K). We call f(0) the starting point and f(1) the ending point, and say that f is a path from f(0) to f(1). Given x, y, let \mathcal C_{x,y}(K) \subseteq \mathcal C([0, 1], K) denote the subset of paths from x to y.

Definition 1. For any f,f' \in \mathcal C_{x,y}(K), we say that f and f' are path-homotopic if there exists a continuous map F : [0,1] \times [0,1] \to K such that

F(\cdot ,0) = f,\quad F( \cdot ,1) = f',\quad F(0, \cdot ) = x,\quad F(1, \cdot ) = y.

The first argument measures the “progress” of a point between x and y, and the second argument measures the “interpolation” between f and f'. In this case, we write f \simeq f'. In particular, f is a loop centered at x if f \in \mathcal C_{x,x}(K).

Lemma 1. Given x, y \in K, \simeq forms an equivalence relation on \mathcal C_{x,y}(K). Hence, [x] = [y] if and only if x \simeq y.

Proof. We need to prove that \simeq is reflexive, symmetric, and transitive.

For reflexivity, fix f \in \mathcal C_{x,y}(K). Define the path-homotopy F : [0,1] \times [0,1] \to K by F(\cdot, t) = f so that f \simeq f.

For symmetry, fix f, f' \in \mathcal C_{x, y}(K) and F be a path-homotopy from f to f'. Then G := F(\cdot, 1-\cdot) defines a path-homotopy from f' to f so that f' \simeq f.

Finally, for transitivity, fix f, f', f'' \in \mathcal C_{x, y}(K) and let F, G be path-homotopies from f to f' and from f' to f'' respectively. Since F(\cdot, 1) = G(\cdot, 0), we can define the path-homotopy H from f to f'' via

H(s, \cdot) := \begin{cases} F(2s, \cdot) & t \leq 1/2, \\ G(2s-1, \cdot) & t \geq 1/2, \end{cases}

so that f \simeq f'', as required.

Lemma 2. Given x, y, z \in K, f \in \mathcal C_{x,y}(K), g \in \mathcal C_{y,z}(K), define g \cdot f \in \mathcal C_{x,z}(K) by

(g \cdot f)(t):= \begin{cases} f(2t), & t \leq 1/2, \\ g(2t-1), & t \geq 1/2. \end{cases}

Then for any f \simeq f' and g \simeq g', g \cdot f \simeq g' \cdot f'. In this case, the definition [g] \cdot [f] := [g \cdot f] is well-defined, since given that [f] = [f'] and [g] = [g'],

[g] \cdot [f] = [g \cdot f] = [g' \cdot f'] = [g'] \cdot [f'].

Proof. Let F be a path-homotopy from f to f' and G be a path-homotopy from g to g'. Define the continuous map H from f \cdot g to f' \cdot g' by, for any s, t \in [0, 1],

H( s, t ) := \begin{cases} F( 2s, t ), & s \leq 1/2, \\ G( 2s-1, t ), & s \geq 1/2. \end{cases}

It is clear that

H(0, \cdot) = F( 0, \cdot) =x, \quad H(1, \cdot) = G(1, \cdot ) = z.

Next, for any s \in [0, 1],

\begin{aligned} H( s, 0) &= \begin{cases} F( 2s, 0 ), & s \leq 1/2, \\ G( 2s-1, 0 ), & s \geq 1/2, \end{cases} \\ &= \begin{cases} f( 2s ), & s \leq 1/2, \\ g( 2s-1 ), & s \geq 1/2, \end{cases} \\ &= (g \cdot f)(s). \end{aligned}

Thus, H(\cdot, 0) = g \cdot f. Similarly, H(\cdot , 1) = g' \cdot f'. Hence, H is a path-homotopy from g \cdot f to g' \cdot f'.

Lemma 3. For any x \in K, define e_x \equiv e_x(\cdot) := x \in \mathcal C_{x,x}(K). For any f \in \mathcal C_{x,y}(K), define f^{-1} := f(1 - \cdot) \in \mathcal C_{y,x}(K).

  • For any f \in \mathcal C_{w,x}(K), g \in \mathcal C_{x,y}(K), h \in \mathcal C_{y,z}(K), (h \cdot g) \cdot f \simeq h \cdot (g \cdot f).
  • For any f \in \mathcal C_{x,y}(K), f \cdot e_x \simeq f \simeq e_y \cdot f.
  • For any f \in \mathcal C_{x,y}(K), f^{-1} \cdot f \simeq e_x, f \cdot f^{-1} \simeq e_y.

Proof. For associativity, we first write out the two functions explicitly as follows. Denoting the left-hand side by \phi_{\mathrm L} and the right-hand side by \phi_{\mathrm R}, we leave it as an exercise to verify that

\begin{aligned} \phi_{\mathrm L}(t) &= \begin{cases}f(4t), & t \in [0,1/4], \\ g(4t-1), & t \in [1/4,1/2], \\ h(2t-1), & t \in [1/2,1], \end{cases} \\ \phi_{\mathrm R}(t) &= \begin{cases} f(2t), & t \in [0,1/2], \\ g(4t-2), & t \in [1/2,3/4], \\ h(4t-3), & t \in [3/4,1]. \end{cases} \end{aligned}

Define the lines s = 0, s = \frac 14(t+1), s = \frac 14(t+2) and s = 1 so that we partition [0, 1] \times [0, 1] into three regions A \cup B \cup C:

\begin{aligned} A &:= \textstyle \{(s, t) : 0 \leq s \leq \frac 14 (t+1)\}, \\ B &:= \textstyle \{(s, t) : \frac 14 (t+1) \leq s \leq \frac 14 (t+2)\}, \\ C &:= \textstyle \{(s, t) : \frac 14 (t+2) \leq s \leq 1\}. \end{aligned}

We will construct the path-homotopy using the “interpolation” and “progress” analogy. Any (s ,t) will belong to at least one of A, B, C and at most two of them (on the shared boundary line). Suppose (s, t) belongs to C for illustrative purposes. Then given t, the values of s will vary from \frac 14(t+2) to 1. The ratio

\displaystyle \frac{s - \frac 14(t+2)}{1 - \frac 14(t+2)} = \frac{4s - (t+2)}{ 2 - t }.

then measures the extent of “progress”, and this quantity we substitute into the corresponding function, which in the case of C would be h. Hence, we define the continuous map H from \phi_{\mathrm L} to \phi_{\mathrm R} as follows:

H(s, t) := \begin{cases} f( 4s/ (t+1) ),\quad (s, t) \in A, \\ g (4s - (t+1)),\quad (s, t) \in B, \\ \displaystyle h \left( \frac{4s - (t+2)}{ 2 - t } \right),\quad (s, t) \in C. \end{cases}

and check that it is a path-homotopy, since for any s \in [0, 1],

\begin{aligned} H(s, 0) &= \begin{cases} f( 4s ),\quad s \in [0, 1/4], \\ g (4s - 1),\quad s \in [1/4, 1/2], \\ h(2s-1),\quad s \in [1/2, 1], \end{cases} \\ &= \phi_{\mathrm L}(s). \end{aligned}

We can use similar calculations to conclude that

H(\cdot ,0 ) = \phi_{\mathrm L},\quad H( \cdot, 1) = \phi_{\mathrm R}, \quad H(0,\cdot) = x,\quad H(1, \cdot) = z.

The proof of the identity and inverse properties follow effectively the same argument, but with correctly bookkept partitions as well as careful definitions of \phi_{\mathrm L} and \phi_{\mathrm R}.

  • For the case f \cdot e_x \simeq f, we partition [0, 1] \times [0, 1] using the lines s = 0, s = \frac 12 (1-t), and s = 1.
  • For the case e_y \cdot f \simeq f, we use the lines s = 0, s = \frac 12 (1+t), and s = 1.
  • For the inverse property, we use the lines s = 0, s = \frac 12 t, s = 1 - \frac 12 t, and s = 1.

Theorem 1. For any x \in K, the set \pi_1(K,x) := \mathcal C_{x,x}(K)/{\simeq} equipped with \cdot forms a group with identity [e_x] and inverse [f]^{-1} = [f^{-1}].

Proof. Lemmas 1–3.

If two points x and y are connected by a path, then the fundamental groups with respect to either point are essentially the same. We can formalise this notion using group isomorphisms:

Lemma 4. Any \alpha \in \mathcal C_{x,y}(K) induces a well-defined group isomorphism \gamma_{[\alpha]} : \pi_1(X,x) \to \pi_1(X, y) given by \gamma_{[\alpha]}([f]) := [\alpha \cdot f \cdot \alpha^{-1}]. Furthermore, for any \beta \in \mathcal C_{y,z}(K),

\gamma_{[\beta \cdot \alpha]} = \gamma_{[\beta]} \circ \gamma_{[\alpha]},

so that for any \alpha \in \mathcal C_{x,y}(K), \gamma_{[a]}^{-1} = \gamma_{[a^{-1}]}.

Proof. We first observe that given f, g \in \mathcal C_{x,x}(K),

\alpha \cdot (g \cdot f) \cdot \alpha^{-1} \simeq (\alpha \cdot g \cdot \alpha^{-1}) \cdot (\alpha \cdot f \cdot \alpha^{-1}),

and we obtain \gamma_{[\alpha]} as a group homomorphism. Bijectivity is also obvious, since

\alpha \cdot f \cdot \alpha^{-1} \simeq \alpha \cdot f' \cdot \alpha^{-1} \quad \Rightarrow \quad f \simeq f'

by Lemma 3. Given another path \beta \in \mathcal C_{y,z}(K), for any f \in \mathcal C_{x,x}(K), the observation

(\beta \cdot \alpha) \cdot f \cdot (\beta \cdot \alpha)^{-1} \simeq \beta \cdot (\alpha \cdot f \cdot \alpha^{-1}) \cdot \beta^{-1}

establishes \gamma_{[\beta \cdot \alpha]} = \gamma_{[\beta]} \circ \gamma_{[\alpha]}. In particular,

\gamma_{[\alpha]} \circ \gamma_{[\alpha^{-1}]} = \gamma_{[e_x]} = \gamma_{[\alpha^{-1}]} \circ \gamma_{[\alpha]}

implies that \gamma_{[\alpha]}^{-1} = \gamma_{[\alpha]}.

Corollary 1. If \pi_1(X,x) is abelian, then for any \alpha, \beta \in \mathcal C_{x,y}(K), \gamma_{[\alpha]} = \gamma_{[\beta]}.

Proof. By Lemma 4,

\begin{aligned} (\gamma_{[\beta]}^{-1} \circ \gamma_{[\alpha]})([f]) &= [\beta]^{-1} \cdot [\alpha]^{-1} \cdot [f] \cdot [\alpha] \cdot [\beta] \\ &= [\beta]^{-1} \cdot [\beta] \cdot [\alpha]^{-1} \cdot [\alpha] \cdot [f] \\ &= [e_x] \cdot [e_x] \cdot [f] = [f], \end{aligned}

so that \gamma_{[\beta]}^{-1} \circ \gamma_{[\alpha]} = \gamma{[e_x]} implies \gamma_{[\alpha]} = \gamma_{[\beta]}.

We have scratched the tip of the tip of the massive iceberg of algebraic topology. Before we explore another application in the context of topological vector spaces, let’s use our algebraic topological notions to prove a crucial theorem in the existence of Nash equilibriums in game theory, called Brouwer’s fixed point theorem.

—Joel Kindiak, 13 Jun 25, 2354H

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