Theorems: Algebraic Topology

General Homotopies

\begin{align*} X\times{\mathbf{R}}^n \simeq X \times{\operatorname{pt}}\cong X .\end{align*}

The ranks of \(\pi_{0}\) and \(H_{0}\) are the number of path components.

Any two continuous functions into a convex set are homotopic.

The linear homotopy. Supposing \(X\) is convex, for any two points \(x,y\in X\), the line \(tx + (1-t)y\) is contained in \(X\) for every \(t\in[0,1]\). So let \(f, g: Z \to X\) be any continuous functions into \(X\). Then define \(H: Z \times I \to X\) by \(H(z,t) = tf(z) + (1-t)g(z)\), the linear homotopy between \(f,g\). By convexity, the image is contained in \(X\) for every \(t,z\), so this is a homotopy between \(f,g\).

Fundamental Group

Definition

Given a pointed space \((X,x_{0})\), we define the fundamental group \(\pi_{1}(X)\) as follows:

  • Take the set \begin{align*} L \coloneqq\left\{{\alpha: S^1\to X \mathrel{\Big|}\alpha(0) = \alpha(1) = x_{0}}\right\} .\end{align*}

  • Define an equivalence relation \(\alpha \sim \beta\) iff \(\alpha \simeq\beta\) in \(X\), so there exists a homotopy

\begin{align*} H: &S^1 \times I \to X \\ & \begin{cases} H(s, 0) = \alpha(s)\\ H(s, 1) = \beta(s) , \end{cases} \end{align*}

  • Check that this relation is

  • Symmetric: Follows from considering \(H(s, 1-t)\).

  • Reflexive: Take \(H(s, t) = \alpha (s)\) for all \(t\).

  • Transitive: Follows from reparameterizing.

  • Define \(L/\sim\), which contains elements like \([\alpha]\) and \([\operatorname{id}_{x_{0}}]\), the equivalence classes of loops after quotienting by this relation.

  • Define a product structure: for \([\alpha], [\beta] \in L/\sim\), define \([\alpha][\beta] = [\alpha \cdot \beta]\), where we just need to define a product structure on actual loops. Do this by reparameterizing: \begin{align*} (\alpha \cdot \beta )(s) \coloneqq \begin{cases} \alpha (2s) & s \in [0, 1/2] \\ \beta (2s-1) & s \in [1/2, 1] . \end{cases} \end{align*}

  • Check that this map is:

    • Continuous: by the pasting lemma and assumed continuity of \(f, g\).

    • Well-defined: ?

  • Check that this is actually a group

    • Identity element: The constant loop \(\operatorname{id}_{x_0}: I\to X\) where \(\operatorname{id}_{x_0}(t) = x_0\) for all \(t\).

    • Inverses: The reverse loop \(\overline{\alpha}(t) \coloneqq\alpha(1-t)\).

    • Closure: Follows from the fact that start/end points match after composing loops, and reparameterizing.

    • Associativity: Follows from reparameterizing.

Elements of the fundamental group are homotopy classes of loops, and every continuous map between spaces induces a homomorphism on fundamental groups.

Conjugacy in \(\pi_{1}\):

  • See Hatcher 1.19, p.28
  • See Hatcher’s proof that \(\pi_{1}\) is a group
  • See change of basepoint map

Calculating \(\pi_1\)

If \(\tilde X \to X\) the universal cover of \(X\) and \(G\curvearrowright\tilde X\) with \(\tilde X/G = X\) then \(\pi_1(X) = G\).

\(\pi_1 X\) for \(X\) a CW-complex only depends on the 2-skeleton \(X^{2}\), and in general \(\pi_k(X)\) only depends on the \(k+2\)-skeleton. Thus attaching \(k+2\) or higher cells does not change \(\pi_k\).

Suppose \(X = U_{1} \cup U_{2}\) where \(U_1, U_2\), and \(U \coloneqq U_{1} \cap U_{2} \neq \emptyset\) are open and path-connected 1

, and let \(x_0 \in U\).

Then the inclusion maps \(i_{1}: U_{1} \hookrightarrow X\) and \(i_{2}: U_{2} \hookrightarrow X\) induce the following group homomorphisms: \begin{align*} i_{1}^*: \pi_{1}(U_{1}, x_0) \to\pi_{1}(X, x_0) \\ i_{2}^*: \pi_{1}(U_{2}, x_0) \to\pi_{1}(X, x_0) \end{align*}

There is a natural isomorphism \begin{align*} \pi_{1}(X) \cong \pi_{1} U \ast_{\pi_{1}(U \cap V)} \pi_{1} V ,\end{align*}

where the amalgamated product can be computed as follows: A pushout is the colimit of the following diagram

Example of a pushout of spaces

For groups, the pushout is realized by the amalgamated free product: if \begin{align*} \begin{cases} \pi_1 U_1 = A = \left\langle{G_{A} {~\mathrel{\Big\vert}~}R_{A}}\right\rangle \\ \pi_1 U_2 = B = \left\langle{G_{B} {~\mathrel{\Big\vert}~}R_{B}}\right\rangle \end{cases} \implies A \ast_{Z} B \coloneqq\left\langle{ G_{A}, G_{B} {~\mathrel{\Big\vert}~}R_{A}, R_{B}, T}\right\rangle \end{align*} where \(T\) is a set of relations given by \begin{align*} T = \left\{{\iota_{1}^*(z) \iota_{2}^* (z) ^{-1} {~\mathrel{\Big\vert}~}z\in \pi_1 (U_1 \cap U_2)}\right\} ,\end{align*} where \(\iota_2^*(z) ^{-1}\) denotes the inverse group element. If we have presentations

\begin{align*} \pi_{1}(U, x_0) &= \left\langle u_{1}, \cdots, u_{k} {~\mathrel{\Big\vert}~}\alpha_{1}, \cdots, \alpha_{l}\right\rangle \\ \pi_{1}(V, w) &=\left\langle v_{1}, \cdots, v_{m} {~\mathrel{\Big\vert}~}\beta_{1}, \cdots, \beta_{n}\right\rangle \\ \pi_{1}(U \cap V, x_0) &=\left\langle w_{1}, \cdots, w_{p} {~\mathrel{\Big\vert}~}\gamma_{1}, \cdots, \gamma_{q}\right\rangle \end{align*}

then \begin{align*} \pi_{1}(X, w) &= \left\langle u_{1}, \cdots, u_{k}, v_{1}, \cdots, v_{m} \middle\vert \begin{cases} \alpha_{1}, \cdots, \alpha_{l} \\ \beta_{1}, \cdots, \beta_{n} \\ I\left(w_{1}\right) J\left(w_{1}\right)^{-1}, \cdots, I\left(w_{p}\right) J\left(w_{p}\right)^{-1} \\ \end{cases} \right\rangle \\ \\ &= \frac{ \pi_{1}(U_1) \ast \pi_{1}(U_2) } { \left\langle{ \left\{{\iota_1^*(w_{i}) \iota_2^*(w_{i})^{-1}{~\mathrel{\Big\vert}~}1\leq i \leq p}\right\} }\right\rangle } \end{align*}


    
  • Construct a map going backwards
  • Show it is surjective
    • “There and back” paths
  • Show it is injective
    • Divide \(I\times I\) into a grid

\(A = {\mathbf{Z}}/4{\mathbf{Z}}= \left\langle{x {~\mathrel{\Big\vert}~}x^4}\right\rangle, B = {\mathbf{Z}}/6{\mathbf{Z}}= \left\langle{y {~\mathrel{\Big\vert}~}x^6}\right\rangle, Z = {\mathbf{Z}}/2{\mathbf{Z}}= \left\langle{z {~\mathrel{\Big\vert}~}z^2}\right\rangle\). Then we can identify \(Z\) as a subgroup of \(A, B\) using \(\iota_{A}(z) = x^2\) and \(\iota_{B}(z) = y^3\). So \begin{align*}A\ast_{Z} B = \left\langle{x, y {~\mathrel{\Big\vert}~}x^4, y^6, x^2y^{-3}}\right\rangle\end{align*} .

\begin{align*} \pi_1(X \vee Y) = \pi_1(X) \ast \pi_1(Y) .\end{align*}

By van Kampen, this is equivalent to the amalgamated product over \(\pi_1(x_0) = 1\), which is just a free product.

Facts

\(H_{1}\) is the abelianization of \(\pi_{1}\).

If \(X, Y\) are path-connected, then \begin{align*} \pi_1 (X \times Y) = \pi_1(X) \times\pi_2(Y) .\end{align*}


    
  • A loop in \(X \times Y\) is a continuous map \(\gamma : I \xrightarrow{} X \times Y\) given by \(\gamma (t) = (f(t), g(t)\) in components.
  • \(\gamma\) being continuous in the product topology is equivalent to \(f, g\) being continuous maps to \(X, Y\) respectively.
  • Similarly a homotopy \(F: I^2 \to X \times Y\) is equivalent to a pair of homotopies \(f_t, g_t\) of the corresponding loops.
  • So the map \([ \gamma ] \mapsto ([f], [g])\) is the desired bijection.

\(\pi_{1}(X) = 1\) iff \(X\) is simply connected.

\(\Rightarrow\): Suppose \(X\) is simply connected. Then every loop in \(X\) contracts to a point, so if \(\alpha\) is a loop in \(X\), \([\alpha] = [\operatorname{id}_{x_{0}}]\), the identity element of \(\pi_{1}(X)\). But then there is only one element in in this group.

\(\Leftarrow\): Suppose \(\pi_{1}(X) = 0\). Then there is just one element in the fundamental group, the identity element, so if \(\alpha\) is a loop in \(X\) then \([\alpha] = [\operatorname{id}_{x_{0}}]\). So there is a homotopy taking \(\alpha\) to the constant map, which is a contraction of \(\alpha\) to a point.

For a graph \(G\), we always have \(\pi_{1}(G) \cong {\mathbf{Z}}^n\) where \(n = |E(G - T)|\), the complement of the set of edges in any maximal tree. Equivalently, \(n = 1-\chi(G)\). Moreover, \(X \simeq\bigvee^n S^1\) in this case.

General Homotopy Theory

A map \(X \xrightarrow{f} Y\) on CW complexes that is a weak homotopy equivalence (inducing isomorphisms in homotopy) is in fact a homotopy equivalence.

Individual maps may not work: take \(S^2 \times{\mathbf{RP}}^3\) and \(S^3 \times{\mathbf{RP}}^2\) which have isomorphic homotopy but not homology.

The Hurewicz map on an \(n-1{\hbox{-}}\)connected space \(X\) is an isomorphism \(\pi_{k\leq n}X \to H_{k\leq n} X\).

I.e. for the minimal \(i\geq 2\) for which \(\pi_{iX} \neq 0\) but \(\pi_{\leq i-1}X = 0\), \(\pi_{iX} \cong H_{iX}\).

Any continuous map between CW complexes is homotopy equivalent to a cellular map.


    
  • \(\pi_{k\leq n}S^n = 0\)
  • \(\pi_{n}(X) \cong \pi_{n}(X^{(n)})\)
\todo[inline]{Theorem}

    
  • \(\pi_{i\geq 2}(X)\) is always abelian.

    • \(X\) simply connected \(\implies \pi_{k}(X) \cong H_{k}(X)\) up to and including the first nonvanishing \(H_{k}\)
  • \(\pi_{k} \bigvee X \neq \prod \pi_{k} X\) (counterexample: \(S^1 \vee S^2\))

    • Nice case: \(\pi_{1}\bigvee X = \ast \pi_{1} X\) by Van Kampen.
  • \(\pi_{i}(\widehat{X}) \cong \pi_{i}(X)\) for \(i\geq 2\) whenever \(\widehat{X} \twoheadrightarrow X\) is a universal cover.

  • \(\pi_{i}(S^n) = 0\) for \(i < n\), \(\pi_{n}(S^n) = {\mathbf{Z}}\)

    • Not necessarily true that \(\pi_{i}(S^n) = 0\) when \(i > n\)!!!
      • E.g. \(\pi_{3}(S^2) = {\mathbf{Z}}\) by Hopf fibration
  • \(S^n / S^k \simeq S^n \vee \Sigma S^{k}\)

    • \(\Sigma S^n = S^{n+1}\)
  • General mantra: homotopy plays nicely with products, homology with wedge products. 2

  • \(\pi_{k}\prod X = \prod \pi_{k} X\) by LES. 3

  • In general, homotopy groups behave nicely under homotopy pull-backs (e.g., fibrations and products), but not homotopy push-outs (e.g., cofibrations and wedges). Homology is the opposite.

  • Constructing a \(K(\pi, 1)\): since \(\pi = \left< S \mathrel{\Big|}R\right> = F(S)/R\), take \(\bigvee^{|S|} S^1 \cup_{|R|} e^2\). In English, wedge a circle for each generator and attach spheres for relations.

Footnotes
1.
Note that the hypothesis that \(U_1 \cap U_2\) is path-connected is necessary: take \(S^1\) with \(U,V\) neighborhoods of the poles, whose intersection is two disjoint components.
2.
More generally, in \(\mathbf{Top}\), we can look at \(A \leftarrow{\operatorname{pt}}\to B\) – then \(A\times B\) is the pullback and \(A \vee B\) is the pushout. In this case, homology \(h: \mathbf{Top} \to \mathbf{Grp}\) takes pushouts to pullbacks but doesn’t behave well with pullbacks. Similarly, while \(\pi\) takes pullbacks to pullbacks, it doesn’t behave nicely with pushouts.
3.
This follows because \(X\times Y \twoheadrightarrow X\) is a fiber bundle, so use LES in homotopy and the fact that \(\pi_{i\geq 2} \in \mathbf{Ab}\).