Homotopy

• Show that any non-surjective map \(f: X \rightarrow S^n\) is homotopic to the constant map.
• Let \(f,g \rightarrow S^n\) be such that \(\forall x\in X, f(x) \neq -g(x)\). Show that \(f \simeq g\).
• Let \(\alpha: S^n \to S^n,~ \alpha(p) = -p\) be the antipodal map on \(S^n\). Show that \(n ~\text{odd} \implies f \simeq \text{id}\).
• Show that \(X\) is homotopy-equivalent to a point \(\iff\) \(\text{id}_X \simeq g\) for some constant map \(g\).
• Show that \(S^1 \times I \simeq M\), the Mobius strip.
• Show that \(\mathbb{R}^3 - S^1 \simeq S^1 \vee S^2\).
• Classify the letters of the alphabet up to homeomorphism, and up to homotopy.
• REVISIT Let \(f,g : S^1 \rightarrow X\), \(P = X \cup_f B^2 \cong X \coprod B^2 / \sim\), where \(x \sim f(x)\), \(Q = X \cup_g B^2\). Show that \(f\simeq g \implies P\simeq Q\).

Fundamental Group

• Show that \(x,y\in X\) path & simply-connected \(\implies\) all paths from \(x\) to \(y\) are homotopic rel \(\{0, 1\}\).
• Show that for \(X\) path connected, \(\pi_1(X) = \mathbb{1} \iff \forall \text{cts.}~f: S^1 \rightarrow X\) \(f\), extends to a continuous map \(F: B^2 \rightarrow X\).
• Show \(\pi_1(X\times Y, (x_0, y_0)) \cong \pi_1(X,x_0) \times \pi_1(Y, y_0)\).
• Show \(\pi_1(S^n) = 1\) for \(n\geq 2\).
• Show that \(S^2 - \{p_0, p_1\} \simeq S^1\).
• Show that \(S^3 - \{p_0, p_1\} \simeq S^2\)
• Show that \(S^2 \not\cong S^3\).
• For each of the following \(f: S^1 \rightarrow S^1\), identify the corresponding \(f_*: \mathbb{Z} \to \mathbb{Z}\):
  • \(z\mapsto z^n\)
  • \(\overline{x} \mapsto -\overline{x}\)
  • \(e^{i\theta} \mapsto e^{2\pi i\sin\theta}\)
• Determine the winding number of the following map: \(f: S^1 \to \mathbb{C}-\{0\}, z\mapsto 8z^4 + 4z^3 + 2z^2 + z^{-1}\)
• Identify \(\pi_1(M, [(1, \frac{1}{2})])\), and identify the class of \(\partial M\).
• Let \(X = S^1\times S^1\) and \(\gamma\) a loop based at \(x_0\). What is the induced map \(\gamma_\sharp\)?

Group Actions

• Show that octagon pasting is homeomorphic to the \(T = \mathbb{R}^2 / \mathbb{Z}^2\).
• Let \(x_0\) be the image of \(0\), show that there is an order 6 homeomorphism \(f: T \to T\) fixing \(x_0\). Find a representation of \(f_*\) as a matrix, and find its determinant.
• Show that \(\pi_1(K)\), the Klein bottle, is given by pairs \((m,n)\) where \((m,n)\star (p,q) = (m+(-1)^np, n+q)\)
  • Show this is torsion-free
  • Show that \(T\) is a double cover of \(K\).
• For each of these actions of \(\mathbb{Z}_2\) on \(S^n\), compute \(\pi_1(S^n/\mathbb{Z}_2)\)
  • \(S^1, z\mapsto -z\)
  • \(S^2, (x,y,z) \mapsto (-x,-y,z)\)
  • \(S^3, (z,w) \mapsto (-z, -w)\)

Applications

• Let \(i: \mathbb{RP}^2 \to \mathbb{RP}^3\), induced by \(S^2 \hookrightarrow S^3\) as the equator. Show that \(i \not\simeq \text{const}\).
• Show that there is no map \(f: S^2 \to S^1\) that commutes with the antipodal map.
• Prove that for any \(f: S^2 \to \mathbb{R}^2\), there exists \(x\in S^2\) such that \(f(x) = f(-x)\).
• Prove the Ham Sandwich theorem.
• Show that \(K\) can not be a topological group.

Van Kampen’s Theorem

• Compute a presentation of \(\pi_1(T)\) and prove it is isomorphic to \(\mathbb{Z}_2\).
• (Images)
• Show that \(T-D^1 := X \simeq S^1 \vee S^1\).
  • Show there does not exist a retraction \(r: X \to \partial X\).
• Images
• IMages
• Images
• Calculate a presentation of \(\pi_1(S^3-K)\)
• Show that all 3 presentations of \(\pi_1(K)\) are isomorphic
  • Square with sides glued
  • Two mobius strips glues along boundary
  • Multiplication rule
• Given a group \(G = <A : R>\), show how to construct a CW-complex \(X\) such that \(\pi_1(X) = G\).
• Write down the fundamental group of the following spaces:
• \(\mathbb{R}^2 - \{0, 1\}\)
• \(\mathbb{R}^2 - I\)
• The symbol \(\oplus \in \mathbb{R}^2\)
• \(S^2 - \{p_i\}_{i=1}^4\)
• \(T - \{p_0\}\)
• \(S^2 / \mathbb{Z}_2\) via the antipodal map
• \(S^2/\mathbb{Z}_3\) via a \(2\pi/3\) rotation about the \(z\)-axis.
• \(S_2 \cup \{(0,0,z) \mathrel{\Big|}-1 \leq z \leq 1 \}\)
• \(\mathbb{R}^3 - \{ (x,y,0) \mathrel{\Big|}x^2 + y^2 = 1\}\)
• \(\mathbb{R}^2 - H\), the Hopf link
• Prove that the homophony group is trivial.

Mayer Vietoris (Sheet 7)

• Compute the homology of:

  • \({\mathbf{RP}}^2 = M \cup_{\partial}D^2\)
  • \(T^2 = S^1 \times S^1 = (S^1\times I)\cup_f (S^1\times I)\) where \((x,0) \sim (x,1) \sim (\overline{x}, 0) \in {\mathbf{C}}\)
  • \(S^1 \cup_{f} B^2\) attached along \({\partial}B^2\) using \(z\mapsto z^n\)

• Show \(\tilde H_i(\Sigma X) \cong \tilde H_{i-1}(X)\)

  • Show \(\Sigma S^n \cong S^{n+1}\)

• For \(f: S^n\circlearrowleft\), show \(\deg f = \deg \Sigma f\)

  • Conclude \(\pi_n(S^n) = {\mathbf{Z}}\)

• Let \(\left\{{A_i}\right\}^n \in \mathbf{Ab}\) be finitely generated, show \(\exists X \mathrel{\Big|}H_i(X) \cong A_i\) for \(i\leq n\) and 0 otherwise.

• Suppose \(X = \cup_i^n A_i\) such that for any \(1\leq k \leq n,~ \cap_i^k A_i\) is either empty or contractible, show \(i\geq n-1 \implies \tilde H_i(X) = 0\) and that this bound is sharp.

• Compute \(H_*(X\times S^n)\) in terms of \(H_*(X)\)

  • Compute \(H_*(T^n)\)

• Let \(M = (S^1 \times B^2) \cup_{\operatorname{id}_{\partial}} (S^1 \times B^2)\) and compute \(H_*(M; {\mathbf{Z}})\)

• Let \(X = S^n\times I\) with its ends glued together by a map \(S^n \circlearrowleft\) of degree \(d\), calculate \(H_*(X)\).

• Compute \(H_*(X)\) for \(X = S^3 - N\), with \(N\) a knotted solid torus and \({\partial}N = T\) its boundary torus

• Let \(CA\) be the cone on \(A\), show that \(\tilde H_*(X \cup CA) \cong \tilde H_*(X, A)\).

• Show that the Mayer-Vietoris sequence is natural, i.e. If \(X\xrightarrow{f} Y\) where \(f(A) \subset C\) and \(f(B) \subset D\), then this commutes: \begin{align*} \begin{CD} H_n(X) @>>> H_n(A\cap B) @>>> H_n(A) \oplus H_n(B) @>>> H_{n-1}(X)\\ @VVf_*V @VVf_*V @VVf_*V @VVf_*V\\ H_n(Y) @>>> H_n(C\cap D) @>>> H_n(C) \oplus H_n(D) @>>> H_{n-1}(Y)\\ \end{CD} \end{align*}

Cellular Homology (Sheet 8)

Compute the homology of these spaces

• \(S_m \vee S_n\)
• \(S^m \times S^n\)
• A hexagon with the identifications \(a+b+c-a-b-c\)
• Orientable surface of genus \(g\)
  • \(g=2\) is given by \(a+b-a-b+c+d-c-d\)
• Nonorientable surface of genus \(g\) Obtain by removing \(g\) discs from \(S^2\) and attaching \(g\) mobius strips
• \(S_1 \vee S_1\) with two discs attached via \((ab)^3\) and \((ab)^6\)
• This identification space: assets/2018-02-11 16_28_51-290.pdf
• This identification space: assets/1518395440173
• This identification space: assets/1518395458668
• Describe a CW complex structure for the lens space \(L(p, 1)\) and compute \(\pi_1, H_*\) for it.

Degree

• Let \(p(x) = \sum_i^na_ix^i\), view \(p: {\mathbf{C}}\cup\infty {\circlearrowleft}\) and determine its topological degree
• Let \(p(z) = \frac{\prod_i^n z-a_i}{\prod_j^m z-b_j}\) with all \(a_i, b_j\) distinct. What is its topological degree?
• Show that if \(f: S^m \to S^n\) and \(\exists U \subset S^m\) such that \({\left.{{f}} \right|_{{U}} } \cong f(U)\), then \(m=n\) and \(f\) is surjective.

Universal Coefficient Theorem (Sheet 10)

• Identify the following groups up to isomorphism
  • \({\mathbf{Z}}_m \otimes{\mathbf{Z}}_n\)
  • \({\mathbf{Z}}_{60}^4 \otimes({\mathbf{Z}}_{24}^3 \oplus {\mathbf{Z}}_8^4 \oplus {\mathbf{Z}}_{120})\)
  • \({\mathbf{Z}}_n \otimes{\mathbf{Q}}\)
  • \(({\mathbf{Z}}\oplus {\mathbf{Z}}_n) \otimes({\mathbf{Q}}/{\mathbf{Z}})\)
• Compute:
  • \(\operatorname{Tor}({\mathbf{Z}}\oplus {\mathbf{Z}}_2 \oplus {\mathbf{Z}}_8, {\mathbf{Z}}\oplus {\mathbf{Z}}_4 \oplus {\mathbf{Z}}_4)\)
  • \(\operatorname{Ext}({\mathbf{Z}}\oplus {\mathbf{Z}}_2 \oplus {\mathbf{Z}}_3, {\mathbf{Z}}\oplus {\mathbf{Z}}_4 \oplus {\mathbf{Z}}_5)\)
• Compute the following directly from chain complexes and check using UCT:
  • \(H_*({\mathbf{RP}}^n; {\mathbf{Z}}_2)\)
  • \(H_*({\mathbf{RP}}^n, {\mathbf{Z}}_3)\)
  • \(H^*({\mathbf{RP}}^n, {\mathbf{Z}}_6)\)
• For any space \(X\), show that \(H^1(X)\) is free abelian
• Show that \(H_*(X; {\mathbf{Q}}) = H_*(X;{\mathbf{Z}})\otimes{\mathbf{Q}}\) \(H^*(X; {\mathbf{Z}}) = \hom(H_*(X; {\mathbf{Z}}), {\mathbf{Q}})\)
• Construct a space \(X\) such that \(H_*(X; {\mathbf{Z}}) = ({\mathbf{Z}}, {\mathbf{Z}}_6, {\mathbf{Z}}_{12}, {\mathbf{Z}}\oplus {\mathbf{Z}}_4, 0 \cdots)\) Compute \(H^*(X; {\mathbf{Z}})\)
• Compute \(H_*({\mathbf{RP}}^2 \times{\mathbf{RP}}^2; {\mathbf{Z}}_2)\)
• Compute \(H_*(\Sigma{\mathbf{RP}}^2 \times{\mathbf{RP}}^2; {\mathbf{Z}})\)
• Compute \(H_*({\mathbf{RP}}^2\times{\mathbf{RP}}^3; {\mathbf{Z}})\)
• Let \(G\) be a topological group. Show that \(H_*(G)\) is an algebra. Show that \(G\curvearrowright H_*(G)\), which factors through the homomorphism \(G \to\pi_0(G)\) yielding a trivial action if \(G\) is path-connected.

Homological Algebra (Sheet 11)

• Show that \(\ker A \to A\otimes{\mathbf{Q}}\) given by \(a \mapsto a\otimes 1\) is the torsion subgroup of \(A\).
• Show that \(A\hookrightarrow B \implies A\otimes{\mathbf{Q}}\hookrightarrow B\otimes{\mathbf{Q}}\)
• Find a free resolution of \({\mathbf{Q}}\) as a \({\mathbf{Z}}\)-module.
• Compute \(\operatorname{Tor}({\mathbf{Q}}, A)\)
  • Compute \(\operatorname{Tor}({\mathbf{Q}}/{\mathbf{Z}}, A)\)
• Let \(R = {\mathbf{Z}}[x,y]\), and \(M = R/(x-y), N = R/(x,y)\). Construct free resolutions of \(M,N\) to compute:
  • \(\operatorname{Ext}_R^*(M, M)\)
  • \(\operatorname{Ext}_R^*(M, N)\)
  • \(\operatorname{Ext}_R^*(N, M)\)
  • \(\operatorname{Ext}_R^*(N, N)\)
• Let \(\Lambda_*\) be the exterior algebra generated by the symbols \(\left\{{dx_i}\right\}^n\) over a field \(k\). Show that letting \(d = \cdot \vee dx_1\) yields a chain complex \(0 \to\Lambda^0 \to\Lambda^1 \to\cdots \to\Lambda^n \to 0\) with trivial homology. Compute what happens when \(dx_1\) is replaced with an arbitrary non-zero element in \(\Lambda^1\).
• Define \(M\) as the group ring \(R = {\mathbf{Z}}[{\mathbf{Z}}_2]\) with the action \((\cdot) \times -1\). Construct a free resolution of \(M\) and compute \(\operatorname{Tor}_R^*(M, M)\).
• Show \(\operatorname{Tor}_R^*(\cdot, \cdot)\) is symmetric in the following way: Given \(M, N\), take free resolutions, view \(M_* \to M\) as a chain map and tensor with \(N_*\) to get a chain map\(\psi: M_* \otimes_R N_* \to M \otimes_R N_*\). Show that \(\psi\) is a quasi-isomorphism using the exact sequence \(0 \to(Z_n, 0) \to(N_n, 0) \to(B_{n-1}, 0) \to 0\), then switch the roles of \(M, N\).
• Prove that for a SES \(0\to A\to B\to C\), the group \(\operatorname{Ext}(C,A)\) classifies extensions of \(C\) by \(A\) up to isomorphism.

Cohomology Ring (Sheet 12)

Todo

Topology Problems: Solutions

Homotopy

• Main Idea: A linear homotopy projected onto the sphere works.

Let \(f: X \to S^n \subset {\mathbf{R}}^{n+1}\) be an arbitrary map that fails to be surjective. Then, by definition, there is at least one point \(s_0 \in S^n - f(X)\).

Then, \(\forall x\in X\), since \(f(x) \neq s_0\), there is a unique geodesic \(C\) connecting \(f(x)\) and \(s_0\). So a variant of the straight line homotopy will work, by interpolating between \(f(x)\) and \(s_0\) along \(C\).

So let \(H:X \times I \to S^n\) be defined by \(H(x, t) = P(ts_0 + (1-t)f(x))\), where \(P: \mathbb{R}^{n+1} \to S^n\) is given by \(P(x) = x/{\left\lVert {x} \right\rVert}\). This is well defined, since the denominator is zero iff \(f(x) = s_0\), which by assumption is not the case. This is a homotopy, since \(H(x, 0) =P(f(x)) = f(x)\) (since \(P\) fixes \(S^n\)) and \(H(x, 1) = P(s_0) = s_0\) (since \(s_0 \in S^n\)).

• Main Idea: Exact same idea as 1, just a more complicated check.

Take \(H(x, t) = P(tf(x) + (1-t)g(x))\). This is well defined; the only case to check is when the denominator is zero. But \({\left\lVert {x} \right\rVert} = 0\) iff \(x =0\), which would imply \(tf(x) +(1-t)g(x) = 0\) and so \(tf(x) = -(1-t)g(x)\).

Taking norms and observing that since \(f,g \in S^n \implies {\left\lVert {f} \right\rVert} = {\left\lVert {g} \right\rVert} = 1\), this forces \(t = 1-t\) and thus \(t=1/2\). But this would force \((1/2)f(x) = (-1/2)g(x)\) and thus \(f(x) = -g(x)\), which we assumed was not the case.

• Main Idea: Linear homotopy fails continuity without the condition from (2), so use complex embedding to avoid the origin at \(t=1/2\).

Suppose \(n\) is odd and define \(f:S^n \to S^n\) to be the antipodal map. Since \(n+1\) is even, we have \(n+1 =2m\) for some \(m\in {\mathbb{N}}\), so identify \(S^n = S^{2m-1} \subset {\mathbf{R}}^{2m} \cong{\mathbf{C}}^m\)

Then \(z\in S^n\) can be written as a vector \(z \in {\mathbf{C}}^m\) such that \({\left\lVert {z} \right\rVert} = 1\).

Then define \(P: {\mathbf{C}}^m \to {\mathbf{C}}^m\) by \(P(z) = z/{\left\lvert {z} \right\rvert}\), the projection onto the complex unit sphere, and define \(H: {\mathbf{C}}^m \times I \to {\mathbf{C}}^m\) by \(H(z, t) = P(e^{i\pi t}z)\).

This is a homotopy, since \(H(z, 0) = P(z) = z\) (since \({\left\lVert {z} \right\rVert} = 1\)), so this is the identity map. We also have \(H(z, 1) = P(-z) = -z\), the antipodal map.

This is well-defined, since \(e^{i\pi t} > 0\) and \(z \neq 0\), so the linear homotopy in ambient \({\mathbf{C}}^m\) avoids the origin and thus the denominator when taking the projection is never zero.

• \(\Leftarrow\): Main Idea: Projection and inclusion are homotopy inverses. One composition is equality, the other is just equality up to homotopy, but that’s all we need!

Suppose \(\operatorname{id}_X\) is nullhomotopic.

Then there exists some constant map \(g: X \to\left\{{x_0}\right\}\) for some \(x_0 \in X\) where \(g(x) = x_0\) and \(g \simeq\operatorname{id}_X\).

This means there is some homotopy \(F: X \times I \to X\) such that \(F(x,0) = \operatorname{id}_X(x) = x\) and \(F(x,1) = g(x) = x_0\) for all \(x \in X\).

So let \(p:X \to\left\{{x_0}\right\}\) be the projection map sending every point to \(x_0\), and \(\iota: \left\{{x_0}\right\} \to X\) be the inclusion. We will show that the two compositions are homotopy inverses, from which it follows that \(X \simeq\left\{{x_0}\right\}\). This means that \(X\) is homotopy-equivalent to a point, and thus by definition contractible.

Then \((p\circ \iota): \left\{{x_0}\right\} \to \left\{{x_0}\right\}\) is given by \(p(\iota(x_0)) = p(x_0) = x_0\), so this is the identity on the target space \(\left\{{x_0}\right\}\).

Similarly, \((\iota \circ p): X \to X\) is given by \(\iota(p(x)) = \iota(x_0) = x_0\), so this is the constant map on \(X\) mapping every point from \(X\) to \(x_0\). But then this map is exactly \(g\), and by assumption this is homotopic to the identity on \(X\)

But then we have \(p\circ \iota \simeq\operatorname{id}_{\left\{{x_0}\right\}}\) and \(\iota \circ p \simeq\operatorname{id}_X\), so they are homotopy inverses.

\(\Rightarrow\): Main Idea: One of the homotopy inverses is just a constant map.

Suppose \(X \simeq\left\{{x_0}\right\}\), then there exist a pair of homotopy inverses

\(f: X \to\left\{{x_0}\right\}\) and \(g: \left\{{x_0}\right\} \to X\) such that \(f\circ g \simeq\operatorname{id}_{\left\{{x_0}\right\}}\) and \(g\circ f \simeq\operatorname{id}_X\).

Since \(\left\{{x_0}\right\}\) is a single point space, \(f\) is necessarily a constant map (i.e. \(f(x) = x_0\) for every \(x\in X\).) But then \((g\circ f)(x) = g(x_0) = y_0\) for some constant \(y_0 \in X\), so \(g\circ f\) is a constant map. By assumption, \(g\circ f \simeq\operatorname{id}_X\), so the identity is homotopic to a constant map.

• Main Idea: Deformation retract \(M\) onto its center circle; two spaces that deformation retract onto a common space are themselves homotopy equivalent.

Claim: \(S^1 \times I \simeq S^1 \times\left\{{*}\right\}\) This is because \(I\) is contractible, so \(I \simeq\left\{{*}\right\}\). (Maybe needs further proof)

Claim: \(M \simeq S^1 \times\left\{{*}\right\}\).

If both of these claims hold, then we will have \(M \simeq S^1 \times I\) as two spaces that deformation retract onto a common space. Identifying \(M = I \times I / \sim\) where \((x, 0) \sim (1-x, 1)\), fix \(x=1/2\).

Then consider the subspace \(U = \left\{{(1/2, y) \mathrel{\Big|}y \in [0,1]}\right\} \subset M\). Claim: \(U \cong \left\{{*}\right\} \times S^1\) for some point \(*\).

\(U\) can be written \(\left\{{1/2}\right\} \times(I/\sim)\), and since \((1/2, 0) \sim (1/2,1)\), we have \(I/ \sim = I /{{\partial}}I \cong S^1\), so \(U \cong \left\{{1/2}\right\}\times S^1\) as desired (taking \(* = \frac{1}{2}\)).

However, we can define a homotopy from \(M\) onto \(U\), in the form of a deformation retract.

Let \(F: M \times I \to M\) be defined by \(F((x,y), t) = F_t(x,y) = ((1-t)x + \frac{1}{2}t, y)\). Then \(F((x,y), 0) = (x,y) = \operatorname{id}_M\), and \(F((x,y), 1) = (\frac{1}{2}, y) \subseteq U\). Moreover, if \((x,y) \in U\), then \((x,y) = (\frac{1}{2}, y)\) and \(F((x,y), t) = ((1-t)\frac{1}{2} + \frac{1}{2}t, y) = (\frac{1}{2} - t\frac{1}{2} + \frac{1}{2}t, y) = (\frac{1}{2}, y) = (x,y)\), so \(F = \operatorname{id}_U\). This makes \(F\) a deformation retract from \(M\) onto \(U\), and so \(M \simeq U\).

But then, summarizing our results, we have \(S^1 \times I \simeq S^1 \times\left\{{*}\right\} \cong S^1 \times\left\{{\frac{1}{2}}\right\} = U \simeq M\), and so \(S^1 \times I \simeq M\) as desired.

• Main Idea: Using a funky deformation retract. See Hatcher, PDF page 55, Example 1.23. Add picture!!

Deformation retract \(\\R^3 - S^1\) onto \(S^2 - U\), where \(U\) is a diameter inside \(S^2\) also passing through the middle of \(S^1\) in the interior. This can be done by moving points outside of \(S^2\) towards the surface, and points inside \(S^2\) just move away from the \(S^1\) inside (either towards \(U\) or towards the surface of \(S^2\), so they don’t hit \(S^1\)).

Then take a geodesic between the endpoints of the diameter on \(S^2\), pick any point \(p\) on the geodesic, and move both diameter points towards it. This yields \(S^2 \vee S^1\) at the point \(p\).

• Main Idea: Nothing to it. Homotopy:

• \(A \simeq\Delta \simeq S^1\)

• \(a \simeq d \simeq o \simeq S^1\)

• \(B \simeq 8 \simeq S^1 \vee S^1\)

• \(b \simeq o \simeq S^1\)

• \(C \simeq*\)

• \(c \simeq l \simeq*\)

• \(D \simeq S^1\)

• \(d \simeq o \simeq S^1\)

• \(E \simeq*\)

• \(e \simeq d \simeq S^1\)

• \(F \simeq*\)

• \(f \simeq*\)

• \(G \simeq*\)

• \(g \simeq 8 \simeq S^1 \vee S^1\)

• \(H \simeq*\)

• \(h \simeq l \simeq*\)

• \(I \simeq*\)

• \(i \simeq\left\{{*_1, *_2}\right\}\)

• \(J \simeq*\)

• \(j \simeq i \simeq\left\{{*_1, *_2}\right\}\)

• \(K \simeq*\)

  • \(k \simeq K \simeq*\)

• \(L \simeq*\)

  • \(l \simeq*\)

• \(M \simeq*\)

  • \(m \simeq*\)

• \(N \simeq*\)

  • \(n \simeq*\)

• \(O \simeq S^1\)

  • \(o \simeq S^1\)

• \(P \simeq D \simeq S^1\)

  • \(p \simeq P \simeq S^1\)

• \(Q \simeq O \simeq S^1\)

  • \(q \simeq p \simeq o \simeq S^1\)

• \(R \simeq D \simeq S^1\).

  • \(r \simeq l \simeq S^1\)

• \(S \simeq*\)

  • \(s \simeq S \simeq*\)

• \(T \simeq*\)

  • \(t \simeq l \simeq*\)

• \(U \simeq*\)

  • \(u \simeq U \simeq*\)

• \(V \simeq*\)

  • \(v \simeq V \simeq*\)

• \(W \simeq*\)

  • \(w \simeq W \simeq*\)

• \(X \simeq*\)

  • \(x \simeq X \simeq*\)

• \(Y \simeq*\)

  • \(y \simeq v \simeq*\)

• \(Z \simeq*\)

  • \(z \simeq Z \simeq*\)

This results in a partition of the alphabet into the following homotopy types:

• \(\left\{{A, D, O, P, Q, R, S^1}\right\}\) \(\cup\left\{{a,b,d,e,g,o,p,q}\right\}\)
• \(\left\{{C, E, F, G, H, I, J, K, L, M, N, S, T, U, V, W, X ,Y, Z, *}\right\}\) \(\cup\left\{{c,f,h,k,l,m,n,r,s,t,u,v,w,x,y,z}\right\}\)
• \(\left\{{B, S^1 \vee S^1}\right\}\)
• \(\left\{{i, j, \left\{{*, *}\right\}}\right\}\)

Homeomorphisms: ignore ligatures!!

• \(\left\{{A, R}\right\}\) Can remove a point to obtain two components homeomorphic to \(\left\{{I, F}\right\}\) respectively.

• \(\left\{{D, O, S^1}\right\}\) These all have no single point that can be removed to disconnect the space.

• \(\left\{{B,S^1 \vee S^1}\right\}\) Remove point at junction

• \(\left\{{C, G, I, J, L, M, N, S, U, V, W, Z, [0,1]}\right\}\) These all have a point that can be removed to yield two components, but no points that yield three. (Intuitively, all can be obtained by twisting a straight wire.)

• \(\left\{{E, F, T, Y, \bigvee_{i=1}^3 [0,1]}\right\}\) These all have a point that can be removed to yield 3 connected components homeomorphic to \(I\). This is the “pasting” point in the vee.

• \(\left\{{H, K, \bigvee_{i=1}^5 [0,1]}\right\}\) Can remove two points to disconnect each into five components.

• \(\left\{{P, Q, S^1 \vee [0,1]}\right\}\) Both contain a nontrivial loop.

• \(\left\{{X, \bigvee_{i=1}^4 [0,1]}\right\}\) Can remove one point to separate into four components.

• Main Idea: Show that both spaces are a deformation retract of the same space. (See Hatcher, Proposition 0.18, p. 25)

Suppose we have the following maps

\begin{align*} f: S^1 \to X\\ g: S^1 \to X \end{align*}

where \(f \simeq g\). Then there exists a homotopy

\begin{align*}H: S^1 \times I \to X\end{align*}

such that \(H(z, 0) = f(z)\) and \(H(z,1) = g(z)\).

Then define \begin{align*} P \coloneqq X \coprod_f B^2\\ Q \coloneqq X \coprod_g B^2 \end{align*}

We want to that \(P\) and \(Q\) are homotopy-equivalent. In order to do so, we will construct a larger space which deformation retracts onto both \(P\) and \(Q\), which is a homotopy equivalence.

With \(H\) in hand, we can define the space \(R = X \coprod_H B^2 \times I\), where we recognize \(S^1 = {{\partial}}B^2\). In particular, \(S^1\) is a subspace of \(B^2\).

Claim: Both \(P\) and \(Q\) are subspaces of \(R\). Since \(H(z, 0) = f(z)\). So considering \(X \coprod_H B^2 \times \left\{{0}\right\} \cong X \coprod_f B^2 = P\). A similar argument holds at the point \(1\in I\). (Not a strong argument)

But note that \(B^2 \times I\) is a solid cylinder, and so can be deformation retracted onto the outer shell plus one of the “lids”. Formally, this would be given by \(S^1 \times I \cup B^2 \times\left\{{p}\right\}\) for some \(p\in [0,1]\).

Claim: choosing \(p=0\) induces a deformation retract of \(R\) onto \(P\), and choosing \(p=1\) induces a deformation retract of \(R\) onto \(Q\).

Proof: ?

Fundamental Group

• Main idea: just algebraic manipulations using the \(\pi_1\) functor and unravelling definitions.

Let \(X\) be path connected and simply connected, and let \(x,y \in X\) be two arbitrary points. Then consider two paths, \(\gamma: I \to X, \gamma(0) = x, \gamma(1) = y\) \(\alpha: I \to X, \alpha(0) = x, \alpha(1) = y\).

We would like to show \(\gamma \simeq\alpha\). Since \(X\) is simply connected, we know that \(\pi_1(X) = 0\). This means that for any \(a,b \in \pi_1(X), a = b = e\), the identity element in this group.

So we construct two loops: one as \(\gamma \overline{\alpha}\), the other as \(\alpha\overline{\gamma}\). Apply the \(\pi_1\) functor yields \([\gamma\overline{\alpha}] = e = [c_x] = [\alpha\overline{\gamma}]\), where \([c_x]\) is the equivalence class of the constant path at \(x\), and equivalently the identity element in \(\pi_1(X)\). Lemma: If \(f\simeq g\), then \(f\circ h \simeq g \circ h\) for any \(h\).

But this says \(\gamma\overline{\alpha }\simeq c_x\) and \(\alpha\overline{\gamma }\simeq c_x\). But \(\gamma \simeq c_x \circ \gamma \simeq(\alpha\overline{\gamma}) \circ \gamma \simeq\alpha\circ (\overline{\gamma }\circ\gamma) \simeq\alpha\), which is what we desired.

• Main Idea Homotopies on maps \(S^1\to X\) are cylinders, find a way to continuously map a cylinder onto a disk given the existence of such a homotopy. Let \(X\) be path connected, \(\pi_1(X) = 0\), and let \(f:S^1 \to X\) be arbitrary. Then \(f(S^1) \subseteq X\) is a path in \(X\), and since \(\pi_1(X) = 0\), this path is homotopic to a point \(x_0\). So \(f\) is homotopic to the constant map \(c_{x_0}: S^1 \to X, z \mapsto x_0\).

So let \(H:S^1 \times I \to X\) be this homotopy. We know that \(H(z, 0) = f(z)\) and \(H(z, 1) = c_{x_0}(z) = x_0\).

Claim: Consider quotient \(\frac{S^1\times I}{S^1 \times\left\{{1}\right\}}\) with the projection map \(p: S^1 \times I \to S^1 \times\left\{{1}\right\}\). Then \(H\) factors through the quotient uniquely (why?), and there exists a unique \(\widehat{H}\) making this diagram commute:

universal1

This follow from the universal property of the quotient in \(\mathbf{Top}\), where it is sufficient that \(H\) is constant on \(S^1 \times\left\{{1}\right\}\) - but this is exactly what was deduced above.

However, the quotient object constructed is homeomorphic to \(D^2\), as per the following diagram

2017-11-24 14_59_29-Untitled page - OneNote

Here, we just recognize that \(S^1 \times I\) is a cylinder, and quotienting at the \(t=1\) point in \(I\) simply collapses the top portion of the cylinder to a point, forming a cone. We then take the flattening map to just project every point on the cone directly downwards onto the base circle, yielding \(D^2\).

(Note: I guess this map can be constructed as \(\Phi: S^1 \times I \to D^2\) where \(\Phi(z, t) = z(1-t)\). Since \(t=1\) on \(S^1 \times\left\{{1}\right\}\), \(\Phi(z, 1) = 0\) and this is exactly the kernel of \(\Phi\). Continuous as product of continuous functions, need to check injective/surjective and show inverse is continuous.)

Need to check injective/surjective, show that kernel is \(S^1 \times 1\), then use first isomorphism theorem.)

But then \(\widehat{H}\) is exactly a continuous map from \(D^2 \to X\), as desired.

• \(\Rightarrow\) Let \([\alpha] \in \pi_1(X\times Y, (x_0, y_0))\) be an arbitrary loop in \(X \times Y\). Then \(\alpha\) is equivalently a map \(S^1 \to X \times Y\). Considering \(S^1\) to be a subset of \({\mathbf{R}}^2\), we can parameterize \(\alpha\) as \(\alpha(z) = \alpha(x+iy) = (\alpha_x(x), \alpha_y(y))\) in components. In particular, since \(\alpha\) is continuous, so are \(\alpha_x, \alpha_y\). Moreover, since \(\alpha(0) = \alpha(0 + i0) = (x_0, y_0)\), we have \(\alpha_x(0) = x_0, \alpha_y(0) = y_0\). (Note: alternatively, given the product, we have projections \(p_X, p_y\), so we can define the map \(\alpha \mapsto (p_X \circ \alpha, p_Y \circ \alpha)\))

But then \(\alpha_x: S^1 \to X\) and \(\alpha_y: S^1 \to Y\) are loops entirely in \(X, Y\) at the respective base points, and so we can define the map \(F: \pi_1(X\times Y, (x_0, y_0)) \to\pi_1(X, x_0) \times\pi_1(Y, y_0)\) by \([\alpha] = [(\alpha_x, \alpha_y)] \mapsto ([\alpha_x], [\alpha_y])\)

This is injective, since \(([a],[b]) = ([c],[d])\) on the RHS means that \([a] = [c], [b] = [d]\) in the fundamental groups, and thus \(a\simeq c, b\simeq d\) in the spaces. We want to show that \([(a,b)] = [(c,d)]\), which would follow if \(\alpha(x+iy) = (a(x),b(y)) \simeq\beta(x+iy) = (c(x),d(y))\) in \(X\ \times Y\). …?

This is surjective, because if \(([a], [b])\) are elements in the right-hand side, then \(a(0) = a(1) = x_0\) and \(b(0) = b(1) = y_0\), so we can consider \((a,b): I \to X \times Y\) where \((a,b)(z) = (a,b)(x+iy) = (a(x), b(y))\). This is then a loop in \(X\times Y\), since \((a,b)(0) = (a(0), b(0)) = (0,0) = (x_0, y_0)\) and similarly \((a,b)(1) = (a(1), b(1)) = (x_0, y_0)\). So this is actually a map \((a,b): S^1 \to X \times Y\), or in other words, a loop in \(X\times Y\) based at \((x_0, y_0)\), which lifts to an element of the fundamental group on the LHS.

Maps in both directions are continuous, since a vector function is continuous iff its component functions are continuous.

This is well-defined, due to the fact that if \(a \simeq b\), then \(p_X \circ a \simeq p_X\circ b\), and \(F = (f_x, f_y)\) is a homotopy iff its components functions are homotopies.

• Let \(A = S^n - \left\{{n_p = \text{North Pole}}\right\}, B = S^n - \left\{{s_p = \text{South Pole}}\right\}\). Then \(A\cup B = S^n\) and \(A\cap B = S^n - \left\{{n_p, s_p}\right\}\). Since \(A,B\) are open and path connected, we can apply van Kampen’s theorem to obtain \(\pi_1(X) = \pi_1(A) * \pi_1(B)\) amalgamated over \(\pi_1(A\cap B)\). But \(A \cong {\mathbf{R}}^{n} \cong B\) via stereographic projection, and since \({\mathbf{R}}^n\) is contractible, \(\pi_1({\mathbf{R}}^n) = 0 = \pi_1(A) = \pi_1(B)\). So \(\pi_1(X) = 0 * 0 = 0\) as desired.

This follow because we can compute \(A \cap B \cong {\mathbf{R}}^n - \left\{{\text{pt}}\right\} \cong S^n{-1}\), and so \(\pi_1(A\cap B) = \pi_1(S^n) \times\pi_1({\mathbf{R}}^1) = 0 \times 0 = 0\), and so has the presentation \(\pi_1(A\cap B) = \left< w \mathrel{\Big|}w^1 = e\right>\). We can then look at the inclusions \(i: A\cap B \to A\) \(j: A\cap B \to B\) and the induced homomorphisms \(I: \pi_1(A\cap B) \to\pi_1(A)\) \(J: \pi_1(A\cap B) \to\pi_1(B)\). But since both sides in both maps are trivial, these are constant maps between identities. We can then present the group \(0 = \pi_1(A) =\left< a\mathrel{\Big|}a^1 = e\right>\) and since \(I(w) J(w)^{-1} = e e^{-1} = e\), we have \(\pi_1(B) = \left< b \mathrel{\Big|}b^1 = e\right>\), so \(\pi_1(A) *_{\pi_1(A\cap B)} \pi_1(B) = \left< a,b \mathrel{\Big|}a^1 =b^1 = e\right>\).

(See https://en.wikipedia.org/wiki/Seifert%E2%80%93van_Kampen_theorem for presentation of amalgamated product)

• WLOG, assume \(p_0, p_1\) are the north and south poles of \(S^2\). We can then form a deformation retract of \(X\) onto the equator of \(S^2\), which is equal to \(S^1\). To do so, just move every point \(x\) along the unique great circle connecting \(x, p_0, p_1\), and proceed at linear speed towards the equator. This is well defined at every point on \(S^2\) except the poles, which are not included in \(X\), and the equator is fixed at every instant. So this forms a deformation retract. Alternatively, use the fact that \({\mathbf{R}}^n -{\operatorname{pt}}\cong S^{n-1} \times{\mathbf{R}}\) via polar coordinates, and \(S^n - {\operatorname{pt}}\cong {\mathbf{R}}^n\) by stereographic projection. So \(S^2 - \left\{{p_0, p_1}\right\} \cong {\mathbf{R}}^2 - \left\{{p_1}\right\} \cong S^{1} \times{\mathbf{R}}\). But since \({\mathbf{R}}\) is contractible, the last one is homotopic to \(S^1 \times\left\{{0}\right\} \cong S^1\). Alternatively: use the lemma, then \(k=2\) and so \(S^2 - \left\{{p_1, p_2}\right\} \simeq\bigvee_{i=1}^{1}S^1 = S^1\).

• Lemma: \(S^n - \left\{{p_i}\right\}_{i=1}^k = \bigvee_{k-1}S^{n-1}\), i.e. \(S^n\) minus \(k\) points is equal to \(k-1\) copies of of \(S^{n-1}\). Proof: \(S^n - \left\{{p_1}\right\} \cong {\mathbf{R}}^n\) by stereographic projection, so \(S^n - \left\{{p_1, p_2 \cdots p_k}\right\} \cong {\mathbf{R}}^n - \left\{{p_2, \cdots p_k}\right\}\). WLOG, suppose none of these points are zero (otherwise, take a translation away from zero. This is affine and continuous.) Then fix 0 as the base point, and form \(k-1\) loops \(\alpha_i\), where the \(i\)th loop encircles \(p_i\). Then \({\mathbf{R}}^n\) deformation retracts onto \(\cup_{i=1}^{k-1} \alpha_i\), which is homeomorphic to \(\bigvee_{i=1}^{k-1} S^1\).

• Theorem: \(\pi_1(\bigvee_{i=1}^k S^1) \cong {\Large{*}}_{i=1}^n{\mathbf{Z}}\), the free product of \(n\) copies of \({\mathbf{Z}}\). Proof: By induction, using Van-Kampen’s theorem. Base case: Take \(i=1\), then \(\pi_1(S^1) = {\mathbf{Z}}\) as proved in Hatcher. Inductive step: Suppose this holds for all \(k < n\), then we have \(X = \bigvee^n S^1 = \left( \bigvee^{n-1}S^1\right) \vee S^1\). Let \(p\) be the point of common intersection, then let \(U = \bigvee^{n-1} S^1\) \(V = S^1 \cup\left\{{p}\right\}\)

Then \(U\cup V = X\), \(U \cap V = \left\{{p}\right\}\), both \(U,V\) are path-connected. Since we have \(\pi_1({\operatorname{pt}}) = 0\), the amalgamated free product reduces to the usual free product. By the IH, we have \(\pi_1(U) = {\Large{*}}^{n-1}{\mathbf{Z}}\), so

\(\pi_1(X) = \pi_1(U\cup V) = \pi_1(U) * \pi_1(V) =_{\text{IH}} ({\Large{*}}^{n-1} {\mathbf{Z}}) * \pi_1(V) = ({\Large{*}}^{n-1} {\mathbf{Z}}) * {\mathbf{Z}}= {\Large{*}}^{n} {\mathbf{Z}}\).

Definition: Let \(F_n \coloneqq{\Large{*}^n} {\mathbf{Z}}\) be the free abelian group on \(n\) generators. Lemma: If \(n\neq m, F_n \not\cong F_m\). Proof: If \(F^n \cong F^m\), then \({\mathbf{Z}}^n \cong {\mathbf{Z}}^m\). But then tensor both sides with \({\mathbf{Z}}_2\) over \({\mathbf{Z}}\), yielding \({\mathbf{Z}}^n \otimes_{\mathbf{Z}}{\mathbf{Z}}_2 \cong Z^m \otimes_{\mathbf{Z}}{\mathbf{Z}}_2\). But the LHS is isomorphic to \(({\mathbf{Z}}/2{\mathbf{Z}})^n\), while the RHS is isomorphic to \(({\mathbf{Z}}/2{\mathbf{Z}})^m\). (Why?) These are both finite groups - there are 2 elements in \({\mathbf{Z}}/2{\mathbf{Z}}\), so the first has \(2^n\) elements and the latter has \(2^m\) elements. But if \(2^n=2^m\), then \(n=m\). The lemma follows from the contrapositive.

Now we have all we need - let \(X = S^2 - \left\{{p_1, p_2}\right\}\) and \(Y = S^3 - \left\{{q_1, q_2}\right\}\). Then by the previous problems, \(X \simeq S^1\) and \(Y \simeq S^2\), so if \(S^2 \cong S^3\) then \(X \simeq Y\) and \(S^1 \simeq S^2\). But \(\pi_1(S^1) = {\mathbf{Z}}\) and \(\pi_1(S^2) = 0\), so \(S^1 \not\simeq S^2\), a contradiction.

• Here we go:

• Let \(\alpha(t) = e^{2\pi it}\) where \(t \in [0, 1]\), be a loop in \(S^1\) parameterized by \(t\), which goes around \(S^1\) exactly once. Then under the map \(f: z \mapsto z^n\), we obtain \(f(\alpha(t)) = e^{2\pi n i t}\) where \(t \in [0,1]\). This resulting loop then goes around \(S^1\) \(n\) times, so the induced homomorphism on \(\pi_1(S^1) = {\mathbf{Z}}\) is the map \(f^*: {\mathbf{Z}}\to{\mathbf{Z}}\) given by \(f^*(a) = na\).

• Define \(\alpha\) as above, and define \(f: S^1 \to S^1\) to be the antipodal map, so \(f(z) = -z\) for \(z \in S^1 \subset {\mathbf{C}}\). We then left \(\alpha\) to the fundamental group, and define \(f_*([\alpha]) = [f \circ \alpha]\). Computing, we have \((f\circ \alpha)(t) = f(\alpha(t)) = -e^{2\pi i t}\). Where \(\alpha(0) = \alpha(1) = 1 + 0i\), we have \((f\circ \alpha)(0) = (f\circ \alpha)(1) = -1 + 0i\). But note that \(\alpha\) was a counter-clockwise loop in \(S^1\), and the image of \(\alpha\) is also a counter-clockwise loop. So this maps the generator \([\alpha] \in \pi_1(S^1, 1)\) to the generator \([\alpha'] \in \pi_1(S^1, -1)\). But since \(S^1\) is path-connected, the fundamental groups at these two base points are isomorphic. Alternatively: the antipodal map on \(S^1\) is homotopic to the identity map (since \(n=1\) is odd), so \([f\circ \alpha] = [f][\alpha] = [\operatorname{id}][\alpha] = [\alpha]\), so the induced homomorphism on \(\pi_1(S^1)\) is the identity map.

• Let \(\alpha(t) = e^{it}\) where \(t\in [0, 2\pi]\) be a counter-clockwise loop in \(S^1\); then \([\alpha]\) generates the fundamental group. Then \(f^*([\alpha]) = [(f\circ \alpha) (t)] = [e^{it} \mapsto e^{2\pi i \sin t}]\). Then just consider how \(\sin\) behaves in each quadrant. In quadrant 1, as \(t\) ranges from \(0, \pi/2\) then \(\sin t\) ranges from 0 to 1, so \(\alpha\) is exactly traced out. In quadrant two, \(\overline{\alpha}\) is traced out, since \(\sin t\) decreases from 1 to 0. This happens again in the bottom quadrants, so we have \(f^*([\alpha]) = [\alpha\overline{\alpha}\alpha\overline{\alpha}] = [\alpha][\alpha]^{-1}[\alpha][\alpha]^{-1} = [\operatorname{id}]\). But the identity element in \({\mathbf{Z}}\) is 0, so the induced homomorphism on \({\mathbf{Z}}\) is \(f^*(a) = 0\), the homomorphism sending everything to 0.

• From complex analysis, \(W(f(\alpha(t))) = Z_f - P_f = 4 - 1 = 3\). No idea how to approach with induced maps on the fundamental group of \(S^1\) or \({\mathbf{C}}- \left\{{0}\right\}\).

• Let \(M\) be the mobius strip, identified as \(I\times I / (t,0) \sim (1-t, 1)\), and let \(x_0 = [(1, \frac{1}{2})] = [(0, \frac{1}{2})]\). Let \(X\) be the line \((t, \frac{1}{2})\) for \(t\in I\); by the identification of the endpoints this is actually a copy of \(I / {{\partial}}I \cong S^1\) inside of \(M\) representing the middle circle of the strip. But then \(M\) deformation retracts onto \(S^1\) by just moving every point in \(I\times I\) horizontally towards this line, so \(M \simeq S^1\) and \(\pi_1(M) \cong {\mathbf{Z}}\), generated by the loop described which we’ll call \(\alpha\).

To see what the boundary curve is, label the corners \(a,b\) with the suitable identification. Then take a path from \(a\) to \(b\) on the right-hand boundary of the square. By sliding this through \(I\times I\), this is homotopic \(\alpha\). But similarly, the path from \(b\) to \(a\) on the LHS of the square is also homotopic to \(\alpha\), so the loop \(a\to b \to a \simeq\alpha^2\), so if \([\alpha] = 1 \in \pi_1(M)\), then \([a\to b\to a] = 2\).

• First note that \(\pi_1(S^1\times S^1) \cong F^2\), the free group on two generators, say \([\alpha], [\beta]\) corresponding to the two nontrivial loops on the torus - say \(\alpha\) is the longitudinal loop, and \(\beta\) is the meridian. Then if \(\gamma\) is a loop on a torus, then you can just count how many times it winds longitudinally and around the meridian, say \(m\) and \(n\) times respectively. Then \(\gamma\) can be homotoped into \(m\) copies of \(\alpha\) and \(n\) copies of \(\beta\) based at \(x_0\). So the induced map is \(f_\sharp: F^2 \to F^2\) given by \(\alpha \mapsto \alpha^m, \beta \mapsto \beta^n\). Since \(F^2 \cong Z\times Z\), we equivalently have \([\alpha] = (1,0), [\beta] = (0,1)\), and then \(f_\sharp : Z^2 \to Z^2\) is given by \((1,0) \mapsto (m,0)\) and \((0,1) \mapsto (0,n)\).

Group Actions

Covering Spaces

• Any covering map \(p: S^1\times S^1 \to{\mathbf{RP}}^2\) would induce an injection on fundamental groups, but \(\pi_1(T) = {\mathbf{Z}}^2\) and \(\pi_1({\mathbf{Z}}_2)\) - but there are no homomorphisms between these groups. Why? One of them has an element of order 2, the other does not.

• Theorem: if \(M_g \twoheadrightarrow M_h\) is an \(n-\)sheeted covering space, then \(g = n(h-1) +1\).

• Draw CW square for \(T\) and cut down the center to see two copies of \(K\).

• Let \(p: \tilde G \twoheadrightarrow G\) be such a covering, \(a,b\in \tilde G\), we then want to show that \(p(a)p(b) = p(a\star b)\) for some group operation \(\star\) which we need to construct.

Pick a basepoint \(x\in G\) and any point \(\tilde x \in p^{-1}(x)\). Since \(\tilde G\) is path connected, pick two paths \(\alpha, \beta\) from \(\tilde x\) to \(a,b\) respectively.

Now define a path \(f: I \to G\) by \(f(t) = (p\circ \alpha)(t) \cdot (p\circ \beta)(t)\), that is, evaluating \(f, g\) at a given time in \(\tilde G\), projecting the results down into \(G\), and multiplying them there. By uniqueness of path lifting, this yields a lift \(\tilde f: I \to\tilde G\)

Then define \(a\star b = \tilde f(1)\), the endpoint of \(\tilde f\) in \(\tilde G\). Then by construction,

\(p(a\star b) = p(\tilde f(1)) = f(1) = (p\circ\alpha)(1)\cdot (p\circ\beta)(1) = p(a)p(b)\). (Need to show this is continuous, and doesn’t depend on \(\alpha,\beta\)?)

• Since \(T^n = \prod_nS^1\), we have \(\pi_1(T^n) = \prod_n \pi_1(S^1) = {\mathbf{Z}}^n\). We can also construct a cover \(p:{\mathbf{R}}^n \to T^n\) by just taking \({\mathbf{R}}\twoheadrightarrow S^1\) the usual cover in each coordinate, yielding the covering space \(\tilde X = {\mathbf{R}}^n\) over \(X = T^n\).

By Hatcher (prop 4.1), the induced maps \(p_*^i: \pi_i(\tilde X) \to\pi_i(X)\) is an isomorphism for \(i \geq 2\). But \(\pi_i({\mathbf{R}}^n) = 0\) for \(i \neq 0\), so by this isomorphism \(\pi_i(T^n) = i \geq 2\).

• General construction: construct a tree \(T\) by picking a basepoint in \(G\) and adding a vertex for every non-backtracking walk in \(G\).

In this case, it’s the infinite 3-valent graph (also called the infinite \(k-\)regular tree)

This is the universal cover, because \(T\) is connected and acyclic (i.e. a tree). This means that \(\pi_1(T) = 0\), so \(T\) is simply connected. Since universal covers are simply connected and unique up to isomorphism, this is it.

• Generators of the subgroups:

• \(\left< ab^{-1}, aba^{-2}, a^3b^{-1}a^{-2}, a^3\right>\)

• \(\left< b, aba^{-1}, a^2ba^{-2},a^3\right>\)

• \(\left<b^2, ba, a^3, aba^{-1}\right>\)

• \(\left<b\right>\)

• \(\left<ba, b^{-1}a\right>\)

Relevant covers:

• 1512964258737

• 1512964650272

• 1512965253808

• 1512965792844

• Let \(T\) be a copy of the Cayley Tree on two on the two generators \(a, b\), then: 1512966232873

• This is just the Cayley graph over \({\mathbf{Z}}\times{\mathbf{Z}}\), or essentially the integer lattice:1512966445331

• It’s helpful to note that \(\left< (1,0), (0, p) \right> \subset \left< (1,0), (0,1) \right> \cong {\mathbf{Z}}\times{\mathbf{Z}}\subset {\mathbf{R}}\times{\mathbf{R}}\) is an index \(p\) subgroup.

Simplicial Homology

• Todo

• Figure 8 1513072050945 Here we have: \(C_3 = \emptyset\) \(C_1 = [12], [23], [13], [34], [35], [45] \cong {\mathbf{Z}}^6\) \(C_0 = [1], [2], [3],[4], [5] \cong {\mathbf{Z}}^5\)

So we have \(C_2 \to C_1 \to C_0 \cong 0\xrightarrow{{\partial}_2} {\mathbf{Z}}^6 \xrightarrow{{\partial}_1} {\mathbf{Z}}^5\xrightarrow{{\partial}_0} 0\)

Computing boundary operators, we have

\({\partial}_1([12]) = [2] - [1]\) \({\partial}_1([23]) = [3] - [2]\) \({\partial}_1([13]) = [3] - [1]\) \({\partial}_1([34]) = [4] - [3]\) \({\partial}_1([35]) = [5] - [3]\) \({\partial}_1([45]) = [5] - [4]\)

\({\partial}_0 = 0\)

And so \(H_0 = \ker {\partial}_0/\operatorname{im}{\partial}_1 = \frac{C_0}{<{\partial}_1([ij])>}\), but from the above calculation we have \([5] = [4] = [3] = [2] = [1]\) in the quotient, so there is just one generator and \(H_0 \cong {\mathbf{Z}}\).

Note that \({\partial}_2\) is an injection from 0 into \(C_1\), since there are no 2-simplices. Moreover, one can generate two 1-cycles, so we have \(H_1 = \frac{\ker {\partial}_1}{\operatorname{im}{\partial}_2} =\frac{<[23]-[31] + [12],~[45] - [35] + [34]>}{0} \cong {\mathbf{Z}}^2\).

One way to see that these are the generators is to pretend there are two 2-simplices, \([123], [345]\) and compute \({\partial}_2\) of both of them. Since \({\partial}_1{\partial}_2 = 0\), anything in the image of \({\partial}_2\) would have to go to zero anyways, and would thus be in the kernel of \({\partial}_1\). Since it’s not actually the boundary of any 2-chain, it doesn’t become trivial in homology.

So we have \(H_2 \to H_1 \to H_0 = 0 \to{\mathbf{Z}}^2 \to{\mathbf{Z}}\).

• \(S^2\) 1513072379449 So we have \(C_0 = 1,2,3,4,5,6\) \(C_1 = 12,14,15,16,23,25,26,34,35,36,45,46\) \(C_2 = 126, 236, 346, 146, 125, 235, 345, 145\)

\(C_3 = \emptyset\)

And \(0 \xrightarrow{{\partial}_3} C_2 \xrightarrow{{\partial}_2} C_1 \xrightarrow{{\partial}_1} C_0 \xrightarrow{{\partial}_0} 0 \cong 0 \xrightarrow{{\partial}_3} {\mathbf{Z}}^{8} \xrightarrow{{\partial}_2} {\mathbf{Z}}^{12} \xrightarrow{{\partial}_1} {\mathbf{Z}}^{6} \xrightarrow{{\partial}_0} 0\) We have \({\partial}_1([ij]) = j-i\) and \({\partial}_2([ijk]) = jk -ik +ij\).

We know in advance we should have \(\prod H_n = (\cdots,0, {\mathbf{Z}}, 0, {\mathbf{Z}})\).

For \(H_0 = \frac{\ker {\partial}_0}{\operatorname{im}{\partial}_1} = \frac{C_0}{\left<\left\{{j-i \mathrel{\Big|}i < j}\right\}\right>}\). In the quotient, we see \(1=6=3=2=5=4\) by just taking the indicated walk on the graph, so there is one generator in the quotient and \(H_0 \cong {\mathbf{Z}}\).

For \(H_1 = \frac{\ker {\partial}_1}{\operatorname{im}{\partial}_2}\), we just note that there are 6 2-cycles, so each are in the kernel of \({\partial}_1\), but each of them comes from a 2-cell, so is in the image of \({\partial}_2\). So both groups in question are \({\mathbf{Z}}^8\), and the quotient is zero. For \(H_3 = \frac{\ker{\partial}_2}{\operatorname{im}{\partial}_3}\), since \(\operatorname{im}{\partial}_3 = 0\), we can just look at \({\partial}_3([123456]) = 23456 - 13456 + 12456 - 12356 +12346 - 12345\). This is an element (and the only one) that goes to zero under \({\partial}_2\), it generates \(\ker{\partial}_2\). So there is one generator, and \(H_3 ={\mathbf{Z}}\).

• \({\mathbf{RP}}^2\)

• \(S^2 \cup_f D^2\), where \(f\) attaches to the equator

• \(T\cup_f D^2\), where \(f\) attaches inside the torus

Mayer Vietoris Problems

\({\mathbf{RP}}^2\)

We start with a few known facts. Let \(A=M\), the Mobius strip, and \(B= D^2\), the solid disk.

• \({\mathbf{RP}}^2 = M {\textstyle\coprod}_{\partial}D^2\)
• \(H_*(M) = H_*(S^1)\), by a deformation retract of \(M\) onto its center circle.
• \(H_*(D^2) = {\mathbf{Z}}\delta_0\)
• \(H_*(S^1)= {\mathbf{Z}}(\delta_0 + \delta_1)\)
• \(M \cap D^2 = {\partial}M = S^1\)

From Mayer-Vietoris, we have

and plugging in what is known yields

where \(i: S^1 \to M\) and \(j: S^1 \to D^2\).

We can then identify all of the induced maps:

• \(i^2: H_2 {\partial}M \to H_2 M \implies i^2: 0 \to 0 \implies i^2 = 0\)
• \(i^1: H_1 {\partial}M \to H_1 M\), i.e. \(i^1: {\mathbf{Z}}\to {\mathbf{Z}}\) where \(1 \mapsto 2\)
  • Since \(M\) deformation retracts onto its center circle, \(H_1 M \cong H_1 S_M\) where \(S_M\) is the center circle (homotopies induce isomorphisms on homology). But \(H_1 {\partial}M\) is generated by a cycle of edges with includes into \({\partial}M\), which retracts onto a cycle that double covers \(S_M\), so this map acts by doubling the generator.
• \(i^0: H_0 {\partial}M \to H_0 M\), i.e. \(i^0: {\mathbf{Z}}\to {\mathbf{Z}}\)
• \(j^2: H_2 {\partial}M \to H_2 D^2 \implies j^2: 0 \to 0 \implies j^2 = 0\)
• \(j^1: H_1 {\partial}M \to H_1 D^2 \implies j^1: {\mathbf{Z}}\to 0 \implies j^1 = 0\)
• \(j^0: H_0 {\partial}M \to H_0 D^2 \implies j_0: {\mathbf{Z}}\to {\mathbf{Z}}\)

So we can that the only nontrivial maps are \(j^0, i^0, i^1\).

Claim: \(H_2({\mathbf{RP}}^2) = 0\):

We consider the portion of the sequence \begin{align*}\cdots 0 \xrightarrow{} H_2{{\mathbf{RP}}^2} \xrightarrow{\delta_2} H_1 {\partial}M \xrightarrow{(i^1, -j^1)} H_1M \oplus H_1 D^2 \cdots\\ \cdots 0 \xrightarrow{} H_2{\mathbf{RP}}^2 \xrightarrow{\delta_2} {\mathbf{Z}}\xrightarrow{(i^1, -j^1)} {\mathbf{Z}}\oplus 0 \cdots\end{align*}

We will show that \(\ker \delta_2 = \operatorname{im}\delta_2 = 0\). By the first isomorphism theorem, we would then have \(\frac{H_2 {\mathbf{RP}}^2}{\ker \delta_2} \cong \operatorname{im}\delta_2\) yielding \(\frac{H_2 {\mathbf{RP}}^2}{0} = H_2 {\mathbf{RP}}^2 \cong 0\).

• Claim: \(\ker \delta_2 = 0\)

  This follows because it is on the left tail of an exact sequence, where \(\ker \delta_2 = \operatorname{im}0 = 0\).

• Claim: \(\operatorname{im}\delta_2 = 0\)

  \begin{align*}(i^1, -j^1): H_1 {\partial}M \to H_1 M \oplus H_1 D^2\end{align*} is injective; explicitly, it is the map \begin{align*}M_2: {\mathbf{Z}}\to {\mathbf{Z}}\oplus 0\\~1\mapsto (2, 0)\end{align*}

  From above, know that \(-j^1\) is a zero map, and that \(i^1\) doubles each generator. By this explicit construction, it is injective since 0 maps to 0.

  But then \(\ker (i^1, -j^1) = \operatorname{im}\delta_2 = 0\) by exactness.

So now we have:

Claim: \(H_1({\mathbf{RP}}^2) = {\mathbf{Z}}_2\)

Here we are examining this portion of the sequence:

\begin{align*} \cdots {\mathbf{Z}}\xrightarrow{x\mapsto (2x, 0)} H_1 M \oplus H_1 D^2 \xrightarrow{l^1 - r^1} H_1 {\mathbf{RP}}^1 \xrightarrow{\delta_1} H_0 {\partial}M \xrightarrow{(i^0, -j^0)} H_0 M \oplus H_0 D^2 \cdots\\ \cdots {\mathbf{Z}}\xrightarrow{x\mapsto (2x, 0)} {\mathbf{Z}}\oplus 0 \xrightarrow{l^1-r^1} H_1 {\mathbf{RP}}^1 \xrightarrow{\delta_1} {\mathbf{Z}}\xrightarrow{(i^0, -j^0)} {\mathbf{Z}}\oplus {\mathbf{Z}}\cdots \end{align*}

In general, we have the first isomorphism theorem: given any map \(f\) we have \(\frac{\mathrm{dom}f}{\ker f} \cong \operatorname{im}f\). Here we will take \(f = l^1 - r^1\) and identify the necessary components to apply this theorem.

• Claim: \(\operatorname{im}l^1 - r^1 = H_1 {\mathbf{RP}}^2\).
  • We use the fact that the maps \((i^*, j^*)\) are all injections, so in particular \(0 = \ker (i^0, j^0) = \operatorname{im}\delta_1\) by exactness. Consequently \(\ker \delta_1 = H_1{\mathbf{RP}}^1 = \operatorname{im}l^1 - r^1\) by exactness.
• What is \(\ker (l^1-r^1)\)?
  • By exactness, \(\ker (l^1 - r^1) = \operatorname{im}(x \mapsto (2x, 0)) = 2{\mathbf{Z}}\oplus 0\)

By the first isomorphism theorem, we have \(\operatorname{im}(l^1-r^1) \cong \frac{\mathrm{dom} (l^1-r^1)}{\ker (l^1-r^1)} = \frac{{\mathbf{Z}}\oplus 0}{2{\mathbf{Z}}\oplus 0} \cong {\mathbf{Z}}_2\).

Note that \(l^1 - r^1\) is a nontrivial homomorphism from \(2{\mathbf{Z}}\cong {\mathbf{Z}}\) to \({\mathbf{Z}}_2\), of which there is only one: the natural quotient map \(x \mapsto x \operatorname{mod}2\).

There is also no nontrivial homomorphism from \({\mathbf{Z}}_2 \to {\mathbf{Z}}\), so \(\delta_1 = 0\).

We now have:

Claim: \(H_0({\mathbf{RP}}^2) = {\mathbf{Z}}\)

Here we examine

\begin{align*} H_1{\mathbf{RP}}^2 \xrightarrow{\delta_1} H_0 {\partial}M \xrightarrow{(i^0, j^0)} H_0 M \oplus H_0 D^2 \xrightarrow{l^0 - r^0} H_0{\mathbf{RP}}^2 \xrightarrow{\delta_0} 0\\ {\mathbf{Z}}_2 \xrightarrow{\delta_1} {\mathbf{Z}}\xrightarrow{(i^0, j^0)} {\mathbf{Z}}\oplus {\mathbf{Z}}\xrightarrow{l^0 + r^0} H_0{\mathbf{RP}}^2 \xrightarrow{\delta_0} 0 \end{align*}

Since there is no nontrivial homomorphism from \({\mathbf{Z}}_2 \to {\mathbf{Z}}\), we have \(\delta_1 = 0\).

We also have \(\delta_0 = 0\) and \(\ker \delta_0 = H_0 {\mathbf{RP}}^2 = \operatorname{im}l^0 + r^0\) making \(l^0 + r^0\) surjective, so by the first isomorphism theorem we have \(H_0 {\mathbf{RP}}^2 \cong \frac{{\mathbf{Z}}\oplus {\mathbf{Z}}}{\ker l^0 + r^0} = \frac{{\mathbf{Z}}\oplus {\mathbf{Z}}}{\operatorname{im}(i^0, j^0)}\)

By a similar argument used earlier, the double covering of the boundary circle \({\partial}M\) over \(S^1\) yields the map \((i^0, j^0): {\mathbf{Z}}\to{\mathbf{Z}}\oplus {\mathbf{Z}}\) given by \(x \mapsto (2x, 2x)\) with

Summary:

With all of this information, we finally have

And so we find \(H_*({\mathbf{RP}}^2) = {\mathbf{Z}}\delta_0 + {\mathbf{Z}}_2\delta_1\)

Cellular Homology

Degree

UCT

Homological Algebra