1
Describe all possible covering maps between \(S^2, T^2, K\)
- \(\tilde X \twoheadrightarrow X\) induces \(\pi_1(\tilde X) \hookrightarrow\pi_1(X)\)
- \(\chi(\tilde X) = n \chi (X)\)
- \(\pi_n(X) = [S^n, X]\)
- \(Y \to X\) with \(\pi_1(Y) = 0\) and \(\tilde X \simeq{\operatorname{pt}}\implies\) every \(Y\xrightarrow{f} X\) is nullhomotopic.
- \(\pi_*(T^2) = [{\mathbf{Z}}\ast {\mathbf{Z}}, 0\rightarrow]\)
- \(\pi_*(K) = [{\mathbf{Z}}\rtimes_{{\mathbf{Z}}_2} {\mathbf{Z}}, 0\rightarrow]\)
- Universal covers are homeomorphic.
- \(\pi_{\geq 2}(\tilde X) \cong \pi_{\geq 2}(X)\)
Spaces
- \(S^2 \twoheadrightarrow T^2\)
- \(S^2 \twoheadrightarrow K\)
- \(K \twoheadrightarrow S^2\)
-
\(T^2 \twoheadrightarrow S^2\)
- All covered by the fact that \begin{align*} {\mathbf{Z}}= \pi_2(S^2) \neq \pi_2(X) = 0 \end{align*} for \(X = T^2, K\).
-
\(K \twoheadrightarrow T^2\)
- Doesn’t cover, would induce \(\pi_1(K) \hookrightarrow\pi_1(T^2) \implies {\mathbf{Z}}\rtimes{\mathbf{Z}}\hookrightarrow{\mathbf{Z}}^2\) but this would be a non-abelian subgroup of an abelian group.
-
\(T^2 \twoheadrightarrow K\)
- ?
\todo[inline]{Not complete!}
2
Show that \({\mathbf{Z}}^{\ast 2}\) has subgroups isomorphic to \({\mathbf{Z}}^{\ast n}\) for every \(n\).
- \(\pi_1(\bigvee^k S^1) = {\mathbf{Z}}^{\ast k}\)
- \(\tilde X \twoheadrightarrow X \implies \pi_1(\tilde X) \hookrightarrow\pi_1(X)\)
- Every subgroup \(G \leq \pi_1(X)\) corresponds to a covering space \(X_G \twoheadrightarrow X\)
- \(A \subseteq B \implies F(A) \leq F(B)\) for free groups.
It is easier to prove the stronger claim that \({\mathbf{Z}}^{\mathbb{N}}\leq {\mathbf{Z}}^{\ast 2}\) (i.e. the free group on countably many generators) and use fact 4 above. Just take the covering space \(\tilde X \twoheadrightarrow S^1 \vee S^1\) defined via the gluing map \({\mathbf{R}}\cup_{{\mathbf{Z}}} S^1\) which attaches a circle to each integer point, taking 0 as the base point. Then let \(a\) denote a translation and \(b\) denote traversing a circle, so we have \(\pi_1(\tilde X) = \left<\cup_{n\in{\mathbf{Z}}}a^nba^{-n}\right>\) which is a free group on countably many generators. Since \(\tilde X\) is a covering space, \(\pi_1(\tilde X) \hookrightarrow\pi_1(S^1 \vee S^1) = {\mathbf{Z}}^{\ast 2}\). By 4, we can restrict this to \(n\) generators for any \(n\) to get a subgroup, and \(A\leq B \leq C \implies A \leq C\) as groups.
3
Construct a space having \(H_*(X) = [{\mathbf{Z}}, 0, 0, 0, 0, {\mathbf{Z}}_4, 0, \cdots]\).
- Construction of Moore Spaces
- \(\tilde H_n(\Sigma X) = \tilde H_{n-1}(X)\), using \(\Sigma X = C_X \cup_X C_X\) and Mayer-Vietoris.
Take \(X = e^0 \cup_{\Phi_1} e^5 \cup_{\Phi_2} e^6\), where \begin{align*} \Phi_1: {\partial}B^5 = S^4 \xrightarrow{z~\mapsto z^0} e^0 \\ \Phi_2: {\partial}B^6 = S^5 \xrightarrow{z~\mapsto z^4} e^5 .\end{align*}
where \(\deg \Phi_2 = 4\).
4
Compute \(H_*\) of the complement of a knotted solid torus in \(S^3\).
- \(H_*(T^2) = [{\mathbf{Z}}, {\mathbf{Z}}^2, {\mathbf{Z}}, 0\rightarrow]\)
- \(N^{(1)} \simeq S^1\), so \(H_{\geq 2}(N) = 0\).
- A SES \(0\to A\to B \to F \to 0\) with \(F\) free splits.
- \(0\to A \to B \xrightarrow{\cong} C \to D \to 0\) implies \(A = D = 0\).
Let \(N\) be the knotted solid torus, so that \({\partial}N = T^2\), and let \(X = S^3 - N\). Then
- \(S^3 = N \cup_{T^2} X\)
- \(N \cap X = T^2\)
and we apply Mayer-Vietoris to the reduced homology of \(S^3\):
We can plug in known information and deduce some maps:
We then deduce:
-
\(H_0(X) = {\mathbf{Z}}\): ? (Appeal to some path-connectedness argument?)
-
\(H_1(X) = {\mathbf{Z}}\) using the SES appearing on the first row: \begin{align*} 0 \to {\mathbf{Z}}^{ \oplus 2} \to {\mathbf{Z}}\oplus H_1(X) \to 0 \end{align*} which is thus an isomorphism.
-
\(H_2(X) = H_3(X) = 0\) by examining the SES spanning lines 3 and 2: \begin{align*} 0 \hookrightarrow H_3(X) \hookrightarrow{\mathbf{Z}}\xrightarrow{\cong_{{\partial}_3}} {\mathbf{Z}}\twoheadrightarrow H_2(X) \twoheadrightarrow 0 \end{align*} Claim: \({{\partial}}_3\) must be an isomorphism. If this is true, \(H_3(X) \cong \ker {{\partial}}_3 = 0\) and \(H_2(X) \cong \operatorname{coker}({{\partial}}_3) \coloneqq{\mathbf{Z}}/\operatorname{im}({{\partial}}_3) \cong {\mathbf{Z}}/{\mathbf{Z}}= 0\).
\todo[inline]{Why is this true?}
5
Compute the homology and cohomology of a closed, connected, oriented 3-manifold \(M\) with \(\pi_1(M) = {\mathbf{Z}}^{\ast 2}\).
Facts used:
- \(M\) closed, connected, oriented \(\implies H_i(M)\cong H^{n-i}(M)\)
- \(H_1(X) = {\mathsf{Ab}}(\pi_1(X))\).
- For orientable manifolds \(H_n(M^n) = {\mathbf{Z}}\)
Homology
- Since \(M\) is connected, \(H_0 = {\mathbf{Z}}\)
- Since \(\pi_1(M) = {\mathbf{Z}}^{\ast 2}\), \(H_1\) is the abelianization and \(H_1(X) = {\mathbf{Z}}^2\)
- Since \(M\) is closed/connected/oriented, Poincaré Duality holds and \(H_2 = H^{3-2} = H^1 = \mathbf{F} H_1 + \mathbf{T}H_0\) by UCT. Since \(H_0={\mathbf{Z}}\) is torsion-free, we have \(H_2(M) = H_1(M) = {\mathbf{Z}}^2\).
- Since \(M\) is an orientable manifold, \(H_3(M) = {\mathbf{Z}}\)
- So \(H_*(M) = [{\mathbf{Z}}, {\mathbf{Z}}^2, {\mathbf{Z}}^2, {\mathbf{Z}}, 0\rightarrow]\)
Cohomology
- By Poincaré Duality, \(H^*(M) = \widehat{H_*(M)} = [{\mathbf{Z}}, {\mathbf{Z}}^2, {\mathbf{Z}}^2, {\mathbf{Z}}, 0, \cdots]\). (Where the hat denotes reversing the list.)
6
Compute \(\operatorname{Ext}({\mathbf{Z}}\oplus {\mathbf{Z}}/2 \oplus {\mathbf{Z}}/3, {\mathbf{Z}}\oplus {\mathbf{Z}}/4 \oplus {\mathbf{Z}}/5)\).
Facts Used: 1
-
Since \({\mathbf{Z}}\) is a free \({\mathbf{Z}}{\hbox{-}}\)module, \begin{align*} \operatorname{Ext}({\mathbf{Z}}, {\mathbf{Z}}/m) = 0 \end{align*}
-
Using the usual projective resolution \(0 \to {\mathbf{Z}}\to {\mathbf{Z}}\to {\mathbf{Z}}/n \to 0\), \begin{align*} \operatorname{Ext}({\mathbf{Z}}/n, {\mathbf{Z}}) = {\mathbf{Z}}/n .\end{align*}
-
\begin{align*} \operatorname{Ext}({\mathbf{Z}}/n, {\mathbf{Z}}/m) = ({\mathbf{Z}}/m) / (n \cdot {\mathbf{Z}}/m) \cong ({\mathbf{Z}}/m) / (d \cdot {\mathbf{Z}}/m) && \\ \text{where } d \coloneqq\gcd(m, n) .\end{align*} General principle: \(\operatorname{Ext}({\mathbf{Z}}/n, G) = G/nG\)
By applying \(\mathop{\mathrm{Hom}}_{\mathbf{Z}}({-}, G)\) to the above resolution:
which can be identified with:
- Contravariant Hom takes coproducts to products: \begin{align*} \operatorname{Ext}(\bigoplus_{i\in I} A_i, \prod_{k\in K} B_k) = \prod_{i\in I} \prod_{k\in K} \operatorname{Ext}(A_i, B_k) .\end{align*}
Write \begin{align*} A_{-}&\coloneqq A_1 \oplus A_2 \oplus A_3 \coloneqq{\mathbf{Z}}\oplus {\mathbf{Z}}/2 \oplus {\mathbf{Z}}/3 \\ B_{-}&\coloneqq B_1 \oplus B_2 \oplus B_3 \coloneqq{\mathbf{Z}}\oplus {\mathbf{Z}}/4 \oplus {\mathbf{Z}}/5 .\end{align*}
We can then define the bicomplex \begin{align*} C_{{-}, {-}} \coloneqq \operatorname{Ext}(A_{-}, B_{-}) = \bigoplus_{0 \leq i, k \leq 3} \operatorname{Ext}(A_i, B_k) ,\end{align*} i.e. \(C_{i, k} \coloneqq \operatorname{Ext}(A_i, B_k)\), which can be organized into the following diagram where we take the Ext at each position and sum them all together:
This equals the following:
Which simplifies to:
So the answer is \({\mathbf{Z}}/2 \oplus {\mathbf{Z}}/2 \oplus {\mathbf{Z}}/3 \cong {\mathbf{Z}}/2 \oplus {\mathbf{Z}}/6\).
7
Show there is no homeomorphism \({\mathbf{CP}}^2 \xrightarrow{f} {\mathbf{CP}}^2\) such that \(f({\mathbf{CP}}^1)\) is disjoint from \({\mathbf{CP}}_1 \subset {\mathbf{CP}}_2\).
- Every homeomorphism induces isomorphisms on homotopy/homology/cohomology.
- \(H^*({\mathbf{CP}}^2) = {\mathbf{Z}}[\alpha] / (\alpha^2)\) where \(\deg \alpha = 2\).
- \([f(X)] = f_*([X])\)
- \(a\frown b = 0 \implies a=0~\text{or}~b=0\) (nondegeneracy).
Supposing such a homeomorphism exists, we would have \([{\mathbf{CP}}^1] \frown[f({\mathbf{CP}}^1)] = 0\) by the definition of these submanifolds being disjoint. But \([{\mathbf{CP}}^1]\frown[f({\mathbf{CP}}^1)] = [{\mathbf{CP}}^1]\frown f_*([{\mathbf{CP}}^1])\), where \begin{align*} f_*: H^*({\mathbf{CP}}^2) \to H^*({\mathbf{CP}}^2) \end{align*} is the induced map on cohomology. Since the intersection pairing is nondegenerate, either \([{\mathbf{CP}}^1] = 0\) or \(f_*([{\mathbf{CP}}^1]) = 0\). We know that \(H^*({\mathbf{CP}}^2) = {\mathbf{Z}}[\alpha] / \alpha^2\) where \(\alpha = [{\mathbf{CP}}^1]\), however, so this forces \(f_*([{\mathbf{CP}}^1]) = 0\). But since this was a generator of \(H^*\), we have \(f_*(H^*({\mathbf{CP}}^2)) = 0\), so \(f\) is not an isomorphism on cohomology.
8
Describe the universal cover of \(X = (S^1 \times S^1) \vee S^2\) and compute \(\pi_2(X)\).
- \(\pi_{\geq 2}(\overline{X} ) \cong \pi_{\geq 2}(X)\) for \(\overline{X}\) the universal cover of \(X\)
- Structure of the universal cover of a wedges
- \(\overline{T^2} = {\mathbf{R}}^2\) and \(\overline{S^2} = S^2\)
- By Mayer-Vietoris, \(H_n(\vee X_i) = \bigoplus H_n(X_i)\).
The universal cover can be identified as \begin{align*} \overline{X} = {\mathbf{R}}^2 \vee_{i, j \in {\mathbf{Z}}^2} S^2 ,\end{align*} i.e. the plane with a sphere wedged onto every integer lattice point. We can then check \begin{align*} \pi_1(X) &\cong \pi_1(\overline{X} ) \\ &= \pi_1( {\mathbf{R}}^2 \vee_{i, j \in {\mathbf{Z}}^2} S^2 ) \\ &= \pi_1( {\mathbf{R}}^2 \vee_{i, j \in {\mathbf{Z}}^2} S^2 ) \\ &= \prod_{i,j \in {\mathbf{Z}}^2} \pi_1({\mathbf{R}}^2) \times\pi_1(S^2) \\ &= 0 ,\end{align*} using that \(\pi_1(S^2) = 0\). Then by Hurewicz, \(\pi_2(X) \cong H_2(X)\), so we can compute \begin{align*} H_2(X) &= H_2( {\mathbf{R}}^2 \vee_{i, j \in {\mathbf{Z}}^2} S^2 ) \\ &= \bigoplus_{i,j \in {\mathbf{Z}}^2} H_2({\mathbf{R}}^2) \oplus H_2(S^2) \\ &= \bigoplus_{i,j \in {\mathbf{Z}}^2} {\mathbf{Z}} .\end{align*}
9
Let \(S^3 \to E \to S^5\) be a fiber bundle and compute \(H_3(E)\).
- Homotopy LES: \(F\to E\to B \leadsto \pi_*F() \to \pi_*(E) \to \pi_*(B)\).
- Hurewicz: \(\pi_{\leq n}(X) = 0, \pi_n(X) \neq 0 \implies \pi_n(X) \cong H_n(X)\).
- \(0\to A\to B \to 0\) exact iff \(A\cong B\)
From the LES in homotopy we have
and plugging in known information yields
where
- Rows 3 and 4 force \(\pi_3(E) \cong {\mathbf{Z}}\),
- Rows 0 and 1 force \(\pi_0(E) = {\mathbf{Z}}\) (todo: not clear if this is true… is it even needed here?)
- The remaining rows force \(\pi_1(E) = \pi_2(E) = 0\).
By Hurewicz, we thus have \(H_3(E) = \pi_3(E) = {\mathbf{Z}}\).
\todo[inline]{Four-corner spectral sequences, only homology in degrees 1,3,5,8. No differentials hit anything!}