1. Describe all possible covering maps between $S^2, T^2, K$

Useful facts:

$\tilde X \twoheadrightarrow X$ 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*} `{=html} 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. - $T^2 \twoheadrightarrow K$ - Covers, #todo

2. Show that ${\mathbf{Z}}^{\ast 2}$ has subgroups isomorphic to ${\mathbf{Z}}^{\ast n}$ for every $n$.

Facts Used

$\pi_1(\bigvee^k S^1) = {\mathbf{Z}}^{\ast k}$
$\tilde X \twoheadrightarrow X \implies \pi_1(\tilde X) \hookrightarrow\pi_1(X)$
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.

Alternatively, just take a covering space of this form: ?? $\hfill\blacksquare$

3. Construct a space having $H_*(X) = [{\mathbf{Z}}, 0, 0, 0, 0, {\mathbf{Z}}_4, 0\rightarrow]$

Facts used:

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 $\deg \Phi_2 = 4$.$\hfill\blacksquare$

4. Compute the complement of a knotted solid torus in $S^3$.

Facts used:

$H_*(T^2) = [{\mathbf{Z}}, {\mathbf{Z}}^2, {\mathbf{Z}}, 0\rightarrow]$
$N^{(1)} \simeq S^1$, 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 $S^3$:

\begin{align*} 4\qquad H_4(T^2) \to H_4(N) \times H_4(X) \to H_4(S^3) \\ 3\qquad H_3(T^2) \to H_3(N) \times H_3(X) \to H_3(S^3) \\ 2\qquad H_2(T^2) \to H_2(N) \times H_2(X) \to H_2(S^3) \\ 1\qquad H_1(T^2) \to H_1(N) \times H_1(X) \to H_1(S^3) \\ 0\qquad H_0(T^2) \to H_0(N) \times H_0(X) \to H_0(S^3) \\ \end{align*}

where we can plug in known information and deduce some maps: \begin{align*} \begin{align} 4\qquad &0 \to &0 \qquad\to &0 \xrightarrow{{\partial}_4} \\ 3\qquad &0 \to &H_3(X) \qquad\to &{\mathbf{Z}}\xrightarrow{{\partial}_3}\\ 2\qquad &{\mathbf{Z}}\to &H_2(X) \qquad\to &0 \xrightarrow{{\partial}_2}\\ 1\qquad &{\mathbf{Z}}^2 \cong &{\mathbf{Z}}\times H_1(X) \qquad\to &0 \xrightarrow{{\partial}_1}\\ 0\qquad &{\mathbf{Z}}\to &{\mathbf{Z}}\times H_0(X) \qquad\to &{\mathbf{Z}}\to 0 \\ \end{align} \end{align*}

We then deduce:

$H_0(X) = {\mathbf{Z}}$ by the splitting of the line 0 SES `

\begin{align*}0 \to {\mathbf{Z}}\to {\mathbf{Z}}\times H_0(X) \to {\mathbf{Z}}\to 0\end{align*} `{=html} yielding $Z\times H_0(X) \cong {\mathbf{Z}}\times {\mathbf{Z}}$. - $H_1(X) = {\mathbf{Z}}$ by the line 1 SES ` \begin{align*}0 \to {\mathbf{Z}}^2 \to {\mathbf{Z}}\times H_1(X) \to 0\end{align*} `{=html} which yields 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*} `{=html} Since ${\partial}_3$ must be an isomorphism, this forces the edge terms to be zero. $\hfill\blacksquare$

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 $\implies H_i(M)\cong H^{n-i}(M)$
$H_1(X) = \pi_1(X) / [\pi_1(X), \pi_1(X)]$
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, Poincare 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 Poincare Duality, $H^*(M) = \widehat{H_*(M)} = [{\mathbf{Z}}, {\mathbf{Z}}^2, {\mathbf{Z}}^2, {\mathbf{Z}}, 0\rightarrow]$. (Where the hat denotes reversing the list.) $\hfill\blacksquare$

6. Compute $\operatorname{Ext}({\mathbf{Z}}\times {\mathbf{Z}}_2 \times {\mathbf{Z}}_3, {\mathbf{Z}}\times {\mathbf{Z}}_4 \times {\mathbf{Z}}_5)$

Facts Used:

$\operatorname{Ext}({\mathbf{Z}}, {\mathbf{Z}}_m) = {\mathbf{Z}}_m$
$\operatorname{Ext}({\mathbf{Z}}_m, {\mathbf{Z}}) = 0$
$\operatorname{Ext}(\prod_i A_i, \prod_j B_j) = \prod_i \prod_j \operatorname{Ext}(A_i, B_j)$

Break it up into a bigraded complex, take Ext of the pieces, and sum over the complex: $\operatorname{Ext}(\downarrow, \rightarrow)$ | ${\mathbf{Z}}$ | ${\mathbf{Z}}_4$ | ${\mathbf{Z}}_5$ ––––––––––––––––|———|———|–––– ${\mathbf{Z}}$ | 0 | 0 | 0 ${\mathbf{Z}}_2$ | ${\mathbf{Z}}_2$ | ${\mathbf{Z}}_2$ | 0 ${\mathbf{Z}}_3$ | ${\mathbf{Z}}_3$ | 0 | 0

So the answer is ${\mathbf{Z}}_2 \times {\mathbf{Z}}_2 \times {\mathbf{Z}}_3 = {\mathbf{Z}}_{12}$. $\hfill\blacksquare$

7. Show there is no homeomorphism ${\mathbf{CP}}^2{\circlearrowleft}_f$ such that $f({\mathbf{CP}}^1)$ is disjoint from ${\mathbf{CP}}_1 \subset {\mathbf{CP}}_2$.

Facts used:

Every homeomorphism induces isomorphisms on homotopy/homology/cohomology.
$H^*({\mathbf{CP}}^2) = {\mathbf{Z}}[\alpha] / (\alpha^2)$ where $\deg \alpha = 2$.
$[f(X)] = f_*([X])$
$a\dot{} b = 0 \implies a=0~\text{or}~b=0$ (nondegeneracy).

Supposing such a homeomorphism exists, we would have $[{\mathbf{CP}}^1] \dot{} [f({\mathbf{CP}}^1)] = 0$ by the definition of these submanifolds being disjoint.

But $[{\mathbf{CP}}^1]\dot{}[f({\mathbf{CP}}^1)] = [{\mathbf{CP}}^1]\dot{} f_*([{\mathbf{CP}}^1])$, 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. $\hfill\blacksquare$

8. Describe the universal cover of $X = (S^1 \times S^1) \vee S^2$ and compute $\pi_2(X)$.

Facts used:

$\pi_{\geq 2}(\tilde X) \cong \pi_{\geq 2}(X)$
Structure of the universal cover of a wedge product
${\mathbf{R}}^2 \twoheadrightarrow_p T^2 = S^1 \times S^1$

$\tilde X = {\mathbf{R}}^2 \cup_{{\mathbf{Z}}^2} S^2$, so $\pi_2(X) \cong \pi_2(\tilde X) = \prod_{i,j \in {\mathbf{Z}}^2} {\mathbf{Z}}= {\mathbf{Z}}^{{\mathbf{Z}}^2} = {\mathbf{Z}}^{\aleph_0}$.$\hfill\blacksquare$

9. Let $S^3 \to E \to S^5$ be a fiber bundle and compute $H_3(E)$.

Facts used:

Homotopy LES
Hurewicz
$0\to A\to B \to 0$ exact iff $A\cong B$

From the LES in homotopy we have \begin{align*} \begin{align} 4\qquad \pi_4(S^3) \to \pi_4(E) \to \pi_4(S^5) \\ 3\qquad \pi_3(S^3) \to \pi_3(E) \to \pi_3(S^5) \\ 2\qquad \pi_2(S^3) \to \pi_2(E) \to \pi_2(S^5) \\ 1\qquad \pi_1(S^3) \to \pi_1(E) \to \pi_1(S^5) \\ 0\qquad \pi_0(S^3) \to \pi_0(E) \to \pi_0(S^5) \\ \end{align} \end{align*}

and plugging in known information yields \begin{align*} \begin{align} 4\qquad &\pi_4(S^3) \to &\pi_4(E) \quad \to 0 \\ 3\qquad &{\mathbf{Z}}\to &\pi_3(E) \quad\to 0 \\ 2\qquad &0 \to &\pi_2(E) \quad\to 0 \\ 1\qquad &0 \to &\pi_1(E) \quad\to 0 \\ 0\qquad &{\mathbf{Z}}\to &\pi_0(E) \quad\to {\mathbf{Z}}\\ \end{align} \end{align*}

where rows 3 and 4 force $\pi_3(E) \cong {\mathbf{Z}}$, rows 0 and 1 force $\pi_0(E) = {\mathbf{Z}}$, and the remaining rows force $\pi_1(E) = \pi_2(E) = 0$.

By Hurewicz, we thus have $H_3(E) = \pi_3(E) = {\mathbf{Z}}$. $\hfill\blacksquare$

Summer 2003

1. Describe all possible covering maps between \(S^2, T^2, K\)

2. Show that \({\mathbf{Z}}^{\ast 2}\) has subgroups isomorphic to \({\mathbf{Z}}^{\ast n}\) for every \(n\).

3. Construct a space having \(H_*(X) = [{\mathbf{Z}}, 0, 0, 0, 0, {\mathbf{Z}}_4, 0\rightarrow]\)

4. Compute the complement of a knotted solid torus in \(S^3\).

5. Compute the homology and cohomology of a closed, connected, oriented 3-manifold \(M\) with \(\pi_1(M) = {\mathbf{Z}}^{\ast 2}\).

6. Compute \(\operatorname{Ext}({\mathbf{Z}}\times {\mathbf{Z}}_2 \times {\mathbf{Z}}_3, {\mathbf{Z}}\times {\mathbf{Z}}_4 \times {\mathbf{Z}}_5)\)

7. Show there is no homeomorphism \({\mathbf{CP}}^2{\circlearrowleft}_f\) such that \(f({\mathbf{CP}}^1)\) is disjoint from \({\mathbf{CP}}_1 \subset {\mathbf{CP}}_2\).

8. Describe the universal cover of \(X = (S^1 \times S^1) \vee S^2\) and compute \(\pi_2(X)\).

9. Let \(S^3 \to E \to S^5\) be a fiber bundle and compute \(H_3(E)\).