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\begin{center}
{\Large Exercises from February 14th} \\[.3em]
{\small 18.904 Spring 2011}
\end{center}
\vskip 1ex
These are exercises from John's and Noah's talks. Solutions are on the following page.
\begin{exercise}[John]
\label{john1}
Consider the embedding $\phi:S^1\vee S^1\rightarrow (S^1)^2$ defined such that the point $x$ of the first $S^1$ maps to $(x,1)$ and $x$ of the second $S^1$ maps to $(1,x)$. Describe the pushforward $\phi_\ast$ (a homomorphism from the free group on two generators to $\mathbb{Z}^2$).
\end{exercise}
\begin{exercise}[John]
\label{john2}
Let $\phi:X\rightarrow Y$ be a continuous map of topological spaces, and call $\phi_{x_0}:(X,x_0)\rightarrow(Y,\phi(x_0))$ the corresponding map of pointed topological spaces. Recall that a path $f:I\rightarrow X$ from $f(0)=x_0$ to $f(1)=x_1$ induces an isomorphism which I call $f_\sharp:\pi_1(X,x_0)\rightarrow\pi_1(X,x_1)$. Prove that $$(\phi_{x_1})_\ast\circ f_\sharp=(\phi\circ f)_\sharp\circ(\phi_{x_0})_\ast$$ for all such $\phi$ and $f$.
\end{exercise}
\begin{exercise}[Noah]
\label{noah1}
In class it was stated that the spaces $\R^3-S^2$ and $S^2\coprod \{0\}$ are of the same homotopy type. Prove this by finding a homotopy equivalence between the two spaces.
\end{exercise}
\begin{exercise}[Noah, 58.2 of Munkres]
\label{noah2}
For each of the following subsets of $\R^2$, determine if the fundamental group is $1$, $\Z$, or the free abelian group with two generators:
\begin{enumerate}
\item $\{x| ||x||>1\}$
\item $S^1\cup (\R_+\times 0)$
\item $S^1\cup (\R_+\times \R)$
\item $S^1\cup (\R\times 0)$
\item $\R^2-(\R_+\times 0)$
\end{enumerate}
\end{exercise}
\newpage
\par\noindent{\it Solution to Exercise~\ref{john1}}:
The fundamental group $\pi_1(S^1\vee S^1)$ is generated by (the homotopy classes of) a loop $f$ around the first $S^1$ and a loop $g$ around the second $S^1$ with no other relations. We may see that $\phi_\ast[f]$ is a loop around $S^1$ in the first coordinate of $(S^1)^2$ and $\phi_\ast[g]$ around the second coordinate of $(S^1)^2$. These are the generators of $\pi_1((S^1)^2)$, and have no other relations except commutativity (i.e., they generate a free abelian group). Thus $\phi_\ast$ is just the abelianization which maps the generators of the free group on two generators to the generators of the free \textit{abelian} group on two generators.
\vskip\baselineskip
\par\noindent{\it Solution to Exercise~\ref{john2}}:
Both $(\phi_{x_1})_\ast\circ f_\sharp$ and $(\phi\circ f)_\sharp\circ(\phi_{x_0})_\ast$ are homomorphisms from $\pi_1(X,x_0)$ to $\pi_1(Y,\phi(x_1))$. Let $g:S^1\rightarrow (X,x_0)$ be a loop based at $x_0$ so that $[g]$ represents an arbitrary element in the domain of the two homomorphisms. $$(\phi_{x_1})_\ast\circ f_\sharp[g]=(\phi_{x_1})_\ast[f^{-1}\cdot g\cdot f]=[\phi\circ(f^{-1}\cdot g\cdot f)]$$ where $f^{-1}$ represents the corresponding path from $x_1$ to $x_0$. On the other hand, $$(\phi\circ f)_\sharp\circ(\phi_{x_0})_\ast[g]=(\phi\circ f)_\sharp[\phi_{x_0}\circ g]=[(\phi\circ f)^{-1}\cdot(\phi\circ g)\cdot(\phi\circ f)]$$ We need to show that these two expressions inside the brackets are homotopic. In fact, it is more generally true that $\phi\circ(f\cdot g)=(\phi\circ f)\cdot(\phi\circ g)$ whenever $f$ and $g$ can be composed ($f(1)=g(0)$). This is because this expression is $\phi(f(2t))$ on $0\leq t\leq\frac{1}{2}$ and $\phi(g(2t-1))$ on $\frac{1}{2}\leq t\leq 1$. Therefore the two expressions act the same on any
$[g]$, and so $$(\phi_{x_1})_\ast\circ f_\sharp=(\phi\circ f)_\sharp\circ(\phi_{x_0})_\ast$$
\vskip\baselineskip
\par\noindent{\it Solution to Exercise~\ref{noah1}}:
Let $X=\R^3-S^2, Y=S^2\coprod \{0\}$, and define the map $\phi: X\to Y$ as follows: $$
\phi(x) = \left\{
\begin{array}{ll}
0 & : |x| <1 \\
x\|x| & : |x|>1
\end{array}
%
\right. $$
We claim that $\phi$ is a homotopy equivalence. Let $\psi: Y\to X$ be as follows $$
\psi(x) = \left\{
\begin{array}{ll}
0 & : x=0 \\
2x & : |x|=1
\end{array}
%
\right. $$
It is straightforward to verify that $\phi, \psi$ are continuous. Also, $\phi\psi=\text{Id}_Y$, so trivially $\phi\psi\simeq\text{Id}_Y$. Then, the family of maps $f_t, t\in I$ given by $$
f_t(x) = \left\{
\begin{array}{ll}
tx & : |x| <1 \\
\frac{2}{|x|}+(1-\frac{2}{|x|})t & : |x|>1
\end{array}
%
\right. $$
is a homotopy between $\phi\psi$ and $\text{Id}_X$ so that $\phi\psi\simeq \text{Id}_X$. This completes the proof. Note that we have in fact proved the stronger statement that the base-pointed spaces $(X,0), (Y,0)$ are homotopy equivalent.
\vskip\baselineskip
\par\noindent{\it Solution to Exercise~\ref{noah2}}:
$1) \Z, 2) \Z, 3) \Z, 4) $ free abelian group with two generators, $5) 1$.
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