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\begin{center}
Math 205B - Topology\\
\emph{ }\\
Dr. Baez\\
\emph{ }\\
March 2, 2007\\
\emph{ }\\
Christopher Walker\\
\emph{ }\\
\end{center}
\begin{exercise}\rm{
Show that the fundamental group of the 3-bouquet of circles is $\Z*\Z*\Z$. Generalize this result to the $n$-bouquet of circles, and show that this is the free group on $n$ generators.
}
\end{exercise}
\begin{proof}
We will use the Seifert-van Kampen Theorem to calculate the fundamental group. Let $U,V\subseteq X$ be as pictured (with the end points being open).
$$\includegraphics[scale=0.3]{Math205B_Bouquet.jpg}$$
Since $U\bigcap V$ is contractible, then its fundamental group is trivial. this makes our calculation easier since we get that our normal subgroup $N$ from the theorem is also trivial. $U$ is homotopy equivalent to the figure eight, so $\pi_1(U,x_0)=\Z *\Z$. Also, $V$ is homotopy equivalent to $S^1$, so $\pi_1(V,x_0)=\Z$. This tells us that the pushout of $U$ and $V$ is $\Z*\Z*\Z$ (since $*$ is associative). Thus $\pi_1$ of the 3-bouquet of circles is $\Z*\Z*\Z$.\\
\indent To generalize we simply set $U$ and $V$ to be the as above, where $V$ is still homotopy equivalent to $S^1$, but $U$ is homotopy equivalent to the $(n-1)$-bouquet of circles. This still gives us that $U\bigcap V$ is contractible, so by induction the fundamental group of the $n$-bouquet is $\underbrace{\Z*\cdots *\Z}_{n \emph{\rm{ times}}}$.\\
\indent We can also show that $\underbrace{\Z*\cdots *\Z}_{n \emph{\rm{ times}}}$ is the free group on $n$ generators, Denote $F_n$. We first use the universal property of a free group. Let $X$ be the set $\{x_1,x_2,...,x_n\}$ and define $f:X\to \underbrace{\Z*\cdots *\Z}_{n \emph{\rm{ times}}}$ as $f(x_i)=z_i$ where $z_i$ is the generator from the $ith$ copy of $\Z$. Since $F_n$ is free we have that there exist a homomorphism $g: F_n\to \underbrace{\Z*\cdots *\Z}_{n \emph{\rm{ times}}}$ such that the following diagram commutes:
$$\begin{diagram}
X & \rTo^i & F_n \\
& \rdTo_f & \dTo_g \\
& & \underbrace{\Z*\cdots *\Z}_{n \emph{\rm{ times}}} \\
\end{diagram}$$
or $f=g\circ i$. Also $g$ is bijective on the generators of each group, so it is bijective on the entire sets. \\
\indent Now define $f_1:\pi_1(U,x_0)=\underbrace{\Z*\cdots *\Z}_{n-1 \emph{\rm{ times}}}\to F_n$ as $f_1(z_i)=x_i$ for $i=1,...,n-1$ where the $x_i$'s are generators of $F_n$. also define $f_2:\pi_1(V,x_0)=\Z\to F_n$ with $f_2(1)=x_n$. Since $\underbrace{\Z*\cdots *\Z}_{n \emph{\rm{ times}}}$ is a pushout then there exist a homomorphism $h:\underbrace{\Z*\cdots *\Z}_{n \emph{\rm{ times}}}\to F_n$ such that $h(\pi_1(j_1))=f_1$ and $h(\pi_1(j_2))=f_2$. On the generators, $g\circ h(z_i)=g(x_i)=z_i$ and $h\circ g(x_i)=h(z_i)=x_i$, So $g$ and $h$ are inverses. Therefore $\underbrace{\Z*\cdots *\Z}_{n \emph{\rm{ times}}}\im F_n$.
\end{proof}
\newpage
\begin{exercise}\rm{
Calculate the fundamental group of the two torus $T^2$ given as the quotient topology of the octogon by identifying edges as follows:
$$\includegraphics[scale=0.3]{Math205B_T2.jpg}$$
}
\end{exercise}
\begin{proof}
First we need to define open sets $U,V\subset T^2$ that meet the conditions of the Seifert-van Kampen Theorem. Let $U,V$ be the subsets of $T^2$ as pictured here.
$$\includegraphics[scale=0.35]{Math205B_T2_2.jpg}$$
This choice of $U$ and $V$ gives what we need. The fundamental group of $U$ is the trivial group, since $U$ is contractible. Also, the fundamental group of $U\bigcap V$ is $\Z$ since it has $S^1$ as a deformation retract. \\
\indent Now we need to consider the fundamental group of $V$. One way to think about this is to consider $V$ as a ``wire'' (i.e. we only consider the edges of the octogon). With this we can ``fold'' $V$ as follows:
$$\includegraphics[scale=0.5]{Math205B_TV.jpg}$$
So we have that $V$ is homotopy equivalent to the bouquet of four circles. By the previous exercise, this tells us that the fundamental group of $V$ is $\Z *\Z *\Z *\Z$. We can now apply the Seifert-van Kampen Theorem, to get the following pushout.
$$\begin{diagram}
& & \Z & & \\
& \ldTo^{\pi_1(i_1)} & & \rdTo^{\pi_1(i_2)} & \\
\{0\} & & & & \Z *\Z *\Z *\Z \\
&\rdTo_{\pi_1(j_1)} & & \ldTo_{\pi_1(j_2)} & \\
& & \pi_1(T^2,x_0) & & \\
\end{diagram}$$
We know that for this to be a pushout, $\pi_1(T^2,x_0)=G*H/N$ where $G=\pi_1(U,x_0)$, $H=\pi_1(V,x_0)$, and $N$ we calculate as follows. To calculate $N$, we need to see what the image of $1$ is under the homomorphisms $\pi_1(i_1)$ and $\pi_1(i_2)$. It is clear that $\pi_1(i_1)$ is the trivial homomorphism map, so $\pi_1(i_1)(1)=0$. As for the other direction, we need to trace the loop of $U\bigcap V$ that generates $\pi_1(U\bigcap V,x_0)$ around $V$ to see what we get. From the picture before we see that this loop corresponds to the element $aba^{-1}b^{-1}dcd^{-1}c^{-1}$. We now take these two elements, and form the normal subgroup $N$ of $G * H$ generated by $1(aba^{-1}b^{-1}dcd^{-1}c^{-1})^{-1}=(aba^{-1}b^{-1}dcd^{-1}c^{-1})^{-1}$.\\
\indent Since $G=\{0\}$ and $H=\Z *\Z *\Z *\Z$ we see that $G*H=\Z *\Z *\Z *\Z$. So $\pi_1(T^2,x_0)=(\Z *\Z *\Z *\Z)/N$, where $N$ is the normal subgroup generated by $aba^{-1}b^{-1}dcd^{-1}c^{-1}$.
\end{proof}
\newpage
\begin{exercise}\rm{
Calculate the fundamental group of the ``Dunce Cap'', which is the quotient space of a triangle by associating the three side with each other as pictured below.
$$\includegraphics[scale=0.3]{Math205B_Dunce.jpg}$$
}
\end{exercise}
\begin{proof}
We will use the same argument as with the double torus (and the same $U$ and $V$). Again, $U$ is contractible, so its fundamental group is trivial. Also $U\bigcap V$ has $S^1$ as a deformation retract, so its fundamental group is $\Z$.\\
\indent In order to find the fundamental group of $V$we again think of it as just the edges. When we do this, we identify all three sides, and then connect the ends. Thus $V$ is homotopy equivalent to the $S^1$ and thus has $\Z$ as a fundamental group. So we get the following pushout:
$$\begin{diagram}
& & \Z & & \\
& \ldTo^{\pi_1(i_1)} & & \rdTo^{\pi_1(i_2)} & \\
\{0\} & & & & \Z \\
& \rdTo_{\pi_1(j_1)} & & \ldTo_{\pi_1(j_2)} & \\
& & \pi_1(X,x_0) & & \\
\end{diagram}$$
Now, in order for this to be a pushout we know $\pi_1(X,x_0)=G*H/N$, but we already have that $G=\{0\}$ and $H=\Z$. So we only need to calculate $N$. From the picture we see that the loop that generates $\pi_1(U\bigcap V,x_0)=\Z$ when considered as a loop in $V$ gives the group element $aaa^{-1}=a$. So the normal subgroup of $G*H=\Z$ generated by $a$ is $\Z$, so $G*H/N=\Z/\Z$. Thus $\pi_1(X,x_0)$ for the dunce cap is trivial.
\end{proof}
\end{document}