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{\Large Exercises from February 16th} \\[.3em]
{\small 18.904 Spring 2011}
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These are exercises from Rafael's and Gabriels's talks. Solutions are on the following page.
\begin{exercise}[Rafael]
\label{rafael1}
Read carefully the following attempt to prove that $\pi_1(S^1\vee S^1) \approx F_2.$
``Let $F_2$ be the free group on two generators. Giving a homomorphism from $F_2$ to any group $G$ is the same as giving two homomorphisms from $\zz$ to $G,$ which is equivalent to giving two elements of $G.$ In the same way, if $(X, x_0)$ is a pointed topological space and if we have two elements of $\pi_1(X, x_0) \ -$ we can think of these two elements as maps from $S^1 \to X \ -$ then we get a map $S^1 \vee S^1 \to X$ and consequently a map $\pi_1(S^1\vee S^1) \to \pi_1(X, x_0).$
Since we just showed that $\pi_1(S^1\vee S^1)$ satisfies the universal property of $F_2$, namely, giving a homomorphism $\pi_1(S^1\vee S^1) \to \pi_1(X,x_0)$ is equivalent to giving two elements of $\pi_1(X, x_0)$, we can conclude that these two groups are isomorphic.''
This seems to be a very nice proof that $\pi_1(S^1\vee S^1) \approx F_2$. However, this proof is not correct. Explain.
(Keep in mind that this is a pre van Kampen proof, and cannot implicitly make use of that theorem or its consequences.)
\end{exercise}
\begin{exercise}[Gabriel]
\label{gabriel1}
Complete the example given in lecture by showing that $\mathbb{Z}/2\mathbb{Z}=\mathbb{Z}/2\mathbb{Z}\ltimes\mathbb{Z}$ where the semi-direct product is taken over the homomorphism family $\phi_i:\mathbb{Z}\to\mathbb{Z}$ defined. $\phi_0(x)=x$, $\phi_{1}(x)=-x$
\end{exercise}
\begin{exercise}[Gabriel]
\label{gabriel2}
Let $A$ be the free product on $a_1...a_n$, $B$ be the free product on $b_1...b_m$, $C$ the free product on $c_1...c_k$ with $k\le n,m$.\par
Take homomorphisms $\varphi:c_i\to a_i$ $\psi:c_i\to b_i$\par
Prove that $A*_CB$ is the free product on $n+m-k$ elements.
\end{exercise}
\newpage
\par\noindent{\it Solution to Exercise~\ref{rafael1}}:
The flaw in this proof is that it assumes that every group occurs as a fundamental group of a topological space. It is true
that every group occurs as a fundamental group, but one needs van Kampen's theorem to prove this result.
To make this proof correct, we also need to show that every group occurs as a fundamental group of a pointed topological space. If we show this, which will be done later in the course, then this fact together with the proof above makes it a very nice (and correct) proof that $\pi_1(S^1\vee S^1) \approx F_2.$
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\par\noindent{\it Solution to Exercise~\ref{gabriel1}}:
We take $\rho_1,\rho_2$ to be differently labeled generators of $\mathbb{Z}/2\mathbb{Z}$. Let
$$\varphi\left\{\begin{array}{l}e\to (0,0)\\\rho_1\to (1,0)\\\rho_2\to (1,1)\end{array}\right.$$ Then this generates a homomorphism from $A=\mathbb{Z}/2\mathbb{Z}*\mathbb{Z}/2\mathbb{Z}\to G=\mathbb{Z}/2\mathbb{Z}\ltimes\mathbb{Z}$. To see why, we observe that the only cancelation in $A$ under multiplication is that $\rho_1^2=e,\rho_2^2=e$, and this cancelation applies under the image as well, since $\phi(\rho_1),\phi(\rho_2)$ are both of order $2$, so that $\phi$ is operation preserving.\par
We then wish to show that this is an isomorphism:\\
Injective: we observe that every reduced word is $\rho_2^a(\rho_1\rho_2)^b\rho_1^c$ for some choice of $a,b,c$ with $a,b\in\{0,1\},\ c\ge 0$ Let $w$ a word in $\rho_1,\rho_2$ reduced, and suppose $\phi(w)=(0,0)$. Then $\phi(\rho_2)^a\phi(\rho_1\rho_2)^b\phi(\rho_1)^c=(0,0)$. $(1,1)^a(0,1)^b(1,0)^c=(0,0)$. But this is $(1,1)^a(0,-1)^b(1,0)^c$. In particular, for all possible choices of $a,c$ this produces elements in $\{(0,-b),(0,1+b),(1,-b),(1,1+b)\}$, equal to $(0,0)$ iff $a,b,c=0$. In particular, this means that $\phi$ is injective, since no non-identity element is taken to the identity.\par
Surjective: In particular, we also have that given $n>0$ $\phi((\rho_1\rho_2)^n)=(0,-n)$, $\phi(\rho_2(\rho_1\rho_2)^{n-1})=(1,n)$, $\phi(\rho_2(\rho_1\rho_2)^{n-1}\rho_1)=(0,n)$, $\phi((\rho_1\rho_2)^n\rho_1)=(1,-n)$, so that we can generate any combination $(m,n)$ for $m\in\{0,1\}$, $n\in\mathbb{Z}$. This shows $\phi$ surjective, so that $\phi$ is a bijective homomorphism, and hence an isomorphism from $A=\mathbb{Z}/2\mathbb{Z}*\mathbb{Z}/2\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}\ltimes\mathbb{Z}$, as desired.
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\par\noindent{\it Solution to Exercise~\ref{gabriel2}}:
The result follows by a fairly direct application of the Fundamental Principle of Amalgamated Free Products. Take the free group generated by $a_1...a_n,b_{k+1}...b_m$ to be $D$. Let $G$ a group, $f:A\to G$, $g: B\to G$ homomorphisms such that the maps $C\to A\to G,C\to B\to G$ agree. Then in particular, we may define a homomorphism on $D$, $h$, by $h(a_i)=f(a_i)=g(b_i)$ for $i\le k$. $h(a_i)=f(a_i)$ and $h(b_i)=g(b_i)$ for $i>k$. Then for each $f,g,G$ satisfying the above we get a unique map from $\to G$. On the other hand, given any group $G$, homomorphism $h:A*_CB\to G$, we can define $f:A\to G$ by $f(a_i)=h(a_i)$, $g(b_i)=h(a_i)$ if $i\le k$ and $g(b_i)=h(b_i)$ otherwise. Then $h(\varphi(c))=h(\varphi(c))$, so that homomorphisms from $D$ to $G$ are equivalent to homomorphisms from $A$ to $G$ and $B$ to $G$ which agree on the image of $C$.\par
In particular, this means that $D$ satisfies the fundamental principle of amalgamated free groups, so that $A*_CB\equiv D\equiv$ the free group on $n+m-k$ elements, as desired.
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