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\centerline{\bf Algebraic Topology}
\centerline{Homework 5: Due Friday, October 1}      % e.g. September 13
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This homework is all about reduced words, free products, and
variations on free products. Note that it's fairly long, and also that
it's due on Friday, not Wednesday -- solutions will be posted when I return
from Korea. Subsequent homeworks will be due on
Friday, October 8 and Wednesday October 20, since we have a midterm on
October 15.

1. Page 74, problems 4.1, 4.3, 4.4, 4.9.

2. Page 77, problem 5.3.

3. Page 81, problem 6.1

4. In class, I sketched an argument for the claim that the set of 
all reduced words is a group. Flesh out this argument by defining the
product of two arbitrary reduced words and showing that this product
is associative. 

5. Suppose we have a group $G$ and several elements $\{g_i\}$ in
$G$. Show that there exists a normal subgroup $K$ of $G$ containing
all the $g_i$'s, such that $K$ is contained in every normal subgroup
that contains the $g_i$'s. This is called the {\it normal subgroup
generated by the $g_i$'s}.  Suppose that $H$ is another group and $f:
G \to H$ is a homomorphism. Let $p: G \to G/K$ be the obvious
projection.  Show that $f$ lifts to a map $\hat f: G/K \to H$ if and only
if every $f(g_i)$ is the identity. (By ``lift to a map'' I mean that
$\hat f$ exists such that $f = \hat f \circ p$.)

6. Let $G_1$ and $G_2$ be two groups, and let $H$ inject in both of them
via injections $i_1$ and $i_2$. The
{\it amalgamated free product of $G_1$ and $G_2$ over $H$}, denoted
$G_1 *_H G_2$, is the quotient of $G_1*G_2$ by the normal subgroup generated
by $\{i_1(h) i_2(h)^{-1}\}$, where $h$ ranges over $H$. It's like 
the free product of $G_1$ and $G_2$, only with $i_1(H)$ identified with 
$i_2(H)$. 
Let $\phi_1: G_1 \to G_1*_H G_2$ and $\phi_2: G_2 \to G_1 *_H G_2$ be the
obvious injections. 

Suppose we have a group $B$ and maps $\psi_1: G_1 \to B$ and $\psi_2: G_2 \to B$
such that $\psi_1 \circ i_1 = \psi_2 \circ i_2$. Show that there exists a 
unique homomorphism $f: G_1 *_H G_2 \to B$ such that $\psi_1 = f \circ \phi_1$
and $\psi_2 = f \circ \phi_2$. 

7. Define a relevant category for which $G_1 *_H G_2$ is the universal object. 
In other words, express the conclusion of problem 6 as a universal property. 
  
8. Repeat problems 6 and 7, only with $i_1$ and $i_2$ no longer
assumed to be injective. For this problem, they're just group
homomorphisms.  Note that $\phi_1$ and $\phi_2$ are no longer necessarily 
injective.
As far as I know, there is no standard term for the thing that
replaces $G_1*_H G_2$ -- let's call it the generalized amalgamated
product and denote it $G_1 \tilde *_H G_2$. (Some authors do use ``amalgamated
free product'' to mean ``generalized amalgamated free product'', and they
denote it $G_1 *_H G_2$, but others reserve the term for the case where $H$
is a subgroup of $G_1$ and also a subgroup of $G_2$.) 

{\it The construction of Problem 8 is extremely important, thanks to
van Kampen's theorem (aka the Seifert-van Kampen theorem), which says
that the the fundamental group of the union of two open sets $U$ and
$V$ is the generalized amalgamated free product $\pi_1(U,x_0)$ and
$\pi_1(V,x_0)$ over $\pi_1(U \cap V, x_0)$. where $U \cap V$ is
assumed path-connected, $x_0 \in U \cap V$, and the maps $i_1$ and
$i_2$ are induced from the inclusions $U \cap V \to U$ and $U \cap V
\to V$. We're going to spend a lot of time trying to understand the topology
of this in Chapter 4. Chapter 3 is all about setting up the necessary algebra.}

9. A free abelian group on 3 generators cannot inject in a free abelian group
on two generators, but non-Abelian free groups are different. 
Let $F_2$ be the free group on two generators $a,b$, and let $F_3$ be
the free group on generators $s_1, s_2, s_3$. Consider the map
$f: F_3 \to F_2$ defined by $f(s_1)=ab$, $f(s_2)=a^2b^2$, $f(s_3)=a^3b^3$. Show
that $f$ is an injection.

10. Generalize the construction of problem 9 to construct a subgroup of $F_2$
that is not finitely generated.  (You can take as given the fact that
a free group on infinitely many generators is not finitely generated.)

\bye
