\chapter{Simplicial $\Z$-Homology: Getting Oriented}

\begin{exercise}
Check that this boundary map is well-defined: it does not depend on the oriented representative chosen for the definition.
\end{exercise}

\oneperpage

\begin{exercise}
Find the boundary of the oriented $2$-simplex $\tau=[v_0v_1v_2]$ and
the boundary of the oriented $3$-simplex $\sigma=[w_0w_1w_2w_3]$.
Repeat the procedure for $-\tau$ and $-\sigma$. What is the
relationship between the boundary of $\tau$ and the boundary of
$-\tau$? What is the relationship between the boundary of $\sigma$
and the boundary of $-\sigma$?
\end{exercise}

\oneperpage

\begin{thm}
For any $n$-simplex $\sigma$
$$
\partial(-\sigma)=-\partial(\sigma).
$$
\end{thm}

\oneperpage

\begin{thm}
For all $n\geq 0$, 
$$\partial_{n}\circ\partial_{n+1}=0.$$
\end{thm}

\oneperpage

\begin{thm}
For any simplicial complex $K$, both $Z_n(K)$ and $B_n(K)$ are subgroups of $C_n(K)$, and $B_n(K)\subset Z_n(K)$.
\end{thm}

\oneperpage

\begin{exercise}
Create a triangulation of a M\"obius band such that the central
circle forms a $1$-cycle $\gamma$. Show that the M\"obius band's boundary $1$-cycle $\alpha$
is equivalent to either $2\gamma$ or $-2\gamma$ (depending on the orientation
you give the two cycles).
\end{exercise}

\oneperpage

\begin{thm} 
For a finite simplicial complex $K$, $H_n(K)$ is a finitely generated abelian group.
\end{thm}

\oneperpage

\begin{thm}
If $K$ is a connected simplicial complex, then $H_0(K)$ is isomorphic to $\Z$. If $K$
has $r$ connected components, then $H_0(K)$ is a free abelian group of rank
$r$.
\end{thm}

\oneperpage

\begin{thm}
If $K$ is a one-point space, $\HH_{n}(K) \cong 0$ for $n \geq 0$ and $\HH_{0}(K) \cong \Z$.
\end{thm}

\oneperpage

\begin{thm}
Let $x$ see a complex $K$, and let $c \in C_{n}(K)$ be a chain. Then
$$
\partial \cone_{x} (c) + \cone_{x} (\partial c)  = c.
$$
\end{thm}

\oneperpage

\begin{cor}
For any complex $K$ and $x$ seeing $K$, the complex $x*K$ is acyclic.
\end{cor}

\oneperpage

\begin{thm}
The complex $K$ consisting of an $n$-simplex together with all its faces is acyclic.
\end{thm}

\oneperpage

\begin{thm}
Let $f:K\to L$ be a simplicial map, and let $f_\#$ be the
induced map $f_\#:C_n(K)\to C_n(L)$.  Then for any chain $c\in
C_n(K)$, $\partial(f_\#(c))=f_\#(\partial(c))$. In other words,
the diagram:
$$
\begin{CD}
C_n(K)   @>{f_\#}>>  C_n(L)    \\
@V{\partial}VV      @VV{\partial}V \\
C_{n-1}(K)   @>{f_\#}>>   C_{n-1}(L)
\end{CD}
$$
commutes.
\end{thm}

\oneperpage

\begin{thm}
Let $f:K\to L$ be a simplicial map. 
Then the induced homomorphism $f_*:H_n(K)\to H_n(L)$ is a well-defined homomorphism.
\end{thm}

\oneperpage

\begin{exercise}
Check that $C_{n}(K,K')$ is a free abelian group.
\end{exercise}

\oneperpage

\begin{thm}
\label{relative-boundary}
There is a boundary map 
$$\partial_{n}: C_{n}(K,K') \to C_{n-1}(K,K')$$
such that $\partial_{n} \circ \partial_{n+1} = 0$ for all $n \geq 0$.
\end{thm}

\oneperpage

\begin{exercise}
Check that if $K'= \emptyset$, the empty set, then $H_{n}(K,K')= H_{n}(K)$ for all $n$, the usual homology groups.
\end{exercise}

\oneperpage

\begin{exercise}
Show that $\tilde H_{n}(K) \cong H_{n}(K)$ for $n>0$ and $H_{0}(K) \cong \tilde H_{0}(K) \oplus \Z$. 
\end{exercise}

\oneperpage

\begin{exercise}
Let $K$ be the complex consisting of a triangle and all its faces.  Determine $H_{n}(K,K')$ for all $n \geq 0$.
\end{exercise}

\oneperpage

\begin{exercise}
Let $K$ be a triangulation of an annulus, and let $K'$ be the subcomplex consisting of the inner and outer edges of the annulus.  Find a relative $1$-cycle in $C_{1}(K,K')$  that is not a relative $1$-boundary.
\end{exercise}

\oneperpage

\begin{exercise}
Let $K$ be a triangulation of a M\"obius band, and let $K'$ be its boundary.  Determine $H_{n}(K,K')$ for $n\geq 0$.
\end{exercise}

\oneperpage

\begin{thm}[Excision]
\label{simplicial-excision}
Suppose $K'$ is a subcomplex of $K$.  Remove an open set $U$ from $K'$ such that what remains is a subcomplex $L'$ of $K'$, and remove $U$ from $K$ so that what remains is a subcomplex $L$ of $K$.  Then
$$
H_{n}(L,L') \cong H_{n}(K,K').
$$
\end{thm}

\oneperpage

\begin{thm}
Given a simplicial map $f:(K,K') \to (L,L')$ there is 
an associated chain map $f_{\#}: C_{n}(K,K') \to C_{n}(L,L')$ 
and induced homomorphism $f_{*}: H_{n}(K,K') \to H_{n}(L,L')$.
\end{thm}

\oneperpage

\begin{exercise}
There are natural maps between chain groups:
$$
\begin{CD}
C_{n}(K')   @>{i}>> C_{n}(K) @>{\pi}>> C_{n}(K,K')   \\
\end{CD}
$$
What are the maps $i$ and $\pi$, and what do you notice about them and their relationship with each other?
\end{exercise}

\oneperpage

\begin{thm}
The boundary map $\partial: C_{n}(K) \to C_{n-1}(K')$ induces a well-defined map 
$$\partial_{*}: H_{n}(K,K') \to H_{n-1}(K').$$
\end{thm}

\oneperpage

\begin{thm}[Long Exact Sequence of a Pair]
If $K'$ is a subcomplex of a simplicial complex $K$, then there is a long exact sequence:
$$
\begin{CD}
\cdots @>{\partial_{*}}>> H_{n}(K')   @>{i_{*}}>> H_{n}(K) @>{\pi_{*}}>> H_{n}(K,K') @>{\partial_{*}}>>  H_{n-1}(K') @>{i_{*}}>> \cdots
\end{CD}
$$
\end{thm}

\oneperpage

\begin{thm}[Zig-Zag Lemma]
Suppose $\mathcal{C} = \{ C_{n}, \partial_{n}^{C}\}, \mathcal{D} = \{ D_{n}, \partial_{n}^{D}\}, \mathcal{E} = \{ E_{n}, \partial_{n}^{E}\}$ are chain complexes,
and $\phi:\mathcal{C}\to\mathcal{D}$ and $\psi:\mathcal{D}\to\mathcal{E}$ are chain maps such that 
$$
\begin{CD}
0 @>>> \mathcal{C}  @>{\phi}>> \mathcal{D} @>{\psi}>> \mathcal{E} @>>> 0
\end{CD}
$$
is a short exact sequence of chain complexes.  Then there is a long exact sequence:
$$
\begin{CD}
\cdots @>{\partial_{*}}>> H_{n}(\mathcal{C})   @>{\phi_{*}}>> H_{n}(\mathcal{D}) @>{\psi_{*}}>> H_{n}(\mathcal{E}) @>{\partial_{*}}>>  H_{n-1}(\mathcal{C}) @>{i_{*}}>> \cdots
\end{CD}
$$
where $\partial_{*}$ is induced by $\partial^{D}$.
\end{thm}

\oneperpage

\begin{thm}
\label{axiom:naturality}
Given the commutative diagram of chain maps $\alpha, \beta, \gamma$ between the chain complexes of two short exact sequences:
$$
\begin{CD}
0 @>>> \mathcal{C}   @>{i}>> \mathcal{D} @>{\pi}>> \mathcal{E} @>>> 0 \\
     @. @V{\alpha}VV       @V{\beta}VV      @V{\gamma}VV  \\
0 @>>> \mathcal{C'}   @>{i}>> \mathcal{D'} @>{\pi}>> \mathcal{E'} @>>> 0 \\
\end{CD}
$$
there are corresponding induced homomorphisms between the associated long exact sequences, such that the following diagram is commutative:
$$
\begin{CD}
\cdots @>{\partial_{*}}>> H_{n}(\mathcal{C})   @>{\phi_{*}}>> H_{n}(\mathcal{D}) @>{\psi_{*}}>> H_{n}(\mathcal{E}) @>{\partial_{*}}>>  H_{n-1}(\mathcal{C}) @>{i_{*}}>> \cdots \\
   @.  @V{\alpha_{*}}VV       @V{\beta_{*}}VV      @V{\gamma_{*}}VV  @V{\alpha_{*}}VV  \\
\cdots @>{\partial_{*}}>> H_{n}(\mathcal{C'})   @>{\phi_{*}}>> H_{n}(\mathcal{D'}) @>{\psi_{*}}>> H_{n}(\mathcal{E'}) @>{\partial_{*}}>>  H_{n-1}(\mathcal{C'}) @>{i_{*}}>> \cdots \\
\end{CD}
$$
\end{thm}

\oneperpage

\begin{lem}[The Five Lemma]
Consider the following commutative diagram of groups and homomorphisms, where the rows are exact.
$$
\begin{CD}
A @>{q}>> B  @>{r}>> C @>{s}>> D @>{t}>>  E  \\
   @V{\alpha}VV @V{\beta}VV       @V{\gamma}VV      @V{\delta}VV  @V{\epsilon}VV  \\
A' @>{q'}>> B'  @>{r'}>> C' @>{s'}>> D' @>{t'}>>  E'  \\
\end{CD}
$$
If the rows are exact and $\alpha, \beta, \delta,\epsilon$ are isomorphisms, then $\gamma$ is also an isomorphism.
\end{lem}

\oneperpage

\begin{exercise}
In the proof of the Five Lemma, not all of $\alpha, \beta, \delta,\epsilon$ are required to be isomorphisms for the conclusion to still hold.  Which isomorphisms can be relaxed?
\end{exercise}

\oneperpage

\begin{exercise}[The Snake Lemma]
Consider the following commutative diagram where the rows are short exact sequences.
$$
\begin{CD}
0 @>{}>> A  @>{}>> B @>{}>> C @>{}>>  0  \\
 @.  @V{\alpha}VV @V{\beta}VV       @V{\gamma}VV    \\
0 @>{}>> A' @>{}>> B' @>{}>> C' @>{}>>  0  \\
\end{CD}
$$
Show there is an exact sequence
$$
0 \to \kernel \alpha \to \kernel \beta \to \kernel \gamma 
\to \cokernel \alpha \to \cokernel \beta \to \cokernel \gamma \to 0
$$
where $\cokernel$ stands for the cokernel of a homomorphism: the quotient of its codomain by its image.
\end{exercise}

\oneperpage

\begin{cor}[Long Exact Sequence of a Pair]
If $K'$ is a subcomplex of a simplicial complex $K$, then there is a long exact sequence:
$$
\begin{CD}
\cdots @>{\partial_{*}}>> H_{n}(K')   @>{i_{*}}>> H_{n}(K) @>{\pi_{*}}>> H_{n}(K,K') @>{\partial_{*}}>>  H_{n-1}(K') @>{i_{*}}>> \cdots
\end{CD}
$$
where the maps are induced by the inclusion maps $i:K' \to K$ and $\pi: (K,\emptyset) \to (K,K')$ and the boundary map $\partial: C_{n}(X) \to C_{n-1}(X)$.
\end{cor}

\oneperpage

\begin{thm}
Given a simplicial map $f:(K,K') \to (L,L')$, there is chain map between the long exact sequences:
$$
\begin{CD}
\cdots @>{\partial_{*}}>> H_{n}(K')   @>{i_{*}}>> H_{n}(K) @>{\pi_{*}}>> H_{n}(K,K') @>{\partial_{*}}>>  H_{n-1}(K') @>{i_{*}}>> \cdots \\
  @. @V{f_{*}}VV @V{f_{*}}VV @V{f_{*}}VV @V{f_{*}}VV   \\
\cdots @>{\partial_{*}}>> H_{n}(L')   @>{i_{*}}>> H_{n}(L) @>{\pi_{*}}>> H_{n}(L,L') @>{\partial_{*}}>>  H_{n-1}(L') @>{i_{*}}>> \cdots \\
\end{CD}
$$
\end{thm}

\oneperpage

\begin{thm}[Mayer-Vietoris]
\index{Mayer-Vietoris Theorem!$\Z$} Let $K$ be a finite simplicial
complex and $A$ and $B$ be subcomplexes such that  $K=A\cup B$.  
Then there is a long exact sequence 
$$
\begin{CD}
\cdots @>{\delta}>> H_n(A\cap B)   @>{\phi_{*}}>> H_n(A)\oplus H_n(B) @>{\psi_{*}}>> H_n(K) @>{\partial_{*}}>>  H_{n-1}(A\cap B) @>{\phi}>> \cdots
\end{CD}
$$
\end{thm}

\oneperpage

\begin{exercise}
Compute the $\Z$-homology of the Klein bottle, and compare it to the the $\Z_{2}$-homology of the Klein bottle.
\end{exercise}

\oneperpage

\begin{exercise}
Compute the $\Z$-homology of every compact, triangulated $2$-manifold.
\end{exercise}

\oneperpage

\begin{thm}
Let $K$ be a finite, connected simplicial complex. Then
$$
H_1(K;\Z)\simeq (\pi_1(K))/[\pi_1(K),\pi_1(K)],
$$
that is, the first
homology group of $K$ is isomorphic to the abelianization of the
fundamental group of $K$.
\end{thm}

\oneperpage

\begin{lem}  
If $f:\Sph^n\to\Sph^n$ is continuous, then $\deg f$ is
well-defined. That is, it does not depend on the way in which we
identify $H_n(\Sph^n)$ with $\Z$.
\end{lem}

\oneperpage

\begin{thm} 
Let $f,g:\Sph^n\to\Sph^n$ be continuous maps.  
\begin{enumerate}
\item If $f$ and $g$ are homotopic, they have the same degree.
\item $\deg(f\circ g)=(\deg f)\cdot(\deg g)$\end{enumerate}
\end{thm}

\oneperpage

\begin{thm} 
The identity map on $\Sph^n$ has degree 1.  The
antipodal map has degree $(-1)^{n+1}$.
\end{thm}

\oneperpage

\begin{thm}[Hairy Ball Theorem]
\index{Hairy Ball Theorem}
There exists a non-vanishing vector field on $\Sph^n$ if and only if $n$ is odd.
\end{thm}

\oneperpage

\begin{exercise}
In the definition above and using a little linear algebra, show that $\tr(h_{free})$ does not depend on the choice of basis for $G_{free}$.
\end{exercise}

\oneperpage

\begin{exercise}
Construct a simple example of a map homotopic to the identity map on the triangulated circle whose induced chain map does not have the same trace as the identity chain map.
\end{exercise}

\oneperpage

\begin{thm}
Suppose $0 \to A \to B \to C \to 0$ is a short exact sequence of finitely generated abelian groups 
and $f_{A}:A\to A$ and $f_{B}:B \to B$ are homomorphisms such that $i \circ f_{A} = f_{B} \circ i$.
Then there is an induced homomorphism $f_{C}: C \to C$ that makes the following diagram commutative:
\begin{center}
$\begin{CD}
0 @>>> A @>i>> B @>\pi>> C @>>> 0\\
@. @Vf_{A}VV @Vf_{B}VV @Vf_{C}VV @.\\
0 @>>> A @>i>> B @>\pi>> C @>>> 0
\end{CD}$
\end{center}
Moreover,
$$
\tr(f_{B}) = \tr(f_{A}) + \tr(f_{C})
$$
\end{thm}

\oneperpage

\begin{thm}(The Hopf Trace Formula)
Let $K$ be a finite simplicial complex and let $f:K \to K$ be a simplicial map.  Then 
$$
\sum (-1)^i \tr(f_{\#n}) = \sum (-1)^i \tr(f_{*n}).
$$
\end{thm}

\oneperpage

\begin{thm} (Lefschetz Fixed Point Theorem)
Let $f:|K| \to |K|$ be a continuous map on a simplicial complex $K$.
If $\Lambda(f) \neq 0$, then $f$ has a fixed point.
\end{thm}

\oneperpage

\begin{exercise}
Compute the Lefschetz number of the ``mirror-reversing'' self-map of a circle: $g:\Sph^1 \to \Sph^1$ defined by 
$g(x,y)=(-x,y)$ viewing $\Sph^1$ as a subset of $\R^2$.  
Argue that any reversing map of a circle must have a fixed point.
\end{exercise}

\oneperpage

