\documentclass[12pt]{amsart}
\title{Lecture 14}
\author{}

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\begin{document}
\maketitle

\PP{14.0}{Category of complexes}
%------------------------------
Let $R$ be a ring and let $\CalA=\Rmod{R}$ be the abelian
category of left $R$--modules. All the constructions and results of
these notes work for {\em any abelian category}. However, to keep
things concrete (i.e, so that we talk about ``elements" of an object), 
we are going to narrow our attention to
$\CalA=\Rmod{R}$.\\

\begin{defn}
The category of cochain complexes over $\CalA$, 
denoted by $\CoComp{\CalA}$, consists of following objects.
\[
(C^\bullet,d^{\bullet}): \qquad \cdots\morph{d^{n-1}} 
C^n\morph{d^n}C^{n+1}\morph{d^{n+1}}\cdots 
\]
where,
\begin{itemize}
\item $C^n\in\CalA$ for every $n\in\Z$.\\

\item $d^{n+1}\circ d^n = 0$ for every $n\in \Z$.\\
\end{itemize}

Morphisms from an object $(C^{\bullet},d^{\bullet}_C)$
to $(D^{\bullet},d^{\bullet}_D)$ consist of $\{f^n:C^n\to D^n\}_{n\in\Z}$
such that $d^n_D\circ f^n = f^{n+1}\circ d^n_C$.\\
\end{defn}

\noindent {\bf Remarks.} \\
\begin{enumerate}
\item The terminology is again borrowed from algebraic topology. For chain
complexes, we put numbers in the subscript and they decrease from left
to right. Their category is often denoted by $\Comp{\CalA}$. There
is no conceptual difference between the two setting - merely notational.\\

\item In some texts, it is assumed that cochain complexes are indexed
by $\Z_{\geq 0}$ (i.e, they are bounded from left). When we get to injective
resolutions, this will be the case, but for now our cochains are
unbounded on both sides.\\

\item {\em Abuse of notation.}
The morphisms $d^n$ in the definition are called {\em differentials}.
It is customary to drop the superscripts and write $d\circ d=0$, if
the index of the domain is implicitly clear. Similarly, we just say
``let $X^\bullet$ be a complex", instead of $(X^{\bullet},d^\bullet_X)$
to save some space.\\
\end{enumerate}

\PP{14.1}{Category $\CoComp{\CalA}$ continued}
%--------------------------------------------

\begin{prop}
The category $\CoComp{\CalA}$ is abelian.
\end{prop}

This proposition is proved using an argument similar to the one
in \S 9.5. We merely record some of the observations made there, in the
context of $\CoComp{\CalA}$.\\

\noindent For two complexes $X^\bullet$ and $Y^\bullet$ in $\CoComp{\CalA}$,
the set of homomorphisms
\[
\Hom_{\CoComp{\CalA}}(X^{\bullet},Y^{\bullet}) \subset
\prod_{n\in\Z} \Hom_{\CalA}(X^n,Y^n),
\]

is defined by the condition $d^{n}_Y\circ \alpha^n = \alpha^{n+1}\circ
d^n_X$. So, it is naturally a subgroup.\\

\noindent For a morphism $\alpha^{\bullet} : X^{\bullet}\to Y^{\bullet}$,
we define its kernel, image etc. component--wise. For instance, 
$K^{\bullet} = \Ker(\alpha^{\bullet})$ is the following complex.
\begin{itemize}
\item $K^n = \Ker(\alpha^n : X^n \to Y^n)$.\\

\item By the commutativity of the following diagram, the differential
of $X^\bullet$ restricts to $K^{\bullet}$:
\[
\xy
(0,20)*{K^n}="k1"; (30,20)*{X^n}="x1"; (60,20)*{Y^n}="y1";
{\ar "k1"; "x1"}; {\ar^{\alpha^n} "x1"; "y1"};
(0,0)*{K^{n+1}}="k2"; (30,0)*{X^{n+1}}="x2"; (60,0)*{Y^{n+1}}="y2";
{\ar "k2"; "x2"}; {\ar^{\alpha^{n+1}} "x2"; "y2"};
{\ar_{d^n_X} "x1"; "x2"};
{\ar^{d^n_Y} "y1"; "y2"};
{\ar@{-->} "k1"; "k2"};
\endxy
\]

This type of argument was implicitly used in \S 9.6, but let us
record it as a lemma for future use:

\begin{lem}
Let $\CalC$ be an abelian category, and assume that we have
a commutative diagram in $\CalC$:
\[
\xy
(0,20)*{A_1}="a1"; (30,20)*{A_2}="a2";
{\ar^a "a1"; "a2"};
(0,0)*{B_1}="b1"; (30,0)*{B_2}="b2";
{\ar_b "b1"; "b2"};
{\ar_f "a1"; "b1"};
{\ar^g "a2"; "b2"};
\endxy
\]
Then, we get a morphism $f':\Ker(a)\to \Ker(b)$ and a morphism
$g':\Coker(a)\to \Coker(b)$ making the following diagram commute.
\[
\xy
(-30,20)*{\Ker(a)}="ka"; (0,20)*{A_1}="a1"; (30,20)*{A_2}="a2";
(60,20)*{\Coker(a)}="ca";
{\ar "ka"; "a1"}; {\ar "a2"; "ca"};
{\ar^a "a1"; "a2"}; 
(-30,0)*{\Ker(b)}="kb"; (0,0)*{B_1}="b1"; (30,0)*{B_2}="b2";
(60,0)*{\Coker(b)}="cb";
{\ar "kb"; "b1"}; {\ar "b2"; "cb"};
{\ar@{-->}_{f'} "ka"; "kb"};
{\ar@{-->}^{g'} "ca"; "cb"};
{\ar_b "b1"; "b2"};
{\ar_f "a1"; "b1"};
{\ar^g "a2"; "b2"};
\endxy
\]
\end{lem}

\begin{pf}
Let us see how $f'$ arises. Consider the composition 
\[
\Ker(a) \to A_1 \morph{f} B_1 \morph{b} B_2.
\]
This composition is zero, since it is same as the following,
since $b\circ f = g\circ a$, which is zero by definition
of the kernel.
\[
\Ker(a)\to A_1 \morph{a} A_2 \morph{g} B_2.
\]
Hence, by the definition of $\Ker(b)$, the composition $\Ker(a)\to A_1\to B_1$ factors
through $\Ker(b)$ giving rise to $f':\Ker(a)\to\Ker(b)$ making the left
square in the diagram above commute. Same argument works for the cokernels.
\end{pf}
\end{itemize}
\ \\

\noindent Direct sum of two complexes $X^{\bullet}$ and $Y^\bullet$ is
also defined term-wise: $(X^{\bullet}\oplus Y^\bullet)^n := X^n \oplus Y^n$,
with differential $(d^n_X, d^n_Y)$.\\

\PP{14.2}{Cocycles, coboundaries and cohomology}
%-----------------------------------------------
Given a complex $X^\bullet$ in $\CoComp{\CalA}$, and an integer $n\in\Z$,
we define:

\[
\mathcal{Z}^n(X^{\bullet}) := \Ker(d^n : X^n\to X^{n+1}) \qquad
\text{$n$--cocyles}.
\]

\[
\mathcal{B}^n(X^{\bullet}) := \Im(d^{n-1}:X^{n-1}\to X^n) \qquad
\text{$n$--coboundaries}.
\]

Note that since $d\circ d=0$, we have:
$\ds\mathcal{B}^n(X^\bullet) \subset \mathcal{Z}^n(X^\bullet)$.
The $n^{\scriptstyle{\text{th}}}$ {\em cohomology} of $X^\bullet$ is then
defined as the quotient:

\[
\boxed{
H^n(X^\bullet) = \mathcal{Z}^n(X^\bullet)/\mathcal{B}^n(X^\bullet)
= \Ker(d^n)/\Im(d^{n-1})
}
\]

\begin{prop}
For every $n\in\Z$, $\mathcal{Z}^n, \mathcal{B}^n, H^n$ are additive, covariant
functors $\CoComp{\CalA} \to \CalA$.
\end{prop}

\begin{pf}
Let us first see it for the functor of cocycles. We know how to
define it on objects. For a morphism $\alpha^\bullet:X^\bullet\to
Y^\bullet$, we have the following commutative diagram (here $n\in\Z$
is fixed).

\[
\xy
(0,20)*{\mathcal{Z}^n(X^\bullet)}="z1"; (30,20)*{X^n}="x1"; (60,20)*{X^{n+1}}="x2";
{\ar "z1"; "x1"}; {\ar^{d^n} "x1"; "x2"};
(0,0)*{\mathcal{Z}^n(Y^\bullet)}="z2"; (30,0)*{Y^n}="y1"; (60,0)*{Y^{n+1}}="y2";
{\ar "z2"; "y1"}; {\ar^{d^n} "y1"; "y2"};
{\ar_{\alpha^n} "x1"; "y1"};
{\ar^{\alpha^{n+1}} "x2"; "y2"};
{\ar@{-->} "z1"; "z2"};
\endxy
\]

This follows from the lemma in \S 14.1 above. The argument using
elements of $X^n$ goes as follows, to show that
$\alpha^n$ restricts to a morphism $\mathcal{Z}^n(X^\bullet)
\to\mathcal{Z}^n(Y^\bullet)$. Let $x\in\mathcal{Z}^n(X^\bullet)$, that is,
$d^n(x) =0$. Thus, $d^n(\alpha^n(x)) = \alpha^{n+1}(d^n(x)) = 0$ proving
that $\alpha^n(x)\in\mathcal{Z}^n(Y^\bullet)$.\\ 

This allows us to define
$\mathcal{Z}^n(\alpha^\bullet)$ as the restriction of $\alpha^n$
to $n$--cocyles, which we continue to denote by $\alpha^n$. 
This description makes it clear that:

\[
\mathcal{Z}^n(\Id_{X^\bullet}) = \Id_{\mathcal{Z}^n(X^{\bullet})},\qquad
\mathcal{Z}^n(\beta^\bullet\circ\alpha^\bullet) = 
\mathcal{Z}^n(\beta^\bullet)\circ\mathcal{\alpha^\bullet}.
\]

That is, $\mathcal{Z}^n$ is a functor. It is also easily verified 
that it is additive on morphisms, since it is merely a restriction
of the domain of the morphisms, which respects addition.\\

The same argument as before works for $\mathcal{B}^n$. Namely, we have
the following diagram that defines $\mathcal{B}^n(\alpha^\bullet)$
as the restriction of $\alpha^n$ to $\Im(d_X^{n-1})\subset X^n$.\\

\[
\xy
(-30,20)*{X^{n-1}}="x"; (0,20)*{\Im(d^{n-1}_X)}="b1"; (30,20)*{X^n}="x1"; 
{\ar^{d^{n-1}_X} "x"; "b1"}; {\ar "b1"; "x1"};
(-30,0)*{Y^{n-1}}="y"; (0,0)*{\Im(d^{n-1}_Y)}="b2"; (30,0)*{Y^n}="y1"; 
{\ar^{d^{n-1}_Y} "y"; "b2"}; {\ar "b2"; "y1"};
{\ar_{\alpha^n} "x1"; "y1"};
{\ar^{\alpha^{n-1}} "x"; "y"};
{\ar@{-->} "b1"; "b2"};
\endxy
\]

(apply Lemma 14.1 twice, first to get a map from $\Coker{d_X^{n-1}}\to
\Coker{d_Y^{n-1}}$, and then with $A_1=X^n\to A_2=\Coker{d^{n-1}_X}$,
$B_1=Y^n\to B_2=\Coker(d^{n-1}_Y)$.)\\

Finally $H^n(\alpha^\bullet)$ is defined via the following diagram,
where the rows are short exact sequences (by definition).

\[
\xy
(-20,20)*{0}="i1";
(0,20)*{\mathcal{B}^n(X^\bullet)}="bx";
(30,20)*{\mathcal{Z}^n(X^\bullet)}="zx";
(60,20)*{H^n(X^\bullet)}="hx"; (80,20)*{0}="t1";
{\ar "i1"; "bx"}; {\ar^{i_1} "bx"; "zx"}; {\ar^{p_1} "zx"; "hx"}; {\ar "hx"; "t1"};
(-20,0)*{0}="i2";
(0,0)*{\mathcal{B}^n(Y^\bullet)}="by";
(30,0)*{\mathcal{Z}^n(Y^\bullet)}="zy";
(60,0)*{H^n(Y^\bullet)}="hy"; (80,0)*{0}="t2";
{\ar "i2"; "by"}; {\ar_{i_2} "by"; "zy"}; {\ar_{p_2} "zy"; "hy"}; {\ar "hy"; "t2"};
{\ar_{\alpha^n} "bx"; "by"};
{\ar^{\alpha^n} "zx"; "zy"};
{\ar@{-->} "hx"; "hy"};
\endxy
\]

This assertion again follows from the lemma in \S 14.1 above.
A direct argument with ``elements" is given here, to see clearly how
the definition of the dashed arrow works.\\

For $x\in H^n(X^\bullet)$, choose $\wt{x}\in \mathcal{Z}^n(X^\bullet)$ such that $p_1(\wt{x})=x$.
Let $\wt{y} = \alpha^n(\wt{x})$ and map $x\mapsto p_2(\wt{y})
\in H^n(Y^\bullet)$. To see that it is well--defined, if we chose another
$\wt{x}'$, then $p_1(\wt{x}-\wt{x}')=0$ implying that $\wt{x} = 
\wt{x}' + i_1(b)$ for some $b\in \mathcal{B}^n(X^\bullet)$. Thus,
$\wt{y} = \wt{y}'+i_2(\alpha^n(b))$. As $p_2\circ i_2=0$, we get
that $p_2(\wt{y}) = p_2(\wt{y}')$. Note that we didn't need $i_1$
and $i_2$ to be injective here.\\

\noindent {\bf Remark.} With the ``concrete proof", it remains
to show that the resulting map is $R$--linear, which is left
to the reader. With the ``hands--off proof", we lose a bit of
clarity, but the gain is that we don't have
to check anything.
\end{pf}
\ \\

\noindent {\bf Exercise.} Show that $\mathcal{Z}^n$ is left exact.\\


\PP{14.3}{Null--homotopic morphisms}
%-----------------------------------
Let $\alpha^\bullet : X^\bullet \to Y^\bullet$ be a morphism of two
complexes. We say $\alpha^{\bullet}$ is {\em null--homotopic} if
there exists morphisms $s^n : X^n \to Y^{n-1}$ such that

\[
\boxed{\alpha = ds+sd}
\qquad
\text{(with appropriate superscripts - see the diagram
below)}
\] 

\[
\xy
(20,20)*{X^n}="x1"; (40,20)*{X^{n+1}}="x2";
(0,0)*{Y^{n-1}}="y1"; (20,0)*{Y^n}="y2";
{\ar^{d^n_X} "x1"; "x2"};
{\ar_{d^{n-1}_Y} "y1"; "y2"};
{\ar_{s^n} "x1"; "y1"};
{\ar^{s^{n+1}} "x2"; "y2"};
{\ar^{\alpha^n} "x1"; "y2"};
\endxy
\]

As usual, two morphisms $\alpha^\bullet, \beta^\bullet:X^\bullet
\to Y^\bullet$ are said to be {\em homotopic} if $\alpha^\bullet-\beta^\bullet$
is null--homotopic. In symbols, we write $\alpha^\bullet\sim\beta^{\bullet}$
to mean $\alpha^\bullet$ and $\beta^\bullet$ are homotopic.\\

\begin{prop}
The set of null--homotopic morphisms
$\Hom^0(X^\bullet,Y^\bullet) \subset \Hom(X^\bullet,Y^\bullet)$
is a subgroup. Moreover,

\begin{enumerate}
\item $\alpha^\bullet\in\Hom^0(X^\bullet,Y^\bullet)$ implies that
$H^n(\alpha^\bullet)=0$ for every $n\in\Z$.\\

\item For every $\ds
\alpha^\bullet\in\Hom^0(X^\bullet,Y^\bullet)$ and
$\ds\beta^\bullet\in\Hom(Y^\bullet,Z^\bullet)$, $\ds
\gamma^\bullet\in\Hom(W^\bullet,X^\bullet)$, we have
\[
\beta^\bullet\circ \alpha^\bullet\in\Hom^0(X^\bullet,Z^\bullet)\qquad
\text{and}\qquad
\alpha^\bullet\circ \gamma^\bullet\in\Hom^0(W^\bullet,Y^\bullet)
\]
(often said in words as ``null--homotopic morphisms form an ideal").
\end{enumerate}
\end{prop}

\begin{pf}
(I am going to drop the superscripts for the easy of reading).\\

\noindent {\em Null--homotopic morphisms form a subgroup.}
Clearly zero morphism is null--homotopic.
Assume that $\alpha,\alpha'\sim 0$ are two null--homotopic morphisms. 
That is, we have $r^n,s^n:X^n\to Y^{n-1}$ (for every $n\in\Z$) such that 
$\alpha=dr+rd$ and $\alpha' = ds+sd$. Then:
\[
\alpha - \alpha' = d(r-s) + (r-s)d\ ,
\]
proving that it is also null--homotopic.\\

\noindent $\alpha\sim 0 \Rightarrow H^n(\alpha)=0$. Fix $n\in\Z$ and
consider the commutative diagram sketched above in the definition of
null--homotopic morphisms. That is, $\alpha^n = d^{n-1}s^n + s^{n+1}d^n$.
Now, if $x\in \Ker(d^n_X)$, then $\alpha^n(x) = d^{n-1}(s^n(x))
\in \Im(d^{n-1}_Y)$, proving that at the level of cohomology, we
get $H^n(\alpha)=0$.\\

\noindent {\em Null--homotopic morphisms form an ideal}.
Let $\alpha:X\to Y$ be null--homotopic, with homotopy $s$ so that
$\alpha = ds+sd$. Let $\beta:Y\to Z$ be arbitrary. We have
\[
\beta\circ \alpha = \beta ds+ \beta s d = d\beta s+ \beta sd
= dr + rd\ ,
\]
where $r^n = \beta^{n-1}\circ s^n : X^n \to Z^{n-1}$. In the equalities
above, we have used the fact that $\beta$ being a morphism, commutes
with $d$. Hence $\beta\alpha\sim 0$ with homotopy $\beta\circ s$.
\end{pf}
\end{document}
