\documentclass[12pt]{amsart}
\title{Lecture 17}
\author{}

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\newtheorem*{rem}{Remark}
\newtheorem*{rems}{Remarks}
\newtheorem*{ex}{Example}

\begin{document}
\maketitle

\PP{17.0}{Injective and projective objects}
%------------------------------------------
Recall that in the last lecture we defined injective and projective
objects in $\CalA=\Rmod{R}$. Our definition was phrased for any abelian
category, and is given as follows.\\

\noindent {\em Injective objects}. An object $Q\in\CalA$ is called
an injective object, if it satisfies (one of the) following
equivalent conditions.\\

\begin{itemize}
\item For any injective morphism $f:Q\hookrightarrow \wt{Q}$ in
$\CalA$, there exists $Q'\in\CalA$ and an isomorphism
$g:Q\oplus Q'\isom \wt{Q}$ such that $g\circ i = f$. Here,
$i:Q\hookrightarrow Q\oplus Q'$ is the natural inclusion.\\

\item For any injective morphism $f:A\to B$, and an arbitrary
morphism $g:A\to Q$, there exists a lift $\wt{g}:B\to Q$ such
that $\wt{g}\circ f = g$.\\

\item $\Hom(-,Q)$ is exact.\\

\item Every short exact sequence $0\to Q\to Y\to Z\to 0$ splits.\\
\end{itemize}

\noindent {\em Projective objects}. An object $P\in\CalA$ is called
a projective object, if it satisfies (one of the) following
equivalent conditions.\\

\begin{itemize}
\item For any surjective morphism $f:\wt{P}\twoheadrightarrow P$ in
$\CalA$, there exists $P'\in\CalA$ and an isomorphism
$g:\wt{P}\isom P\times P'$ such that $\pi\circ g = f$. Here,
$\pi:P\times P'\to P$ is the natural surjection.\\

\item For any surjective morphism $f:B\to C$, and an arbitrary
morphism $g:P\to C$, there exists a lift $\wt{g}:P\to B$ such
that $f\circ \wt{g} = g$.\\

\item $\Hom(P,-)$ is exact.\\

\item Every short exact sequence $0\to X\to Y\to P\to 0$ splits.\\
\end{itemize}

Today we are going to discuss injective and project {\em resolutions}.\\

\PP{17.1}{Resolutions}
%--------------------
Let $M\in\CalA$.\\

\begin{defn}
An {\em injective resolution} of $M$ is a {\em cochain} complex in $\CoComp{\CalA}$,
\[
Q^{\bullet}\ :\qquad
0\to Q^0\morph{d^0} Q^1 \morph{d^1} Q^2 \morph{d^2} \cdots\ ,
\]
which is exact at each $Q^n$ ($n\geq 1$), and $\Ker(d^0)\cong M$.
Here, $Q^{-\ell} = 0$ for every $\ell\geq 1$.\\

A {\em projective resolution} of $M$ is a {\em chain} complex
in $\Comp{\CalA}$,
\[
P_\bullet\ :\qquad
\cdots \morph{d_2} P_2 \morph{d_1} P_1\morph{d_0} P_0\to 0,
\]
which is exact at each $P_n$ ($n\geq 1$), and $\Coker(d_0) \cong M$.
Here $P_{-\ell}=0$ for every $\ell\geq 1$.
\end{defn}

The existence of such resolutions is a technical requirement
imposed on the category in question.\\

\PP{17.2}{Enough injectives/projectives}
%--------------------------------------
Let $\CalC$ be an arbitrary abelian category. We say that
$\CalC$ has {\em enough injectives} (resp. {\em enough projectives})
if for every $X\in\CalC$, there exists an injective morphism
$X\hookrightarrow Q$
(resp. surjective morphism $P\twoheadrightarrow X$) where
$Q$ is injective (resp. $P$ is projective).\\

\noindent {\em Building injective resolutions}. If $\CalC$ has enough 
injectives, then for every $X\in\CalC$,
an injective resolution $Q^\bullet$ of $X$ can be constructed
as follows.
\begin{itemize}
\item Start from an injective morphism $f:X\hookrightarrow Q^0$,
where $Q^0$ is injective.\\

\item Let $\pi^0 : Q^0\twoheadrightarrow X^0 = \Coker(f)=Q^0/X$.
Find an injective morphism $f^0 : Q^0/X \hookrightarrow Q^1$
to an injective object $Q^1$. Define $d^0 = f^0\circ \pi^0$.\\

\item Repeat the previous step indefinitely...

\[
\xy
{\ar@{^(->}^f (0,20)*{X\ }="x"; (20,20)*{Q^0}="q0"};
{\ar^{d^0} "q0"; (50,20)*{Q^1}="q1"};
{\ar@{->>}_{\pi^0} "q0"; (35,0)*{\ Q^0/X}="q0b"};
{\ar@{^(->}_{f^0} "q0b"; "q1"};
{\ar^{d^1} "q1"; (80,20)*{Q^2}="q2"};
{\ar@{->>}_{\pi^1} "q1"; (65,0)*{\ Q^1/\Im(d^0)}="q1b"};
{\ar@{^(->}_{f^1} "q1b"; "q2"};
(90,20)*{\cdots};
\endxy
\]
\ \\

\item $\ds Q^\bullet\ := \qquad 0\to Q^0\morph{d^0} Q^1
\morph{d^1} Q^2 \to\cdots$ It is clear from its construction that
the complex is exact at each spot, except $\Ker(d^0) = X$.\\
\end{itemize}


\noindent {\em Building projective resolutions}. This construction
is entirely analogous to the one given above. Namely,

\begin{itemize}
\item Start from a surjection from a projective object $f:P_0\twoheadrightarrow
X$.\\

\item Let $i_0 : K_0 = \Ker(f) \hookrightarrow P_0$. Find a surjection
from a projective object $f_0 : P_1\twoheadrightarrow K_0$.
Define $d_0 = i_0\circ f_0 : P_1 \to P_0$.\\

\item Repeat.

\[
\xy
{\ar@{->>}^{f} (-30,20)*{P_0}="p0"; (0,20)*{X}="x"};
{\ar^{d_0} (-60,20)*{P_1}="p1"; "p0"};
{\ar@{->>}_{f_0} "p1"; (-45,0)*{K_0}="k0"};
{\ar@{^(->}_{i_0} "k0"; "p0"};
{\ar^{d_1} (-90,20)*{P_2}="p2"; "p1"};
{\ar@{->>}_{f_1} "p2"; (-75,0)*{K_1}="k1"};
{\ar@{^(->}_{i_1} "k1"; "p1"};
(-100,20)*{\cdots};
\endxy
\]
\ \\

\item $\ds P_\bullet\ :=\qquad \cdots\to P_2\morph{d_1}P_1\morph{d_0}P_0\to 0$.
Again, this is an exact (chain) complex at each spot, except $\Coker(d_0)=X$.\\
\end{itemize}

We record these constructions as a proposition.\\

\begin{prop}
Assuming an abelian category $\CalC$ has enough injectives (resp. projectives),
every $X\in\CalC$ admits an injective (resp. projective) resolution.\\
\end{prop}

\begin{rems}\hfill
\begin{enumerate}
\item Injective/projective resolutions are not unique, but only up to homotopy,
as we will prove below.\\

\item In practice, projective resolutions are easier to work with, than
injective ones.\\

\item We will have to prove later that $\CalA=\Rmod{R}$ has enough
injectives and projectives. The projective case is easy, but it will
take some work to show that there are enough injective $R$--modules.\\
\end{enumerate}
\end{rems}

\PP{17.3}{Uniqueness (up to homotopy) of resolutions}
%----------------------------------------------------
Let $M,N\in\CalA$ and let $I^\bullet$ and $J^\bullet$ be injective
resolutions of $M$ and $N$ respectively. Recall that this means
the following two sequences are exact:
\[
\xy
{\ar (0,20)*{0}; (20,20)*{M}="m"};
{\ar^i "m"; (40,20)*{I^0}="i0"};
{\ar^{d^0_I} "i0"; (60,20)*{I^1}="i1"};
{\ar^{d^1_I} "i1"; (80,20)*{I^2}="i2"};
{\ar^{d^2_I} "i2"; (100,20)*{\ \cdots}};
{\ar (0,0)*{0}; (20,0)*{N}="n"};
{\ar^j "n"; (40,0)*{J^0}="j0"};
{\ar^{d^0_J} "j0"; (60,0)*{J^1}="j1"};
{\ar^{d^1_J} "j1"; (80,0)*{J^2}="j2"};
{\ar^{d^2_J} "j2"; (100,0)*{\ \cdots}};
\endxy
\]

\begin{thm}
Given a morphism $f:M\to N$, there exists a lift $f^\bullet : I^\bullet
\to J^\bullet$. That is, there are morphisms $\{f^n:I^n\to J^n\}_{n\geq 0}$
such that all the squares in the following diagram are commutative:

\[
\xy
{\ar (0,20)*{0}; (20,20)*{M}="m"};
{\ar^i "m"; (40,20)*{I^0}="i0"};
{\ar^{d^0_I} "i0"; (60,20)*{I^1}="i1"};
{\ar^{d^1_I} "i1"; (80,20)*{I^2}="i2"};
{\ar^{d^2_I} "i2"; (100,20)*{\ \cdots}};
{\ar (0,0)*{0}; (20,0)*{N}="n"};
{\ar_j "n"; (40,0)*{J^0}="j0"};
{\ar_{d^0_J} "j0"; (60,0)*{J^1}="j1"};
{\ar_{d^1_J} "j1"; (80,0)*{J^2}="j2"};
{\ar_{d^2_J} "j2"; (100,0)*{\ \cdots}};
{\ar_f "m"; "n"};
{\ar_{f^0} "i0"; "j0"};
{\ar_{f^1} "i1"; "j1"};
{\ar_{f^2} "i2"; "j2"};
\endxy
\]

Moreover, if $f^\bullet$ and $g^\bullet$ are two lifts of $f$, then
$f^\bullet$ is homotopic to $g^\bullet$.\\
\end{thm}

We will prove this theorem in the next two sections. For now, we
record its important consequence.\\

\begin{cor}
Injective resolutions are unique up to homotopy.
\end{cor}
\begin{pf}
Let $M\in\CalA$, and let $I^\bullet$ and $J^\bullet$ be two injective
resolutions. Take $M=N$ and $f=\Id_M$ in the previous theorem, to
get $\alpha:I^\bullet\to J^\bullet$ and $\beta:J^\bullet\to I^\bullet$.
Then $\alpha\circ\beta:J^\bullet\to J^\bullet$ and 
$\beta\circ \alpha:I^\bullet\to I^\bullet$ are both lifts
of $\Id_M: M\to M$. But so are identity morphisms $\Id_{J^\bullet}$
and $\Id_{I^\bullet}$. By uniqueness part of the theorem, we conclude
that $\alpha\circ\beta\sim \Id_{J^\bullet}$ and $\beta\circ
\alpha\sim\Id_{I^\bullet}$. Hence $I^\bullet$ and $J^\bullet$
are isomorphic (up to homotopy).
\end{pf}

\PP{17.4}{Proof of Theorem 17.3: existence of lifts}
%---------------------------------------------------
We begin by constructing $f^0:I^0\to J^0$. For this we need
to use the fact that $i$ is an injective morphism and $J^0$ is an
injective object:
\[
\xy
{\ar^{i} (0,0)*{M}="a"; (30,0)*{I^0}="b"};
{\ar_{j\circ f} "a"; (0,-20)*{J^0}="q"};
{\ar (-20,0)*{0}; "a"};
{\ar@{-->}^{\exists f^0} "b"; "q"};
\endxy
\]
Assume that we have constructed $\{f^\ell:I^\ell\to J^\ell\}_{0\leq
\ell\leq n}$. Let us construct $f^{n+1}$. For this, we use the
induced morphism at the level of cokernels:
\[
\xy
(0,20)*{I^{n-1}}="a1"; (30,20)*{I^n}="a2";
(60,20)*{I^n/\Im(d_I^{n-1})}="ca";
{\ar "a2"; "ca"};
{\ar^{d_I^{n-1}} "a1"; "a2"}; 
(0,0)*{J^{n-1}}="b1"; (30,0)*{J^n}="b2";
(60,0)*{J^n/\Im(d_J^{n-1})}="cb";
{\ar "b2"; "cb"};
{\ar@{-->}^{\ol{f^n}} "ca"; "cb"};
{\ar_{d_J^{n-1}} "b1"; "b2"};
{\ar_{f^{n-1}} "a1"; "b1"};
{\ar^{f^n} "a2"; "b2"};
\endxy
\]

As $\Im(d_I^{n-1}) = \Ker(d^n_I)$, we get an injective morphism
$\ul{d_I^n}:I^n/\Im(d_I^{n-1}) \hookrightarrow I^{n+1}$ coming from $d^n$.
Similarly we have a morphism, since $\Im(d_J^{n-1})\subset \Ker(d_J^n)$:
\[\ul{d^n_J} : J^n/\Im(d^{n-1}_J) \to J^{n+1}.\]

Now we can use the fact that $J^{n+1}$ is injective, to get $f^{n+1}$
as:
\[
\xy
{\ar^{\ul{d^n}} (0,0)*{I^n/\Im(d_I^{n-1})}="a"; (30,0)*{I^{n+1}}="b"};
{\ar_{\ul{d^n_J}\circ\ol{f^n}} "a"; (0,-20)*{J^{n+1}}="q"};
{\ar (-20,0)*{0}; "a"};
{\ar@{-->}^{\exists f^{n+1}} "b"; "q"};
\endxy
\]

\begin{rem}
Note that in the proof above, we only used the following assumptions:
\begin{itemize}
\item $\ds 0\to M\to I^0\to I^1\to \cdots$ is exact. We did not use
the hypothesis that $I^\ell$ is injective ($\ell\geq 0$).\\

\item $\ds 0\to N\to J^0\to J^1\to \cdots$ is a complex, where
each $J^\ell$ is injective. We did not use exactness of this complex.\\
\end{itemize}
\end{rem}


\PP{17.5}{Uniqueness of lift}
%---------------------------
Note that it suffices to show that if $f^\bullet:I^\bullet\to J^\bullet$
is a lift of the zero morphism $M\morph{0}N$, then $f^\bullet$ is
null--homotopic. That is, there exists $\ds \{s^n:I^n\to J^{n-1}\}_{n\geq 0}$
(with $s^0 = 0$ since $J^{-1}=0$), such that
\[
f^n = d^{n-1}_J\circ s^n + s^{n+1}\circ d_I^n : I^n\to J^n,\ \forall\ 
n\geq 0.
\]

Let us begin by exhibiting how $s^1 : I^1\to J^0$ shows up. For this,
we notice that from the commutative diagram
\[
\xy
{\ar_0 (0,20)*{M}="m"; (0,0)*{N}="n"};
{\ar^i "m"; (30,20)*{I^0}="i0"};
{\ar_j "n"; (30,0)*{J^0}="j0"};
{\ar^{f^0} "i0"; "j0"};
\endxy
\]
we get that $M\subset\Ker(f^0)$. Thus, $f^0$ factors through
$\ul{f^0} : I^0/M \to J^0$. Moreover, $M = \Im(i) = \Ker(d^0_I)$
gives an injective morphism $\ul{d^0_I}: I^0/M\hookrightarrow I^1$.
Now we are in the position to use the fact that $J^0$ is injective
to get $s^1:I^1\to J^0$ as follows.
\[
\xy
{\ar^{\ul{d^0_I}} (0,0)*{I^0/M}="a"; (30,0)*{I^1}="b"};
{\ar_{\ul{f^0}} "a"; (0,-20)*{J^0}="q"};
{\ar (-20,0)*{0}; "a"};
{\ar@{-->}^{\exists s^1} "b"; "q"};
\endxy
\]
It is easy to see that we get $f^0 = s^1\circ d^0_I$ so that
we have built the required homotopy up to $I^1$:

\[
\xy
(0,20)*{0}="z1"; (20,20)*{I^0}="i0"; (40,20)*{I^1}="i1";
{\ar "z1"; "i0"};
{\ar^{d^0_I} "i0"; "i1"};
(0,0)*{0}="z2"; (20,0)*{J^0}="j0";
{\ar "z2"; "j0"};
{\ar_{s^0=0} "i0"; "z2"};
{\ar^{s^1} "i1"; "j0"};
{\ar_{f^0} "i0"; "j0"};
\endxy
\]

Assuming we have successfully built homotopies $\{s^0,\ldots,s^n\}$,
where $n\geq 1$. We proceed with the construction of $s^{n+1}$. For this, we have
to focus on the following part of the picture:
\[
\xy
(20,20)*{I^{n-1}}="i1"; (40,20)*{I^n}="i2"; (60,20)*{I^{n+1}}="i3";
{\ar^{d^{n-1}_I} "i1"; "i2"};
{\ar^{d^{n}_I} "i2"; "i3"};
(0,0)*{J^{n-2}}="j1"; (20,0)*{J^{n-1}}="j2"; (40,0)*{J^{n}}="j3";
{\ar_{d^{n-2}_J} "j1"; "j2"};
{\ar_{d^{n-1}_J} "j2"; "j3"};
{\ar_{s^{n-1}} "i1"; "j1"};
{\ar^{f^{n-1}} "i1"; "j2"};
{\ar^{s^{n}} "i2"; "j2"};
{\ar^{f^{n}} "i2"; "j3"};
\endxy
\]

Define $g=f^n - d_J^{n-1}\circ s^n : I^n \to J^n$. The following
calculation shows that $g\circ d_I^{n-1}=0$:
\[
g\circ d_I^{n-1} = f^n d_I^{n-1} - d_J^{n-1} s^n d_I^{n-1}
= d_J^{n-1}(f^{n-1} - s^n d^{n-1}_I) = d_J^{n-1} d_J^{n-2} s^{n-1} = 0.
\]

Here, we have used the following facts:
\begin{itemize}
\item $f^\bullet$ is a morphism of complexes, so $f^nd_I^{n-1}
=d_J^{n-1}f^{n-1}$.\\

\item $f^{n-1} = d_J^{n-2}s^{n-1} + s^nd_I^{n-1}$.\\

\item $d_J^{n-1}d_J^{n-2}=0$.\\
\end{itemize}

Thus, we conclude that $\Im(d^{n-1}_I) = \Ker(d^n_I) \subset \Ker(g)$.
This allows us to factor $g$ through $\ds
\ul{g} : I^n/\Ker(d^n) \to J^n$. Again we rely on injectivity
of $I^{n+1}$ in the following diagram to get $s^{n+1}$:
\[
\xy
(0,20)*{0}="z1"; (20,20)*{I^n/\Ker(d^n)}="i0"; (50,20)*{I^{n+1}}="i1";
{\ar "z1"; "i0"};
{\ar^{\ul{d^n_I}} "i0"; "i1"};
(20,0)*{J^n}="j0";
{\ar@{-->}^{s^{n+1}} "i1"; "j0"};
{\ar_{\ul{g}} "i0"; "j0"};
\endxy
\]
\end{document}
