# Extensions, Injectives, and Co image

This is the first post for the second semester of Ogus alg. geo.  The first part of this semester was focused on introducing the derived categories and derived functors.  I’m not going to present all the details here.  Mostly I’ll mention the major results and probably will skip proofs (a more complete and very readable treatment is in the first couple chapters of Huybrechts, but even Huybrechts doesn’t include all the details).

### Ext(A,C)

For a little bit I’ll work in $\mathscr{Ab}$, the category of abelian groups.  For any object $A$ there are the covariant and contravariant functors $\hom(A,- )$ and $\hom(- , A)$ from $\mathscr{Ab}$ to itself, and more generally the contravariant functor $h\colon \mathscr{Ab} \to \mbox{Func}(\mathscr{Ab}, \mathscr{Ab})$ sending $A$ to  $\hom(A,-)$.

For a general $A$, $\hom(A,-), \hom(-,A)$ are only left exact.  And so arises the notions of projective ($\hom(P, -)$ exact ) and injective objects ($\hom(-, I)$ exact).  As not all objects are projective or injective, so comes extensions.  In general, and extension of A by C is an exact sequence

$0 \to C \to B \to A \to 0$

A category can be formed taking s.e.s as objects and morphisms to be morphisms between s.e.s where the morphisms $C \to C$ and $A \to A$, and the 5-lemma implies the middle morphism is also an iso.  Ogus calls this category $\mbox{EXT}(A,C)$.  It has a group structure.  Given two objects

$o \to C \to B \to A \to 0$

$0 \to C \to B' \to A \to 0$

Form the sequence $0 \to C \oplus C \to B \times_A \times B' \to A$.  Let $B''$ by the pushforward of the two maps out of $C \oplus C$

$C \oplus C \xrightarrow{sum} C$

$C \oplus C \to B\times_A B'$

Then it can be checked that B” fits into a long exact sequence

$0 \to C \to B'' \to A \to 0$

As it happens, the identity element can be identified with the trivial extension

$0\to C \to C \oplus A \to A \to 0$

Setting $\mbox{Ext}(A,C)$ to be isomrophism classes of elements in $\mbox{EXT}(A,C)$.  Now given a s.e.s $0 \to B' \to B \to B'' \to 0$ and a map $A \to B''$, you can take the fiber product $B \times_{B''} A$ can infact get an extension of $A$ by $B'$:

$0 \to B' \to B \times_{B''} A \to A$

and in this way $\mbox{Ext}(A, -)$ continues the exact seqence started by $\hom(A, -)$ etc.

### Injectives

For abelian groups, or $\mathbb{Z}$ modules, it turns out that projective modules are just free modules.  For injectives, there is the following

(Commutative Baer’s Criterion) Thm: In the category of $R$ modules, $I$ is injective iff for any ideal $J \subet R$ and any module $J \to I$ there is an extension $R \to I$

proof: If $I$ is injective, then by definition any map from and ideal of $R$ extends to all of $R$.  Conversely, suppose there is an injection $0 \to A \to B$ and a map $f \colon A \to B$.  Consider all extension $(A',f')$ of $f$, these form a partially ordered set and chains have maximal elements: given $(A_1, f_1) \subset (A_2, f_2) \subset ...$ set $A_m = \cup_i A_i$ and for $x \in A_m$ it must be x \in A_i&s=-1$for some i so define$latexf_m(x) = f_i(x)&s=-1\$, then it clear $(A_m, f_m)$ is maximal.  Zorn’s lemma says there’s a maximal element $(A', f')$, it remains to show $A' = B$.  Proceeding by contradiction let $b \in B - A'$.  Let $J = \{ r \in R| rb \in A'\}$, this is certainly an ideal, and I have a map $J \to A' \to B$, so by assumption there is an extension $g\colon R \to B$, now set $A'' = A' + bR$, and define $f'' \colon A'' \to B$ by $f''(a' + rb) = f'(a') + g(r)$.  Can check that this is well defined, i.e. if $b \in A' \cap Rb = Jb$ then $f''(b) = f'(0) + g(b) = f'(b)$ so this contradiction establishes the result.

QED

The following theorem is in fact just a corollary of the above theorem, in the case $R = \mathbb{Z}$ or more generally a principal ideal domain, but in any case Ogus proved it using essentially the same proof as above.

Thm: $I$ is an injective object iff $I$ is divisible, i.e. for any $a \in I, 0 < n \in \mathbb{N}$ there is $a/n \in I$

proof: First, if $I$ is injective, the consider for any element $g \in I$ the map $\mathbb{Z} \to I$ sending 1 to $g$.  Injectiveness means there is a commutative diagram

$\begin{array}{ccc} \mathbb{Z} & \xrightarrow{n} & \mathbb{Z} \\ \downarrow & \swarrow & \mbox{} \\ I & \mbox{} & \mbox{} \end{array}$

Shows that $g/n \in I$. The converse is longer and involves Zorn’s lemma.

Consider an injection $0 \to B \to B'$ and let $\mathscr{C}$ be the set of all $B\subset B'' \subset B'$ for which there is a diagram

$\begin{array}{ccc} B & \to & B'' \\ \downarrow & \swarrow & \mbox{} \\ I & \mbox{} & \mbox{} \end{array}$

There is an ordering $(B_1, h_1) \le (B_2, h_2)$ if $B_1 \subset B_2$ and $h_2|_{B_1} = h_1$.  Chains have a maximal element etc. so Zorn says there is a maximal element.  It remains to show I get a contradiction if $B_{max} \ne B'$.

Suppose $b \in B' - B_{max}$.  If $bn$ is not in $B_{max}$ for all n then $B'' = B_{max} \oplus (b)$ with extension information $h''(b) = 0$ is greater than $B_{max}$, contradiction.  Hence $nb \in B''$ for some $n$.  Note then that $B''/B_{max} \cong \mathbb{Z}/n$.  Hence there is the following commutative diagram:

$\begin{array}{ccccccccc} 0 & \to & B_{max} & \to & B'' & \to & \mathbb{Z}/n & \to & 0 \\ \mbox{} & \mbox{} & \downarrow & \mbox{} & \downarrow & \mbox{} & \downarrow & \mbox{} & \mbox{} \\ 0 & \to & I & \to & pB & \to & \mathbb{Z}/n & \to & 0 \end{array}$

Where pB is the pushforward of the maps coming out of B_max and where the first  vertical map is the maximal map of the definition of B_max, $h: B_{max} \to I$.  So $h(nb) = a' \in I$, by assumption $a' = na$ for some $a \in I$.  So  $(b, -a) \in pB$ has the property that $n\cdot (b, -a) = (nb, -na) = (nb, -h(nb)) = 0$, so this element gives a section $\mathbb{Z}/n \to pB$ hence the bottom s.e.s splits which consequently gives a map $B'' \to I$ which extends the the map from $B_{max}$, contradiction.

QED

Its an immediate consequence of this theorem that any quotient of an injective object is injective (i.e. if there is a surjection $I \to J$ then J is injective).  There is also the following corollary

Cor. Any abelian group can be an imbedded into an injective one.

proof: Let B be an abelian group.  Choose a set of generators to produce a surjection $F \to B$ where F is free.  There there is a commutative diagram

$\begin{array}{ccccccccc} 0 & \to & \mbox{Ker} & \to & F & \to & B & \to & 0 \\ \mbox{} & \mbox{} & || & \mbox{} & \downarrow & \mbox{} & \downarrow & \mbox{} & \mbox{} \\ 0 & \to & \mbox{Ker} & \to & F \otimes \mathbb{Q} & \to & (F \otimes \mathbb{Q})/\mbox{Ker} & \to & 0 \end{array}$

All the vertical maps are injective and $F \otimes \mathbb{Q}$ is an injective object by the theorem, hence so is its quotient  by Ker, hence B injects into an injective object.

QED

### Coimage vs Image

Recall an category $\mathscr{A}$ is additive if

1. $\hom(A,B)$ has the structure of ab. group for all objects $A,B \in \mathscr{A}$
2. $\hom(B,C) \times \hom(A,B) \to \hom(A,C)$ is a bilinear pairing
3. Products exists (which in particular make finite coproducts equal to finite products , i.e. $A \oplus B \cong A \times B$)
4. Kernels and Cokernels exists (from which if follows that fibers and co-fibers exist)

So given $A \xrightarrow{\theta} B$, Ogus defined

$\mbox{im} (\theta) = \ker (B \to \mbox{coker} \theta)$

$\mbox{coim}(\theta) = \mbox{coker}(\ker \theta \to A)$

From which I can declare $\mathscr{A}$ to be abelian if $\mbox{im} \theta \cong \mbox{coim} \theta$.  Note a seemingly more common definition of abelian category (for example in Weibel) is to require every $\theta$ to have a kernel and cokernel (required for ogus def. of addititve), every injection (or monic) is the ker of its cok, and every surjection (epi) is cok of its ker.  The equivalence of these definitions is a small commutative diagram exercise.

Its true that not all additive categories are abelian.  Here is an example based on the Ogus definitions. Let $\mathscr{A}$ be the category of vector spaces with filtrations indexed by $\mathbb{Z}$, so an objects is a vector space $V$ with a filtration

$... F^1 \subset F^0 \subset F^{-1} \subset ... \subset V$

Morphisms of objects are morphisms that are compatible with the filtration.  For a vector space $V$, consider the filtration $F^i = 0$ for $i \ge 1$ and $F^0 = V$ for $i \le 0$.  Also consider the filtration $F^i = 0$ for $i \ge 2$ and $F^i = V$ for $i \le 1$.  Let these represents objects $W,W'$ respectively.  There is a map $W \xrightarrow{\alpha} W'$ which at the point of interest looks like

$\begin{array}{ccccccccc} 0 & \subset & 0 & \subset & V & \subset & V & \subset & V \\ \downarrow & \mbox{} & \downarrow & \mbox{} & \downarrow & \mbox{} & \downarrow & \mbox{} & \downarrow \\ 0 & \subset & V & \subset & V & \subset & V & \subset V & V \end{array}$

If I understand things correctly, it seems in this case $\mbox{coim} (\alpha) = W$ and $\mbox{im} (\alpha) = W'$ so there is a map from the coimage to the image of but its not an isomorphism.