Injective Resolutions

This section is the heart of all what follows. All the constructions in next (or probably every) section from now onwards will be depending on the fact that in the relevant category, injection resolutions do exist. We fix $\mathscr{C}$, an abelian category.

Definition 9   $M\in \mathscr{C}$ is called injective, if corresponding (covariant) $Hom(-,M)$ functor is exact. In other words, if $\mathscr{Ab}$ denote the category of abelian groups, then the following functor is exact

$$\begin{array}{ccc} \mathscr{C} & \longrightarrow & \mathscr{Ab} \\ A & \longmapsto & Hom(A,M) \end{array}$$

This is just another way of saying that if $B$ is subobject of some object $A$, then every map from $B$ to $M$ can be extended to map from $A$ to $M$.

Definition 10   A morphism $A\rightarrow M$ (beware, $M$ is not assumed to be injective object here) is called injective envelop of $A$, if it is injective morphism and for every injective morphism $B\rightarrow C$ and some morphism $B\rightarrow A$, one can find another $C\rightarrow M$, making following diagram commutative.

$$\xymatrix{ B \ar[rr] \ar[dd] & & C \ar[dd] \\ & & \\ A \ar[rr] & & M }$$

To prove main result, Theorem 3, we need a lemma stating to check whether some object is injective, it is enough to check for generator (if it exists)

Lemma 2   Assume $\mathscr{C}$ admits a generator, say $U$. Then $M\in \mathscr{C}$ is injective object if and only if for every subobject $V$ of $U$ and a morphism $V\rightarrow M$, there is a morphism $U\rightarrow M$, which extends it.

PROOF:
It suffices to prove sufficiency of this condition. Let $B$ be subobject of $A$ and we are given morphism $u : B\rightarrow M$, where $M$ is an object satisfying condtion of lemma. We need to find extension of $u$ $A\rightarrow M$ to prove that $M$ is injective. Let $\mathscr{P}$ be set of subobjects of $A$ for which such extension exists. Then by Zorn's lemma (this set is inductive, by axiom (AB5)), we can find maximal element (say $B',u')$) of $\mathscr{P}$. Assume that $B'\not= A$. Then there exists $j: U \rightarrow A$ such that $Im(j)\not\subset B'$. Let $V = j^{-1}(B')$ and consider the following:

$$\xymatrix{ V \ar[rr]^{u'j} \ar@{^{(}->}[dd] & & M \\ & & \\ U \ar[uurr]_{\tilde{u}} & & }$$

where $\tilde{u}$ is extension, which exists by assumption on $M$. Now let $B'' = B' + j(U)$. We have the following diagram (the row is exact and dotted map is to be constructed):

$$\xymatrix{ V \ar[rr]^{(\iota, -j)} & & U \oplus B' \ar[rr]^{(j,\iota)} \ar[dd]^{(\tilde{u},u')} & & B''\ar[rr] && 0 \\ &&&&&& \\ && M && && }$$

Thus $B''$ can be identified with quotient of $U\oplus B'$ and to define map from $B''$ to $M$, we only need to check that $Ker(U\oplus B'\rightarrow B'')\subseteq Ker(U\oplus B' \rightarrow M)$, where the first term is same as $V$. This is obvious since $V\rightarrow U\oplus B' \rightarrow M$ composition is zero map. Thus we have extended $u$ to $B'' \not= B'$ and containing $B'$ which contradicts maximality of $B'$.

QED

Theorem 3   If $\mathscr{C}$ satisfies (AB5) and admits a generator, then for every $A\in \mathscr{C}$, there is an injective morphism of $A$ into some injective object $M$.²