The Stacks project

19.11 Injectives in Grothendieck categories

The existence of a generator implies that given an object $M$ of a Grothendieck abelian category $\mathcal{A}$ there is a set of subobjects. (This may not be true for a general “big” abelian category.)

Lemma 19.11.1. Let $\mathcal{A}$ be an abelian category with a generator $U$ and $X$ and object of $\mathcal{A}$. If $\kappa $ is the cardinality of $\mathop{\mathrm{Mor}}\nolimits (U, X)$ then

  1. There does not exist a strictly increasing (or strictly decreasing) chain of subobjects of $X$ indexed by a cardinal bigger than $\kappa $.

  2. If $\alpha $ is an ordinal of cofinality $> \kappa $ then any increasing (or decreasing) sequence of subobjects of $X$ indexed by $\alpha $ is eventually constant.

  3. The cardinality of the set of subobjects of $X$ is $\leq 2^\kappa $.

Proof. For (1) assume $\kappa ' > \kappa $ is a cardinal and assume $X_ i$, $i \in \kappa '$ is strictly increasing. Then take for each $i$ a $\phi _ i \in \mathop{\mathrm{Mor}}\nolimits (U, X)$ such that $\phi _ i$ factors through $X_{i + 1}$ but not through $X_ i$. Then the morphisms $\phi _ i$ are distinct, which contradicts the definition of $\kappa $.

Part (2) follows from the definition of cofinality and (1).

Proof of (3). For any subobject $Y \subset X$ define $S_ Y \in \mathcal{P}(\mathop{\mathrm{Mor}}\nolimits (U, X))$ (power set) as $S_ Y = \{ \phi \in \mathop{\mathrm{Mor}}\nolimits (U,X) : \phi )\text{ factors through }Y\} $. Then $Y = Y'$ if and only if $S_ Y = S_{Y'}$. Hence the cardinality of the set of subobjects is at most the cardinality of this power set. $\square$

By Lemma 19.11.1 the following definition makes sense.

Definition 19.11.2. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $M$ be an object of $\mathcal{A}$. The size $|M|$ of $M$ is the cardinality of the set of subobjects of $M$.

Lemma 19.11.3. Let $\mathcal{A}$ be a Grothendieck abelian category. If $0 \to M' \to M \to M'' \to 0$ is a short exact sequence of $\mathcal{A}$, then $|M'|, |M''| \leq |M|$.

Proof. Immediate from the definitions. $\square$

Lemma 19.11.4. Let $\mathcal{A}$ be a Grothendieck abelian category with generator $U$.

  1. If $|M| \leq \kappa $, then $M$ is the quotient of a direct sum of at most $\kappa $ copies of $U$.

  2. For every cardinal $\kappa $ there exists a set of isomorphism classes of objects $M$ with $|M| \leq \kappa $.

Proof. For (1) choose for every proper subobject $M' \subset M$ a morphism $\varphi _{M'} : U \to M$ whose image is not contained in $M'$. Then $\bigoplus _{M' \subset M} \varphi _{M'} : \bigoplus _{M' \subset M} U \to M$ is surjective. It is clear that (1) implies (2). $\square$

Proposition 19.11.5. Let $\mathcal{A}$ be a Grothendieck abelian category. Let $M$ be an object of $\mathcal{A}$. Let $\kappa = |M|$. If $\alpha $ is an ordinal whose cofinality is bigger than $\kappa $, then $M$ is $\alpha $-small with respect to injections.

Proof. Please compare with Proposition 19.2.5. We need only show that the map (19.2.0.1) is a surjection. Let $f : M \to \mathop{\mathrm{colim}}\nolimits B_\beta $ be a map. Consider the subobjects $\{ f^{-1}(B_\beta )\} $ of $M$, where $B_\beta $ is considered as a subobject of the colimit $B = \bigcup _\beta B_\beta $. If one of these, say $f^{-1}(B_\beta )$, fills $M$, then the map factors through $B_\beta $.

So suppose to the contrary that all of the $f^{-1}(B_\beta )$ were proper subobjects of $M$. However, because $\mathcal{A}$ has AB5 we have

\[ \mathop{\mathrm{colim}}\nolimits f^{-1}(B_\beta ) = f^{-1}\left(\mathop{\mathrm{colim}}\nolimits B_\beta \right) = M. \]

Now there are at most $\kappa $ different subobjects of $M$ that occur among the $f^{-1}(B_\alpha )$, by hypothesis. Thus we can find a subset $S \subset \alpha $ of cardinality at most $\kappa $ such that as $\beta '$ ranges over $S$, the $f^{-1}(B_{\beta '})$ range over all the $f^{-1}(B_\alpha )$.

However, $S$ has an upper bound $\widetilde{\alpha } < \alpha $ as $\alpha $ has cofinality bigger than $\kappa $. In particular, all the $f^{-1}(B_{\beta '})$, $\beta ' \in S$ are contained in $f^{-1}(B_{\widetilde{\alpha }})$. It follows that $f^{-1}(B_{\widetilde{\alpha }}) = M$. In particular, the map $f$ factors through $B_{\widetilde{\alpha }}$. $\square$

slogan

Lemma 19.11.6. Let $\mathcal{A}$ be a Grothendieck abelian category with generator $U$. An object $I$ of $\mathcal{A}$ is injective if and only if in every commutative diagram

\[ \xymatrix{ M \ar[d] \ar[r] & I \\ U \ar@{-->}[ru] } \]

for $M \subset U$ a subobject, the dotted arrow exists.

Proof. Please see Lemma 19.2.6 for the case of modules. Choose an injection $A \subset B$ and a morphism $\varphi : A \to I$. Consider the set $S$ of pairs $(A', \varphi ')$ consisting of subobjects $A \subset A' \subset B$ and a morphism $\varphi ' : A' \to I$ extending $\varphi $. Define a partial ordering on this set in the obvious manner. Choose a totally ordered subset $T \subset S$. Then

\[ A' = \mathop{\mathrm{colim}}\nolimits _{t \in T} A_ t \xrightarrow {\mathop{\mathrm{colim}}\nolimits _{t \in T} \varphi _ t} I \]

is an upper bound. Hence by Zorn's lemma the set $S$ has a maximal element $(A', \varphi ')$. We claim that $A' = B$. If not, then choose a morphism $\psi : U \to B$ which does not factor through $A'$. Set $N = A' \cap \psi (U)$. Set $M = \psi ^{-1}(N)$. Then the map

\[ M \to N \to A' \xrightarrow {\varphi '} I \]

can be extended to a morphism $\chi : U \to I$. Since $\chi |_{\mathop{\mathrm{Ker}}(\psi )} = 0$ we see that $\chi $ factors as

\[ U \to \mathop{\mathrm{Im}}(\psi ) \xrightarrow {\varphi ''} I \]

Since $\varphi '$ and $\varphi ''$ agree on $N = A' \cap \mathop{\mathrm{Im}}(\psi )$ we see that combined the define a morphism $A' + \mathop{\mathrm{Im}}(\psi ) \to I$ contradicting the assumed maximality of $A'$. $\square$

Theorem 19.11.7. Let $\mathcal{A}$ be a Grothendieck abelian category. Then $\mathcal{A}$ has functorial injective embeddings.

Proof. Please compare with the proof of Theorem 19.2.8. Choose a generator $U$ of $\mathcal{A}$. For an object $M$ we define $\mathbf{M}(M)$ by the following pushout diagram

\[ \xymatrix{ \bigoplus _{N \subset U} \bigoplus _{\varphi \in \mathop{\mathrm{Hom}}\nolimits (N, M)} N \ar[r] \ar[d] & M \ar[d] \\ \bigoplus _{N \subset U} \bigoplus _{\varphi \in \mathop{\mathrm{Hom}}\nolimits (N, M)} U \ar[r] & \mathbf{M}(M). } \]

Note that $M \to \mathbf{M}(N)$ is a functor and that there exist functorial injective maps $M \to \mathbf{M}(M)$. By transfinite induction we define functors $\mathbf{M}_\alpha (M)$ for every ordinal $\alpha $. Namely, set $\mathbf{M}_0(M) = M$. Given $\mathbf{M}_\alpha (M)$ set $\mathbf{M}_{\alpha + 1}(M) = \mathbf{M}(\mathbf{M}_\alpha (M))$. For a limit ordinal $\beta $ set

\[ \mathbf{M}_\beta (M) = \mathop{\mathrm{colim}}\nolimits _{\alpha < \beta } \mathbf{M}_\alpha (M). \]

Finally, pick any ordinal $\alpha $ whose cofinality is greater than $|U|$. Such an ordinal exists by Sets, Proposition 3.7.2. We claim that $M \to \mathbf{M}_\alpha (M)$ is the desired functorial injective embedding. Namely, if $N \subset U$ is a subobject and $\varphi : N \to \mathbf{M}_\alpha (M)$ is a morphism, then we see that $\varphi $ factors through $\mathbf{M}_{\alpha '}(M)$ for some $\alpha ' < \alpha $ by Proposition 19.11.5. By construction of $\mathbf{M}(-)$ we see that $\varphi $ extends to a morphism from $U$ into $\mathbf{M}_{\alpha ' + 1}(M)$ and hence into $\mathbf{M}_\alpha (M)$. By Lemma 19.11.6 we conclude that $\mathbf{M}_\alpha (M)$ is injective. $\square$


Comments (8)

Comment #5053 by Sandeep on

Very minor typo: In the third line of the proof of Lemma 19.11.4, in the second direct sum, it should be , not .

Comment #7169 by Liu on

In the proof of Theorem 079H, below the commutative diagram: "Note that M→M(N) is a functor and ..." should be M→M(M) I guess.

Comment #8602 by nkym on

According to tag 0014, categories in Stacks Project are small unless stated otherwise. That's why I think sets of subobjects are always available for a general abelian category in the sense of Stacks Project.

Comment #9170 by on

@#8602: Yes, you are correct. However when we use the phrase Grothendieck abelian category we typically mean a "big" one and should be on the list. So strictly speaking, in the definitions, lemmas, propositions, theorems, and remarks of Sections 19.10, 19.11, 19.12, 19.13, 19.14, and 19.15 should mention the "bigness" as we did correctly for example in Categories, Section 4.25.

Maybe we can leave this alone for now and the reader can assume that a lemma or such discussing Grothendieck abelian categories is one about a suitable "big" category.

Please discuss!

Comment #9501 by on

Typo: In “note that is a functor” I think it should be instead. This was also pointed out in #7169.

In the proof, after “pick any ordinal whose cofinality is greater than ,” I think we later use that is actually a limit ordinal when we say “then we see that factors through for some by Proposition 19.11.5.” Maybe one should mention some of this? ( will always be a limit ordinal since and by Sets, Comment #9498),

Also, in the transfinite recursion, we are not only defining for each but also a natural transformation for each , right? And such that is injective for all and for . In other words, we are getting a functor , where is the totally ordered class of ordinal numbers and is the wide subcategory of where the morphisms are the natural transformations whose components are all injective maps. After thinking for a while, the following is what I came up with.

For each ordinal number , denote to the set of ordinal numbers (this is , the successor of ). For each ordinal number , we want to define a functor such that for all . The desired functor will be obtained by taking the union of the functors .

We do so by transfinite recursion in .

  • Base step. For , define as .

  • case. Suppose such a functor is defined. For an ordinal , define as where is the natural transformation whose component at is the morphism in the cocartesian square that defines , and in the second-to-last equality we are using the whiskering notation from Categories, Section 4.28. Since is injective, so is the morphism in the second-to-last equality; hence also the morphism in the last equality is injective.

  • Limit case. Suppose is a limit ordinal and that is defined for all . We define , on the one hand, by setting to be the union of the functors for , and on other hand, by For , the map is defined to be the leg at of the limiting cocone \eqref{colim}. Since is AB5, by the Lemma in Comment #9497, is injective.

Comment #9502 by on

(Since the previous doesn't currently compile well on my browser, I assume it won't either in other people's browsers, so you may look at the comment's plain text here.)


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