Lemma 19.11.6 (079G)—The Stacks project

To check that an object is injective, one only needs to check that lifting holds for subobjects of a generator.

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 $A'$ . $\square$

Comments (1)

Comment #1091 by Alex Youcis on October 17, 2014 at 07:07

Suggested slogan: To check that an object is injective, one only needs to check that lifting holds for subobjects of a generator.

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