# The Stacks Project

## Tag 079G

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 $$A' = \mathop{\rm colim}\nolimits_{t \in T} A_t \xrightarrow{\mathop{\rm colim}\nolimits_{t \in T} \varphi_t} I$$ 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 $\chi|_{\mathop{\rm Ker}(\psi)} = 0$ we see that $\chi$ factors as $$U \to \mathop{\rm Im}(\psi) \xrightarrow{\varphi''} I$$ Since $\varphi'$ and $\varphi''$ agree on $N = A' \cap \mathop{\rm Im}(\psi)$ we see that combined the define a morphism $A' + \mathop{\rm Im}(\psi) \to I$ contradicting the assumed maximality of $A'$. $\square$

The code snippet corresponding to this tag is a part of the file injectives.tex and is located in lines 1467–1483 (see updates for more information).

\begin{lemma}
\label{lemma-characterize-injective}
\begin{slogan}
To check that an object is injective, one only needs to check that lifting
holds for subobjects of a generator.
\end{slogan}
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.
\end{lemma}

\begin{proof}
Please see Lemma \ref{lemma-criterion-baer} 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
$$A' = \colim_{t \in T} A_t \xrightarrow{\colim_{t \in T} \varphi_t} I$$
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
$\chi|_{\Ker(\psi)} = 0$ we see that $\chi$ factors as
$$U \to \Im(\psi) \xrightarrow{\varphi''} I$$
Since $\varphi'$ and $\varphi''$ agree on $N = A' \cap \Im(\psi)$
we see that combined the define a morphism $A' + \Im(\psi) \to I$
contradicting the assumed maximality of $A'$.
\end{proof}

Comment #1091 by Alex Youcis on October 17, 2014 a 7:07 am UTC

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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