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 (3)

Comment #3065 by Noah Olander on

This comment is totally pedantic since the axiom of choice is used all over the place to prove this, but still I think it's more elegant to say "let \alpha be the least ordinal with cofinality greater than |U|" rather than "choose an ordinal \alpha with cofinality greater than |U|."

Comment #3066 by Noah Olander on

More importantly though it seems that the second double direct sum should have the same indices as the first (compare with section 2).

There are also:

• 2 comment(s) on Section 19.11: Injectives in Grothendieck categories

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