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. $\square$

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).

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