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

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

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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (3)

Comment #3065 by Noah Olander on

Comment #3066 by Noah Olander on

Comment #3167 by Johan on