Lemma 108.57.1. There exists a “big” abelian category $\mathcal{A}$ whose $\mathop{\mathrm{Ext}}\nolimits $-groups are proper classes.

## 108.57 A big abelian category

The purpose of this section is to give an example of a “big” abelian category $\mathcal{A}$ and objects $M, N$ such that the collection of isomorphism classes of extensions $\mathop{\mathrm{Ext}}\nolimits _\mathcal {A}(M, N)$ is not a set. The example is due to Freyd, see [page 131, Exercise A, Freyd].

We define $\mathcal{A}$ as follows. An object of $\mathcal{A}$ consists of a triple $(M, \alpha , f)$ where $M$ is an abelian group and $\alpha $ is an ordinal and $f : \alpha \to \text{End}(M)$ is a map. A morphism $(M, \alpha , f) \to (M', \alpha ', f')$ is given by a homomorphism of abelian groups $\varphi : M \to M'$ such that for *any* ordinal $\beta $ we have

Here the rule is that we set $f(\beta ) = 0$ if $\beta $ is not in $\alpha $ and similarly we set $f'(\beta )$ equal to zero if $\beta $ is not an element of $\alpha '$. We omit the verification that the category so defined is abelian.

Consider the object $Z = (\mathbf{Z}, \emptyset , f)$, i.e., all the operators are zero. The observation is that computed in $\mathcal{A}$ the group $\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {A}(Z, Z)$ is a proper class and not a set. Namely, for each ordinal $\alpha $ we can find an extension $(M, \alpha + 1, f)$ of $Z$ by $Z$ whose underlying group is $M = \mathbf{Z} \oplus \mathbf{Z}$ and where the value of $f$ is always zero except for

This clearly produces a proper class of isomorphism classes of extensions. In particular, the derived category of $\mathcal{A}$ has proper classes for its collections of morphism, see Derived Categories, Lemma 13.27.6. This means that some care has to be exercised when defining Verdier quotients of triangulated categories.

**Proof.**
See discussion above.
$\square$

## 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Your email address will not be published.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 (0)