Lemma 109.61.1. There exists a “big” abelian category $\mathcal{A}$ whose $\mathop{\mathrm{Ext}}\nolimits $-groups are proper classes.

## 109.61 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.
$\square$

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