$\operatorname{Ext}^{n}$ as the class of Yoneda extensions of degree $n$. Given an abelian category $\mathcal{A}$, we can define  $\operatorname{Ext}^{n}(A,B)$ as the class of extensions of degree $n$ of $A$ by $B$.  How can one prove that  $\operatorname{Ext}^{n}(A,B)$, is a set?
In Weibel's book (Introduction to Homology) he says that this is a consequence of Freyd's Embedding Theorem. I cannot see how you can use this for an arbitrary abelian category.
 A: This is in fact not true for a general (locally small) abelian category.  To get a counterexample, let $R=\mathbb{Z}[x_\alpha]_{\alpha\in Ord}$ be a (proper class-sized) polynomial ring over $\mathbb{Z}$ with one variable $x_\alpha$ for each ordinal $\alpha$, and let $\mathcal{A}$ be the category of (set-sized) $R$-modules on which all but a set of the variables act trivially (this last condition allows the category $\mathcal{A}$ to actually be defined as a class in ZFC, if you care about such technicalities).  Let $I\subset R$ be the ideal generated by all the variables and consider $A=B=R/I$, i.e. $A=B=\mathbb{Z}$ with every variable acting trivially.  For any $\alpha\in Ord$, let $I_\alpha\subseteq R$ be the ideal generated by $x_\alpha^2$ and $x_\beta$ for each ordinal $\beta\neq \alpha$, and let $M_\alpha=R/I_\alpha$.
Then for each $\alpha$, we have an extension $0\to B\to M_\alpha\to A\to 0$ where the map $B\to M_\alpha$ sends $1$ to $x_\alpha$ and the map $M_\alpha\to A$ sends $1$ to $1$.  It is not hard to verify that these extensions are all inequivalent.  Thus in $\mathcal{A}$, $\operatorname{Ext}^1(A,B)$ is a proper class.
