Construction of Ext group

Question

Let $A$ and $B$ abelian groups. A extension of $A$ by $B$ is a short short exact sequence $0\longrightarrow A\longrightarrow G \longrightarrow B \longrightarrow 0 $.

Two extensions of $A$ by $B$ are said equivalent if exist $\varphi$ making the diagram commute:

$0\longrightarrow A\longrightarrow G \longrightarrow B \longrightarrow 0 $

$\,\ \,\ \,\ \,\ \,\ \,\ \downarrow \,\ \,\ \,\ \,\ \, \downarrow \varphi \,\ \,\ \,\ \,\ \downarrow$

$0\longrightarrow A\longrightarrow G' \longrightarrow B \longrightarrow 0 $

I denote [G] the equivalence class of extension $0\longrightarrow A\longrightarrow G \longrightarrow B \longrightarrow 0 $.

I know that ${\rm Ext}(B,A):=\{[G]:0\longrightarrow A\longrightarrow G \longrightarrow B \longrightarrow 0 $ is a extension of $A$-by-$B \}$ can be structured as abelian group but i ask why ${\rm Ext}(B,A)$ is a set and no a proper class?

Answer 1

Let $s: B\to G$ be a (set-theoretic) section of $p: G\to B$, i.e. a function whose composite with $G\to B$ is the identity of $B$.

Because $G\to B$ is surjective, you can find such an $s$ - it will, in general, not be a group morphism, just a set function.

Then the following (set-theoretic) function is a bijection $ G\to A\times B, g\mapsto (a, p(g))$ where $a $ is the only antecedent in $A$ of $g-sp(g)$. This is an easy exercise.

Now, up to isomorphism, there is only a set of groups with a bounded cardinality (here $A\times B$), and for each of them, only a set of pairs of morphisms $A\to G, G\to B$, so that overall, up to isomorphism, there is only a set of extensions.

More precisely, you could take "group structures on the underlying set of $A\times B$" and that would be enough.

Answer 2

$\newcommand{\Ext}{\operatorname{Ext}_\mathbb Z^1} \newcommand{\cl}[1]{[\![ #1 ]\!]}$Just to humour the suggestion in the comments, consider the class $\cl\xi$ of an extension of $N$ by $M$, as in your post $$\xi : 0\to N\to E\to M \to 0,$$

and choose a presentation of $M$ of the form

$$0\to F_1\to F_0\to M\to 0.$$

Define $\Ext(M,N) = \operatorname{coker}(\hom(F_0,N)\to \hom(F_1,N))$. Obviously, this is a set, so it suffices to show there is a bijection between elements in this kernel and the collection of classes $\mathcal E(M,N)$. Define a map $$\Phi: \Ext(M,N) \to \mathcal E(M,N)$$ as follows: given a class of maps $\varphi: F_1\to N$, form two pushouts with the presentation above to get $\require{AMScd}$ \begin{CD} 0 @>>> F_1 @>>> F_0 @>>> M @>>> 0 \\ {} @VVV @VVV @VVV \\ 0 @>>> N @>>> E @>>> M' @>>> 0 \end{CD} The module $M'$ is isomorphic to $M$ via the last map, which you can in fact take to be the identity, as the pushout of $N\leftarrow F_1\rightarrow M$ is just $M$ (since $F_1\to M$ is the zero map). This also proves that the bottom sequence is at least a complex, and you can show it is exact.

Thus, you get from $\cl{\varphi}$ a class $\Phi\cl \varphi = \cl \xi$. This is well defined, for if $\psi$ factors through, you can compute the pushout in two steps, and check that you indeed get $M\times N$ and the resulting SEC is (isomorphic to one which is) trivial. You can prove this by computing the first pushout in two steps (using the map factors).

At the same time, define $\Psi : \mathcal E(M,N) \to \Ext(M,N)$ by appending $F_0\to M$ at the end of $\xi$ and taking two pullbacks. Since pullbacks do not affect kernels, you will at the end get a map $\varphi: F_1\to N$, and you can look at its class in $\Ext(M,N)$.

It is a bit time consuming (but otherwise doable!) to show that $\Phi$ and $\Psi$ are inverse bijections.