how to show that $E(A,B_1\oplus B_2)\cong E(A,B_1)\times E(A,B_2)$

Question

Show that $E(A,B_1\oplus B_2)\cong E(A,B_1)\times E(A,B_2)$.$E(A,B)$ here means the set of equivalence classes of extensions of A by B.It's a exercise from GTM 4 ,Chapter 3,but I don't know how to prove it.

Answer 1

First note that if $F$ is any covariant functor between two categories with products, then there is always a natural comparison map $F(A\times B) \to FA \times FB$ induced by $F(p_A)$ and $F(p_B)$. We need to check that in the case of $E(A,-):{_\Lambda\textrm{Mod}}\to \textrm{Set}$ the comparison map is an isomorphism.

Denote the two projection maps of the biproduct $B_1\oplus B_2$ by $p_1$ and $p_2$ and let $\phi: E(A,B_1\oplus B_2) \to E(A,B_1) \times E(A,B_2)$ be the function $\phi(S) = (p_{1*}S, p_{2*}S)$. We like to construct an inverse $\psi$. I found the right formula through experimenting a bit. If $S_1$ and $S_2$ are two exact sequences $S_1= 0 \to B_1 \to E_1 \to A \to 0$ and $S_2 = 0 \to B_2 \to E_2 \to A \to 0$, then denote by $S_1 \oplus S_2$ the exact sequence $0\to B_1\oplus B_2\to E_1\oplus E_2 \to A\oplus A \to 0$. Define \begin{align*} \psi(S_1,S_2) = E(\Delta_A, B_1\oplus B_2)(S_1\oplus S_2) \end{align*} where $\Delta_A: A \to A\oplus A$ denotes the diagonal map $a \mapsto (a,a)$. Let's check that $\psi\phi = id$. Given a short exact seqnece $S$, write $\phi S = (S_1,S_2)$. By definition there are diagrams

for $k = 1,2$ with exact rows and such that the left hand square is a pushout. Now the diagram

commutes and the right hand square is a pullback (for example because of lemma II.1.3 in Hilton and Stammbach's A course in Homological Algebra). This means that indeed $E(\Delta_A,B_1\oplus B_2)(E(A,p_1)S \oplus E(A,p_2)S) = S$. In the same way one checks that $\phi\psi = id$. A good reference are the first sections of chapter 3 in the book by Hilton and Stammbach.