The Stacks project

12.6 Extensions

By abuse of language we often omit mention of the morphisms $A \to E$ and $E \to B$, although they are definitively part of the structure of an extension.

This definition works, because by our conventions $\mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ is a set, and hence $\mathop{\mathrm{Ext}}\nolimits _\mathcal {A}(B, A)$ is a set. In any of the cases of “big” abelian categories listed in Categories, Remark 4.2.2 one can check by hand that $\mathop{\mathrm{Ext}}\nolimits _\mathcal {A}(B, A)$ is a set as well. Also, we will see later that this is always the case when $\mathcal{A}$ has either enough projectives or enough injectives. Insert future reference here.

Actually we can turn $\mathop{\mathrm{Ext}}\nolimits _\mathcal {A}(-, -)$ into a functor

as follows:

1. Given a morphism $B' \to B$ and an extension $E$ of $B$ by $A$ we define $E' = E \times _ B B'$. By Lemma 12.5.12 we have the following commutative diagram of short exact sequences

   $$\xymatrix{ 0 \ar[r] & A \ar[r] \ar[d] & E' \ar[r] \ar[d] & B' \ar[r] \ar[d] & 0 \\ 0 \ar[r] & A \ar[r] & E \ar[r] & B \ar[r] & 0 }$$

   The extension $E'$ is called the pullback of $E$ via $B' \to B$.

2. Given a morphism $A \to A'$ and an extension $E$ of $B$ by $A$ we define $E' = A' \amalg _ A E$. By Lemma 12.5.13 we have the following commutative diagram of short exact sequences

   $$\xymatrix{ 0 \ar[r] & A \ar[r] \ar[d] & E \ar[r] \ar[d] & B \ar[r] \ar[d] & 0 \\ 0 \ar[r] & A' \ar[r] & E' \ar[r] & B \ar[r] & 0 }$$

   The extension $E'$ is called the pushout of $E$ via $A \to A'$.

To see that this defines a functor as indicated above there are several things to verify. First of all functoriality in the variable $B$ requires that $(E \times _ B B') \times _{B'} B'' = E \times _ B B''$ which is a general property of fibre products. Dually one deals with functoriality in the variable $A$. Finally, given $A \to A'$ and $B' \to B$ we have to show that

as extensions of $B'$ by $A'$. Recall that $A' \amalg _ A E$ is a quotient of $A' \oplus E$. Thus the right hand side is a quotient of $A' \oplus E \times _ B B'$, and it is straightforward to see that the kernel is exactly what you need in order to get the left hand side.

Note that if $E_1$ and $E_2$ are extensions of $B$ by $A$, then $E_1\oplus E_2$ is an extension of $B \oplus B$ by $A\oplus A$. We push out by the sum map $A \oplus A \to A$ and we pull back by the diagonal map $B \to B \oplus B$ to get an extension $E_1 + E_2$ of $B$ by $A$.

The extension $E_1 + E_2$ is called the Baer sum of the given extensions.