## 12.6 Extensions

Definition 12.6.1. Let $\mathcal{A}$ be an abelian category. Let $A, B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. An *extension $E$ of $B$ by $A$* is a short exact sequence

\[ 0 \to A \to E \to B \to 0. \]

An *morphism of extensions* between two extensions $0 \to A \to E \to B \to 0$ and $0 \to A \to F \to B \to 0$ means a morphism $f : E \to F$ in $\mathcal{A}$ making the diagram

\[ \xymatrix{ 0 \ar[r] & A \ar[r] \ar[d]^{\text{id}} & E \ar[r] \ar[d]^ f & B \ar[r] \ar[d]^{\text{id}} & 0 \\ 0 \ar[r] & A \ar[r] & F \ar[r] & B \ar[r] & 0 } \]

commutative. Thus, the extensions of $B$ by $A$ form a category.

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.

Definition 12.6.2. Let $\mathcal{A}$ be an abelian category. Let $A, B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. The set of isomorphism classes of extensions of $B$ by $A$ is denoted

\[ \mathop{\mathrm{Ext}}\nolimits _\mathcal {A}(B, A). \]

This is called the *$\mathop{\mathrm{Ext}}\nolimits $-group*.

This definition works, because by our conventions $\mathcal{A}$ 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

\[ \mathcal{A} \times \mathcal{A}^{opp} \longrightarrow \textit{Sets}, \quad (A, B) \longmapsto \mathop{\mathrm{Ext}}\nolimits _\mathcal {A}(B, A) \]

as follows:

Given a morphism $B' \to B$ and an extension $E$ of $B$ by $A$ we define $E' = E \times _ B B'$ so that 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 $B' \to B$*.

Given a morphism $A \to A'$ and an extension $E$ of $B$ by $A$ we define $E' = A' \amalg _ A E$ so that 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

\[ A' \amalg _ A (E \times _ B B') \cong (A' \amalg _ A E)\times _ B B' \]

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 pull back by the diagonal map $B \to B \oplus B$ and we push out by the sum map $A \oplus A \to A$ to get an extension $E_1 + E_2$ of $B$ by $A$.

\[ \xymatrix{ 0 \ar[r] & A \oplus A \ar[r] \ar[d]^{\sum } & E_1 \oplus E_2 \ar[r] \ar[d] & B \oplus B \ar[r] \ar[d] & 0 \\ 0 \ar[r] & A \ar[r] & E' \ar[r] & B \oplus B \ar[r] & 0\\ 0 \ar[r] & A \ar[r] \ar[u] & E_1 + E_2 \ar[r] \ar[u] & B \ar[r] \ar[u]^{\Delta } & 0 } \]

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

Lemma 12.6.3. The construction $(E_1, E_2) \mapsto E_1 + E_2$ above defines a commutative group law on $\mathop{\mathrm{Ext}}\nolimits _\mathcal {A}(B, A)$ which is functorial in both variables.

**Proof.**
Omitted.
$\square$

Lemma 12.6.4. Let $\mathcal{A}$ be an abelian category. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a short exact sequence in $\mathcal{A}$.

There is a canonical six term exact sequence of abelian groups

\[ \xymatrix{ 0 \ar[r] & \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(M_3, N) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(M_2, N) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(M_1, N) \ar[lld] \\ & \mathop{\mathrm{Ext}}\nolimits _\mathcal {A}(M_3, N) \ar[r] & \mathop{\mathrm{Ext}}\nolimits _\mathcal {A}(M_2, N) \ar[r] & \mathop{\mathrm{Ext}}\nolimits _\mathcal {A}(M_1, N) } \]

for all objects $N$ of $\mathcal{A}$, and

there is a canonical six term exact sequence of abelian groups

\[ \xymatrix{ 0 \ar[r] & \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(N, M_1) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(N, M_2) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(N, M_3) \ar[lld] \\ & \mathop{\mathrm{Ext}}\nolimits _\mathcal {A}(N, M_1) \ar[r] & \mathop{\mathrm{Ext}}\nolimits _\mathcal {A}(N, M_2) \ar[r] & \mathop{\mathrm{Ext}}\nolimits _\mathcal {A}(N, M_3) } \]

for all objects $N$ of $\mathcal{A}$.

**Proof.**
Omitted. Hint: The boundary maps are defined using either the pushout or pullback of the given short exact sequence.
$\square$

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