## Tag `010I`

## 12.6. Extensions

Definition 12.6.1. Let $\mathcal{A}$ be an abelian category. Let $A, B \in \mathop{\rm Ob}\nolimits(\mathcal{A})$. An

extension $E$ of $B$ by $A$is a short exact sequence $$ 0 \to A \to E \to B \to 0. $$ Anmorphism of extensionsbetween 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 $A, B \in \mathop{\rm Ob}\nolimits(\mathcal{A})$. The set of isomorphism classes of extensions of $B$ by $A$ is denoted $$ \mathop{\rm Ext}\nolimits_\mathcal{A}(B, A). $$ This is called the

$\mathop{\rm Ext}\nolimits$-group.This definition works, because by our conventions $\mathcal{A}$ is a set, and hence $\mathop{\rm Ext}\nolimits_\mathcal{A}(B, A)$ 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 $\mathop{\rm Ext}\nolimits_\mathcal{A}(B, A)$ 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 $\mathop{\rm Ext}\nolimits_\mathcal{A}(-, -)$ into a functor $$ \mathcal{A} \times \mathcal{A}^{opp} \longrightarrow \textit{Sets}, \quad (A, B) \longmapsto \mathop{\rm Ext}\nolimits_\mathcal{A}(B, A) $$ 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 sumof 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 $\mathop{\rm Ext}\nolimits_\mathcal{A}(B, A)$ 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 $$ \xymatrix{ 0 \ar[r] & \mathop{\rm Hom}\nolimits_\mathcal{A}(M_3, N) \ar[r] & \mathop{\rm Hom}\nolimits_\mathcal{A}(M_2, N) \ar[r] & \mathop{\rm Hom}\nolimits_\mathcal{A}(M_1, N) \ar[lld] \\ & \mathop{\rm Ext}\nolimits_\mathcal{A}(M_3, N) \ar[r] & \mathop{\rm Ext}\nolimits_\mathcal{A}(M_2, N) \ar[r] & \mathop{\rm Ext}\nolimits_\mathcal{A}(M_1, N) } $$ for all objects $N$ of $\mathcal{A}$, and
- there is a canonical six term exact sequence of abelian groups $$ \xymatrix{ 0 \ar[r] & \mathop{\rm Hom}\nolimits_\mathcal{A}(N, M_1) \ar[r] & \mathop{\rm Hom}\nolimits_\mathcal{A}(N, M_2) \ar[r] & \mathop{\rm Hom}\nolimits_\mathcal{A}(N, M_3) \ar[lld] \\ & \mathop{\rm Ext}\nolimits_\mathcal{A}(N, M_1) \ar[r] & \mathop{\rm Ext}\nolimits_\mathcal{A}(N, M_2) \ar[r] & \mathop{\rm Ext}\nolimits_\mathcal{A}(N, M_3) } $$ 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. $\square$

The code snippet corresponding to this tag is a part of the file `homology.tex` and is located in lines 1089–1290 (see updates for more information).

```
\section{Extensions}
\label{section-extensions}
\begin{definition}
\label{definition-extension}
Let $\mathcal{A}$ be an abelian category.
Let $A, B \in \Ob(\mathcal{A})$.
An {\it extension $E$ of $B$ by $A$} is a short
exact sequence
$$
0 \to A \to E \to B \to 0.
$$
An {\it 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.
\end{definition}
\noindent
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.
\begin{definition}
\label{definition-ext-group}
Let $\mathcal{A}$ be an abelian category.
Let $A, B \in \Ob(\mathcal{A})$.
The set of isomorphism classes of extensions
of $B$ by $A$ is denoted
$$
\Ext_\mathcal{A}(B, A).
$$
This is called the {\it $\Ext$-group}.
\end{definition}
\noindent
This definition works, because by our conventions
$\mathcal{A}$ is a set, and hence
$\Ext_\mathcal{A}(B, A)$ is a set.
In any of the cases of ``big'' abelian categories
listed in Categories, Remark \ref{categories-remark-big-categories}
one can check by hand that $\Ext_\mathcal{A}(B, A)$
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.
\medskip\noindent
Actually we can turn $\Ext_\mathcal{A}(-, -)$ into a
functor
$$
\mathcal{A} \times \mathcal{A}^{opp} \longrightarrow \textit{Sets}, \quad
(A, B) \longmapsto \Ext_\mathcal{A}(B, A)
$$
as follows:
\begin{enumerate}
\item 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 {\it pullback of $E$ via
$B' \to B$}.
\item 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 {\it pushout of $E$ via
$A \to A'$}.
\end{enumerate}
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.
\medskip\noindent
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 {\it Baer sum} of the
given extensions.
\begin{lemma}
\label{lemma-baer-sum}
The construction $(E_1, E_2) \mapsto E_1 + E_2$
above defines a commutative group
law on $\Ext_\mathcal{A}(B, A)$ which is
functorial in both variables.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-six-term-sequence-ext}
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 $\mathcal{A}$.
\begin{enumerate}
\item There is a canonical six term exact sequence of abelian groups
$$
\xymatrix{
0 \ar[r] &
\Hom_\mathcal{A}(M_3, N) \ar[r] &
\Hom_\mathcal{A}(M_2, N) \ar[r] &
\Hom_\mathcal{A}(M_1, N) \ar[lld] \\
& \Ext_\mathcal{A}(M_3, N) \ar[r] &
\Ext_\mathcal{A}(M_2, N) \ar[r] &
\Ext_\mathcal{A}(M_1, N)
}
$$
for all objects $N$ of $\mathcal{A}$, and
\item there is a canonical six term exact sequence of abelian groups
$$
\xymatrix{
0 \ar[r] &
\Hom_\mathcal{A}(N, M_1) \ar[r] &
\Hom_\mathcal{A}(N, M_2) \ar[r] &
\Hom_\mathcal{A}(N, M_3) \ar[lld] \\
& \Ext_\mathcal{A}(N, M_1) \ar[r] &
\Ext_\mathcal{A}(N, M_2) \ar[r] &
\Ext_\mathcal{A}(N, M_3)
}
$$
for all objects $N$ of $\mathcal{A}$.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted. Hint: The boundary maps are defined using either the pushout
or pullback of the given short exact sequence.
\end{proof}
```

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