The Stacks project

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:

  1. 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 $B' \to B$.

  2. 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 $\mathcal{A}$.

  1. 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

  2. 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. $\square$

Comments (6)

Comment #372 by Fan on

It might be better to point out that in proof of functorality of Ext, the exactness of the new sequence produced follows from Lemma 3.24.

Comment #383 by on

@#372: Well, I think here the discussion assumes the reader has read the basic facts about short exact sequences (including the snake lemma, etc). If you want to make it more precise, then please code up a lemma with complete proof (but I think we should leave a perhaps shortened informal discussion there).

Comment #2437 by Raymond Cheng on

When turning into a bifunctor, you may wish to swap the and : write rather than to be consistent with the result of the page.

Comment #5934 by Bach on

I am not sure this is important, but in the construction of Baer sum, when I read: "we pull back...and we push out..."makes me think that we start by pulling back the diagonal map, but the following diagram is in the different order.

Comment #6123 by on

@#5934. Maybe others can chime in here.

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