Ext (math)

Ext is a bifunctor that plays a central role in homological algebra .

definition

Let be an Abelian category , for example the category of the modules of a ring , which is the standard example according to Mitchell's embedding theorem . To two objects and from is the class of the short exact sequences of the form ${\displaystyle {\mathcal {A}}}$ ${\displaystyle X}$ ${\displaystyle Z}$ ${\displaystyle {\mathcal {A}}}$ ${\displaystyle {\mathcal {E}}}$

${\displaystyle 0\rightarrow X\rightarrow Y\rightarrow Z\rightarrow 0.}$

On is now an equivalence relation defined. Two exact sequences and are equivalent if there is a morphism , so the diagram ${\displaystyle {\mathcal {E}}}$ ${\displaystyle 0\rightarrow X\rightarrow Y\rightarrow Z\rightarrow 0}$ ${\displaystyle 0\rightarrow X\rightarrow Y'\rightarrow Z\rightarrow 0}$ ${\displaystyle g\colon Y\to Y'}$

${\displaystyle {\begin{matrix}0&\to &X&\to &Y&\to &Z&\to &0\\&&\downarrow \operatorname {id} &&\downarrow g&&\downarrow \operatorname {id} \\0&\to &X&\to &Y'&\to &Z&\to &0\end{matrix}}}$

commutes. It is the identical morphism. ${\displaystyle \operatorname {id} }$

From the five lemma it follows immediately that if there is such a morphism , it must be an isomorphism . The class modulo of this equivalence relation is a set and is denoted by. A group structure can be defined on this set. ${\displaystyle g}$ ${\displaystyle {\mathcal {E}}}$ ${\displaystyle \mathrm {Ext} (Z,X)}$

Functoriality

Morphisms in the Abelian category induce morphisms between the Ext groups in the following way, so that becomes a two-digit functor. ${\displaystyle \mathrm {Ext} }$

You can push-out to and from the sequence : ${\displaystyle g\colon X\to X'}$ ${\displaystyle 0\rightarrow X\rightarrow Y\rightarrow Z\rightarrow 0}$

${\displaystyle {\begin{matrix}0&\to &X&\to &Y&\to &Z&\to &0\\&&\downarrow g&&\downarrow &&\\0&\to &X'&\to &Y'\end{matrix}}}$

Because of the universal property of the push-out there is an induced epimorphism from Y 'to Z, so that the following diagram commutes:

${\displaystyle {\begin{matrix}0&\to &X&\to &Y&\to &Z&\to &0\\&&\downarrow g&&\downarrow &&\downarrow \operatorname {id} \\0&\to &X'&\to &Y'&\to &Z&\to &0\end{matrix}}}$

The bottom line is also exact and its equivalence class is therefore an element in . ${\displaystyle \mathrm {Ext} (Z,X')}$

If the equivalence class of is mapped to the equivalence class of , a well-defined group homomorphism is obtained . ${\displaystyle 0\rightarrow X\rightarrow Y\rightarrow Z\rightarrow 0}$ ${\displaystyle 0\rightarrow X'\rightarrow Y'\rightarrow Z\rightarrow 0}$ ${\displaystyle \mathrm {Ext} (Z,X)\rightarrow \mathrm {Ext} (Z,X')}$

This also works dual with morphisms from Z 'to Z. The following pull-back can be made to the sequence : ${\displaystyle g\colon Z'\to Z}$ ${\displaystyle 0\rightarrow X\rightarrow Y\rightarrow Z\rightarrow 0}$

${\displaystyle {\begin{matrix}&&&&Y'&\to &Z'&\to &0\\&&&&\downarrow &&\downarrow g\\0&\to &X&\to &Y&\to &Z&\to &0\end{matrix}}.}$

Because of the universal property of the pull-back, there is an induced monomorphism from X to Y ', so that the following diagram commutes:

${\displaystyle {\begin{matrix}0&\to &X&\to &Y'&\to &Z'&\to &0\\&&\downarrow \operatorname {id} &&\downarrow &&\downarrow g\\0&\to &X&\to &Y&\to &Z&\to &0\end{matrix}}}$

The top line is also exact and thus defines an element in . ${\displaystyle \mathrm {Ext} (Z',X)}$

If the equivalence class of is mapped onto the equivalence class of , one again obtains a well-defined group homomorphism . ${\displaystyle 0\rightarrow X\rightarrow Y\rightarrow Z\rightarrow 0}$ ${\displaystyle 0\rightarrow X\rightarrow Y'\rightarrow Z'\rightarrow 0}$ ${\displaystyle \mathrm {Ext} (Z,X)\rightarrow \mathrm {Ext} (Z',X)}$

Ext as the derivative of the Hom functor

Another way of defining it uses the derived functors of Hom . The construction defined above can be identified with the first right derivative of the Hom functor.

More precisely, one considers an Abelian category with a sufficient number of projective objects (i.e. each object is the quotient of a projective object) and defines the contravariant functor ${\displaystyle \mathrm {Hom} (-,X)}$

${\displaystyle \mathrm {Ext} ^{n}(Z,X):=R_{n}\mathrm {Hom} (-,X)(Z)}$,

that is, one forms the -th legal derivative of and applies the functor thus created . ${\displaystyle n}$ ${\displaystyle \mathrm {Hom} (-,X)}$ ${\displaystyle Z}$

More specifically, this means the following: Let it be and ${\displaystyle n\geq 1}$

${\displaystyle {\begin{array}{ccccc}\ldots \rightarrow &P_{n}&\rightarrow &P_{n-1}&\rightarrow \ldots \rightarrow Z\rightarrow 0\\&\lambda _{n}\downarrow &\nearrow \kappa _{n}\\&K_{n}\end{array}}}$

a projective resolution of having an epimorphism and a monomorphism such that . Next is the induced homomorphism ${\displaystyle Z}$ ${\displaystyle \lambda _{n}:P_{n}\rightarrow K_{n}}$ ${\displaystyle \kappa _{n}:K_{n}\rightarrow P_{n-1}}$ ${\displaystyle (P_{n}\rightarrow P_{n-1})=\kappa _{n}\circ \lambda _{n}}$ ${\displaystyle \kappa _{n}^{*}=\mathrm {Hom} (\kappa _{n},X)}$

${\displaystyle \kappa _{n}^{*}:\mathrm {Hom} (P_{n-1},X)\rightarrow \mathrm {Hom} (K_{n},X),\,f\mapsto f\circ \kappa _{n}}$.

Then

${\displaystyle \mathrm {Ext} ^{n}(Z,X)\cong \mathrm {coker} (\kappa _{n}^{*})=\mathrm {Hom} (K_{n},X)/\kappa _{n}^{*}(\mathrm {Hom} (P_{n-1},X))}$.

The elements from are therefore certain equivalence classes of elements from . ${\displaystyle \mathrm {Ext} ^{n}(Z,X)}$ ${\displaystyle \mathrm {Hom} (K_{n},X)}$

Finally, it should be noted that one can swap the roles of and also one receives ${\displaystyle X}$ ${\displaystyle Z}$

${\displaystyle \mathrm {Ext} ^{n}(Z,X)\cong R_{n}\mathrm {Hom} (Z,-)(X)}$.

Relationship between Ext and Ext ¹

This section aims to explain how the constructs and defined above are related. We construct a picture . ${\displaystyle \mathrm {Ext} }$ ${\displaystyle \mathrm {Ext} ^{1}}$ ${\displaystyle \mathrm {Ext} (Z,X)\rightarrow \mathrm {Ext} ^{1}(Z,X)}$

Be a short exact sequence that defines an element from . Next is a short exact sequence with projective . Using the projectivity of , one can create a commutative diagram ${\displaystyle 0\rightarrow X\rightarrow Y\rightarrow Z\rightarrow 0}$ ${\displaystyle \mathrm {Ext} (Z,X)}$ ${\displaystyle 0\rightarrow K\rightarrow P\rightarrow Z\rightarrow 0}$ ${\displaystyle P}$ ${\displaystyle P}$

${\displaystyle {\begin{array}{ccccccc}0\rightarrow &K&\rightarrow &P&\rightarrow &Z&\rightarrow 0\\&\downarrow \psi &&\downarrow \varphi &&\Vert \\0\rightarrow &X&\rightarrow &Y&\rightarrow &Z&\rightarrow 0\end{array}}}$

to construct. Then there is a homomorphism whose equivalence class defines an element as described above . ${\displaystyle \psi \in \mathrm {Hom} (K,X)}$ ${\displaystyle \mathrm {Ext} ^{n}(Z,X)}$ ${\displaystyle \mathrm {Ext} ^{1}(Z,X)}$

If one forms the equivalence class of in the equivalence class of in starting, we obtain a well-defined figure , it can be shown from that it is a group isomorphism is. ${\displaystyle 0\rightarrow X\rightarrow Y\rightarrow Z\rightarrow 0}$ ${\displaystyle \mathrm {Ext} (Z,X)}$ ${\displaystyle \psi }$ ${\displaystyle \mathrm {Ext} ^{1}(Z,X)}$ ${\displaystyle \mathrm {Ext} (Z,X)\rightarrow \mathrm {Ext} ^{1}(Z,X)}$

Therefore one can identify with , that is , in this sense it can be defined as the first legal derivative of the -function. ${\displaystyle \mathrm {Ext} }$ ${\displaystyle \mathrm {Ext} ^{1}}$ ${\displaystyle \mathrm {Ext} }$ ${\displaystyle \mathrm {Hom} }$

Long exact sequence

The Hom functor is left exact, that is, for a short exact sequence

${\displaystyle 0\rightarrow X\rightarrow Y\rightarrow Z\rightarrow 0}$

and another object (module) has an exact sequence ${\displaystyle A}$

${\displaystyle 0\rightarrow \mathrm {Hom} (A,X)\rightarrow \mathrm {Hom} (A,Y)\rightarrow \mathrm {Hom} (A,Z)}$,

and in general this cannot be continued exactly with 0. Because of left-hand precision, the 0th derivative of the Hom functor agrees with Hom, that is, if one extends the above definition of to , one has . The long exact sequence for derived additive functors therefore gives the following exact sequence ${\displaystyle \mathrm {Ext} ^{n}}$ ${\displaystyle n=0}$ ${\displaystyle \mathrm {Ext} ^{0}=\mathrm {Hom} }$

${\displaystyle 0\rightarrow \mathrm {Hom} (A,X)\rightarrow \mathrm {Hom} (A,Y)\rightarrow \mathrm {Hom} (A,Z)}$

${\displaystyle \rightarrow \mathrm {Ext} ^{1}(A,X)\rightarrow \mathrm {Ext} ^{1}(A,Y)\rightarrow \mathrm {Ext} ^{1}(A,Z)\rightarrow \mathrm {Ext} ^{2}(A,X)\rightarrow \ldots }$.

A long exact sequence is obtained analogously

${\displaystyle 0\rightarrow \mathrm {Hom} (Z,A)\rightarrow \mathrm {Hom} (Y,A)\rightarrow \mathrm {Hom} (X,A)}$

${\displaystyle \rightarrow \mathrm {Ext} ^{1}(Z,A)\rightarrow \mathrm {Ext} ^{1}(Y,A)\rightarrow \mathrm {Ext} ^{1}(X,A)\rightarrow \mathrm {Ext} ^{2}(Z,A)\rightarrow \ldots }$.

In this sense, the Ext functors close the gap created by the lack of accuracy of the Hom functor.

Individual evidence

1. ↑ Sergei I. Gelfand & Yuri Ivanovich Manin: Homological Algebra , Springer, Berlin, 1999, ISBN 978-3-540-65378-3
2. ^ Charles A. Weibel: An introduction to homological algebra , Cambridge Studies in Advanced Mathematics, 38, Cambridge University Press, 1999, ISBN 978-0-521-55987-4
3. ^ Peter Hilton: Lectures in Homological Algebra , American Mathematical Society (2005), ISBN 0-8218-3872-5 , sentence 3.13
4. ↑ Peter Hilton: Lectures in Homological Algebra , American Mathematical Society (2005), ISBN 0-8218-3872-5 , Theorem 4.5
5. ↑ Saunders Mac Lane : Homology , Springer Grundlehren der Mathematischen Wissenschaften Volume 114 (1967), chap. III, Theorem 3.4 and Theorem 9.1