braid lemma


Diagram chasing lemmas

Homological algebra

homological algebra

(also nonabelian homological algebra)



Basic definitions

Stable homotopy theory notions



diagram chasing

Homology theories





(braid lemma)

Given a commuting diagram of abelian groups of the following form

A B C D E F G H I J K. \array{ && A && \longrightarrow && B && \longrightarrow && C && \longrightarrow && D \\ & \nearrow && \searrow && \nearrow && \searrow && \nearrow && \searrow && \nearrow \\ E && && F && && G && && H \\ & \searrow && \nearrow && \searrow && \nearrow && \searrow && \nearrow \\ && I && \longrightarrow && J && \longrightarrow && K } \,.

Consider the following four sequences inside the diagram

  1. EABGKE\to A \to B \to G \to K;

  2. EIJGCDE \to I \to J \to G \to C \to D;

  3. AFJKHDA \to F \to J \to K \to H \to D;

  4. IFBCHI \to F \to B \to C \to H.

Then: if the first three of these are long exact sequences and the fourth is a chain complex, then the fourth is also long exact.

(Munkres, §26, Exc.)


Long exact sequence of a triple in homology

Given a generalized homology theory H H_\bullet, then by definition for every inclusion AXA \hookrightarrow X of topological spaces, there is an long exact sequence

H n+1(X,A)δH n(A)H n(X)H n(X,A). \cdots \to H_{n+1}(X,A) \overset{\delta}{\longrightarrow} H_n(A) \longrightarrow H_n(X)\longrightarrow H_n(X,A) \to \cdots \,.

For H H_\bullet an unreduced generalized homology theory and for two consecutive inclusions ZYXZ \hookrightarrow Y \hookrightarrow X (a “triple” (X,Y,Z)(X, Y, Z)) there is a long exact sequence of the form

H n+1(X,Y)H n(Y,Z)H n(X,Z)H n(X,Y). \cdots H_{n+1}(X,Y) \to H_{n}(Y,Z) \to H_n(X,Z) \to H_n(X,Y) \to \cdots \,.

Dually for generalized (Eilenberg-Steenrod) cohomology.


Consider the following braid diagram

(graphics from this Maths.SE comment)

The blue, purple and the black sequence are the exact sequences of the pairs (X,Y)(X,Y), (Z,Y)(Z,Y) and (Z,X)(Z,X), respectively. The orange sequence is a chain complex by these exact sequences and using the commutativity of the diagram, and because the diagonals H (Y,Z)H (X,Z)H (X,Y)H_\bullet(Y,Z) \to H_\bullet(X,Z) \to H_\bullet(X,Y) factor through H (Y,Y)=0H_\bullet(Y,Y) = 0. Hence the braid lemma, prop. , implies the claim.


  • Munkres, p. 148 of Elements of algebraic topology

Created on April 12, 2016 at 04:48:16. See the history of this page for a list of all contributions to it.