Talk:Eilenberg–Steenrod axioms

Revision of 3 April 2006

In the additivity axiom, is the $\vee$ symbol supposed to represent disjoint union, or one-point union, or what? The notation should be named in plain English. If it's disjoint union, then isn't $\sqcup$ a more popular (and less overloaded) choice? Joshua Davis 22:39, 3 April 2006 (UTC)

It's supposed to be the coproduct, which is disjoint union in this case. The \vee has been changed to \coprod . Marc Harper 14:47, 11 August 2006 (UTC)

Uniqueness

There should be a discussion of the Eilenberg-Steenrod Uniqueness Theorem Jfdavis (talk) 08:47, 11 November 2009 (UTC)

Natural transformation

Can someone explain how exactly $\partial : H_{i}(X, A) \to H_{i-1}(A)$ is a natural transformation? I think its supposed to be called the informal "natural map" instead of transformation because $\ H_{i}$ is evaluated at (X,A) whereas $\ H_{i-1}$ is evaluated at (A,Ø), a transformation is suppose to have both functor evaluate at the same point. Money is tight (talk) 05:45, 12 January 2011 (UTC)

Yes, I was asking myself the same question and found an answer in Eilenberg's book (Samuel Eilenberg, Norman E. Steenrod, Foundations of algebraic topology, Princeton University Press, Princeton, New Jersey, 1952.) in chapter IV.7 (page 113). There a functor T from the category of topological pairs to the category Top, such that $T(X,A) = A$ and $T(f) = f|A$, is defined. Then the (correct) statement is, that the boundary operator $\partial$ is a natural transformation from $H_i$ to the composite functor $H_{i-1} T$. --Quiet photon (talk) 11:25, 28 April 2011 (UTC)