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One of the Eilenberg-Steenrod axioms for unreduced cohomology is excision, which states that $H^n(X,A)\cong H^n(X\setminus U,A\setminus U)$, for good subspaces such as when $\overset{\circ}U\subseteq \bar A.$ An equivalent statement is that if $(X,A,B)$ is an excisive triad (with $X=\overset{\circ}A\cup\overset{\circ}B$) then $H^n(X,B)\cong H^n(B,A\cap B).$

The excision isomorphism can be used to construct the Mayer-Vietoris long exact sequence for a cohomology theory. Notice that exactness of the Mayer-Vietoris sequence says that $\ker(H^n(A)\oplus H^n(B)\to H^n(A\cap B))=H^n(A)\underset{H^n(A\cap B)}\times H^n(B)=\text{im}(H^n(A\cup B)\to H^n(A)\oplus H^n(B))$

A hypothesis in a common version of the Brown representability theorem is that a contravariant functor $F$ on the homotopy category of spaces satisfy what in this context is called the Mayer-Vietoris axiom, also called an excisive functor, if the map $F(A\cup B)\to F(A)\underset{F(A\cap B)}\times F(B)$ is surjective. That is, for every elements $x\in F(A)$ and $y\in F(B)$ whose restrictions to $A\cap B$ are equal, there is some element $z\in F(A\cup B)$ which restricts on $A$ to $x$ and on $B$ to $y$. So it is a sheaf theoretic gluing axiom, without the uniqueness requirement.

Clearly the classical Mayer-Vietoris exact sequence implies that a cohomology theory is an excisive functor in the sense of Brown. Are the two notions equivalent? Why did Brown choose this seemingly unique form of the criterion? Does anyone phrase the Eilenberg-Steenrod axioms in this kind of sheaf language?

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What's $\overset{\circ}U$? – Sanath K. Devalapurkar Jun 17 '14 at 2:30
@SanathDevalapurkar: It denotes the interior of $ U $. – Berrick Caleb Fillmore Jun 17 '14 at 2:33

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