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I have some conditional questions. First, in the category of topological pairs (as used in cohomology theories), are these only pairs of the form $(X,A)$ with $A \subset X$?

If the answer to this question is yes, then my next question is: is a morphism $(X,A) \to (Y,B)$ a map $f:X \to Y$ with $f(A) \subset B$ or pairs of maps $f_1: X \to Y$ and $f_2: A \to B$?

Thanks!

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    Thanks Akhil. I asked because I've seen the Thom isomorphism phrased in terms of $H^*(E, E - B)$ and $H^*(D(E), S(E))$, where $E \to B$ is a vector bundle, $D(E)$ is the unit disk and $S(E)$ the sphere. I wanted the pairs $(E,E-B)$ and $(D(E), S(E))$ to be homotopic. But I'm guessing the long exact sequence in cohomology and the five-lemma give that their cohomology groups are naturally isomorphic.2011-02-06

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As Akhil points out it is the first. But you could also think of morphisms in the category of pairs as commutative squares. This is just the typical definition in the category where objects are morphisms and morphisms are squares.

Here is response to your comment: They are in fact homotopic in the category of pairs. The disk bundle is homotopic to the whole of the total space, and $E-B$ really means $E$ with out the image of the zero section which is equivalent to $B$. Think about what happens when you remove $0$ from each fiber. You get something homotopic to the sphere in that vector space.

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    I see that $E-B$ is homotopic to $S(E)$ and that $D(E)$ is homotopic to $E$, but I don't see how there can be a homotopy $(E, E-B)$ to $(D(E), S(E))$.2011-02-07