# Relative Cohomology Isomorphic to Cohomology of Quotient

Given a topological space (with nice enough conditions, maybe Hausdorff, compactly generated, or CW complex, I'm not sure) $X$ and a subspace $A\subset X$, is it true that $H^n(X,A)\cong H^n(X\backslash A)$ whenever there is an open set containing $A$ which can be retracted to $A$? Can this be shown easily using the Mayer-Vietoris Sequence, or something else similar to that? (I'm basically assuming simplicial or singular cohomology here, with integer coefficients.)

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That looks like X minus A I guess. I want it to be X quotient A. –  Jon Beardsley Nov 29 '10 at 19:51

It should say reduced cohomology of $X/A$, I believe.
In general, it is always true that the reduced cohomology of $X \cup CA$ (for $CA$ the cone) is isomorphic to the relative cohomology. To see this, one removes a small contractible piece near the top.
The exact sequence shows that $\widetilde{H}^*(X \cup CA) \simeq \widetilde{H}^*(X \cup CA, C A)$ because the cone is contractible. Then, excise the top of the cone. The pair then becomes homotopy equivalent to the pair $(X, A)$. So this proves the claim.
The next fact that one needs to show is that $X/A$ has the same cohomology as $X \cup CA$. This is true if the inclusion $A \to X$ is a cofibration. In general, it is a fact that if $B \subset Y$ is a cofibration, and $B$ is contractible, then $Y$ and $Y/B$ have the same homotopy type. (Apply this with $Y = X \cup CA, B = CA$ to see that $X \cup CA$ and $X/A$ have the same homotopy type under cofibration conditions.)
To see this, one takes a contracting homotopy $B \times I \to B$ of the identity to a point, and extends it to a homotopy $\phi: Y \times I \to Y$ by the cofibration property. This sends $B \times \{1\}$ into a point. So $\phi(., 1)$ is a map $Y/B \to Y$. The composite of $\phi(.,1)$ and $Y \to Y/B$ is homotopic to $1_{Y/B}$ by $\phi$ (since $\phi(.,t)$ always sends $B$ into itself).
The main point I have learned is that $X / A$ is bad homotopically. It just isnt the right thing to work with for the purposes of homotopy theory. But $X \cup CA$ is good, and behaves the way you would have hoped $X / A$ would. –  Sean Tilson Dec 2 '10 at 6:24