Computing some cohomology groups Let $D^n$ be a $n$-dimensional disk, $n>1$, and $X$ a path connected topological space.
I need to show that $H^{n+k}(D^n\times X/(\partial D^n)\times X)\cong H^{k}(X)$.
Any ideas?
I've tried the Mayer-Vietoris sequence but didn't succeed. Also, I've tried thinking about relative cohomology.
 A: I am going to consider the homeomorphic space $X\times I^n/(X \times \partial I^n)$ instead, for reasons which will become apparent. First some terminology might help.
For $(X,x_0)$ a pointed topological space, define the reduced suspension
$$ \Sigma X = X \times I/\sim $$
where $(x,v)\sim (x',v')$ iff either $1)$ $v = v' = 1$, $2)$ $v = v' = 0$, or $3)$ $x = x' = x_0$. You can visualize this by taking the suspension as usual and then collapsing the subspace $\{x_0\} \times I$. Note that in particular the entire subspace $X \times \partial I$ is identified with the basepoint, and this definition is equivalent to $X\times I /(X \times \partial I \cup x_0 \times I)$. It is a fact that if $X$ is "well-pointed" (i.e. the inclusion of the basepoint has the homotopy extension property) then the reduced suspension is homotopy-equivalent to the usual suspension.
Since $\Sigma X$ is still pointed by the equivalence class of $x_0$, we can iterate this process and define $\Sigma^n X = \Sigma(\Sigma^{n-1}X)$ for $n >1$. By induction you can prove the following formula:
$$ \Sigma^n X \cong X \times I^n / (X\times \partial I^n \cup x_0 \times I^n). $$
This looks very similar to our quotient $X\times I^n/(X\times \partial I^n)$ but in our case we don't identify the subspace with any point in $X$, and $X$ was never assumed to be pointed to begin with. However, if $X$ is any space (not necessarily pointed) we can form the well-pointed space $X_+ = X \sqcup \{+\}$ with disjoint basepoint "$+$", and my claim, which you should prove, is that (as long as $X$ is non-empty)
$$X\times I^n / (X \times \partial I^n) \cong \Sigma^n(X_+).$$ 
Since $X_+$ is well-pointed it follows that the reduced suspension is homotopy equivalent to the usual suspension, so we still have a suspension isomorphism. In particular if $k\geq 0$ and $n>0$ then
$$ H^k(X) \cong \tilde{H}^k(X_+) \cong \tilde{H}^{k+n}(\Sigma^n X_+) \cong H^{k+n}(X\times I^n/(X\times \partial I^n)) $$
(where the last group can be written unreduced because $k+n > 0$).

Although the argument outlined above is very geometric in nature and doesn't require a lot of algebraic machinery, I think the argument using the cohomology cross product given in "Characteristic Classes" by Milnor and Stasheff (found starting on page 265 in Appendix A) is very enlightening from an algebraic point of view, and feels less "tricky".
