Let $\Delta : S^n \longrightarrow S^n \times S^ n$ be the diagonal map given by $\Delta (x) = (x,x),$ $x \in S^n.$ Calculate $H_{*} (C(\Delta)),$ the homology of the mapping cone of $\Delta.$
I have tried to use Mayer-Vietoris sequence to solve the problem but couldn't quite able to do it. Is there any specific way to approach such problems? Any help will be greatly appreciated.
Thanks!
If you cover the cone $X := C(\Delta)$ as in the picture, you will have $X = int(A) \cup int(B).$ Note that $A$ contracts to a point, $A \cap B$ is homotopically equivalent to $S^n,$ and $B$ deformation-retracts on $S^n \times S^n.$
Thus, for $k \not= 0,n$ we have $H_k(A \cap B) = 0$ and the Mayer-Vietoris sequence looks as follows:
$$\ldots \to 0 \to H_{n+1}(S^n \times S^n) \to H_{n+1}(X) \xrightarrow{\partial} H_n(S^n) \xrightarrow{\Delta_*} H_n(S^n \times S^n) \to H_n(X) \to 0 \to H_{n-1}(S^n \times S^n) \to H_{n-1}(X) \to 0 \to \ldots$$
Since $\Delta_*$ is injective, $\partial$ is also zero, and thus $H_k(X) \simeq H_k(S^n \times S^n) $ for $k \not= n,$ wheareas $H_n(X)$ is the cokernel of $\Delta_*$.
[Intuitively, you get rid of one of the $n$-holes of $S^n \times S^n$ by glueing a cell along a cycle. It is possible to picture on a torus though I'm not sure how helpful that would be for you. This intuition can of course be made into another formal proof.]