# Join of topological spaces; Mayer-Vietoris

let $X$ and $Y$ be topological spaces, $X\star Y:=\frac{X\times Y\times [0,1]}{\sim}$, where $\sim$ is genereted by $(x,y_1,1)\sim (x,y_2,1)$ and $(x_1,y,0)\sim (x_2,y,0)$ for $x, x_1,x_2\in X$, $y, y_1,y_2\in Y$. $X\star Y$ is the join of X and Y.
http://en.wikipedia.org/wiki/Join_%28topology%29
I want to use Mayer-Vietoris to show that there is a short exact sequence
$$0\to\tilde{H}_k(X\star Y)\to \tilde{H}_{k-1}(X\times Y)\to \tilde{H}_{k-1}(X)\oplus\tilde{H}_{k-1}(Y)\to 0$$ ($\tilde{H}_k(.)$ means the reduced singular homology)
The big problem: I don't understand the wikipedia-picture and how to imagine the join. Here Homology group of the join I find how you can choose $A$ and $B$ such that you can aply Mayer-Vietoris, but $A$ and $B$ should be open subsets of $X\star Y$? And why is for example $X$ a deformation retraction of A? Help would be greatly appreciated. Regards

• Imagine the join as building a cube out of your space X, Y and a interval I. Then on {0} of the interval you contract X to a point, and on {1} you contract Y to a point. – Loreno Heer Jan 13 '15 at 22:04

## 2 Answers

Here we have:

$$X*Y=X\times Y\times [0,1]/\sim\\ A=X \times Y \times [0,1) /\sim \\ B=X \times Y \times (0,1] /\sim \\$$

where clearly $A$ and $B$ are subspaces of $X*Y$ with $A\cap B=X\times Y\times (0,1)$.

Then $A$ deformation retracts on to $X$ (or rather, the subspace $X\times Y\times \{0\}/\sim$ which is clearly homeomorphic to $X$) via:

$$F([(x,y,s)],t)=[(x,y,(1-t)s)]$$

$B$ deformation retracts on to $Y$ in a similar way and $X\times Y\times (0,1)$ deformation retracts on to $X\times Y$, since $(0,1)$ is contractible. That should be enough for you to work out what's going on.

The join can be viewed as the sum of two cones like this:

$X\star Y\cong C(X)\times Y\cup_{X\times Y} C(Y)\times X$

Then note that the cone is contractible to a point.

You get that

$$H_n(C(X)\times Y\cap C(Y)\times X) \to H_n(X) \oplus H_n(Y) \to H_n(X\star Y) \to \cdots$$

EDIT: check if this makes sense. Also show that $H_n(X) \oplus H_n(Y) \to H_n(X\star Y)$ is the zero map.