By using the Mayer-Vietoris sequence in reduced homology :
$…\overset{\Delta_{n+1}}{\longrightarrow} \tilde{H_n}(A)\overset{E_{n}}{\longrightarrow} \tilde{H_n}(X_1)\times \tilde{H_n}(X_2)\overset{S_{n}}{\longrightarrow} \tilde{H_n}(X)\overset{\Delta_{n}}{\longrightarrow} \tilde {H}_{n-1}(A)\overset{E_{n-1}}{\longrightarrow} \tilde{H}_{n-1}(X_1)\times \tilde{H}_{n-1}(X_2) \overset{S_{n-1}}\longrightarrow … $
i have to find the singulary homological groups $H_n(X)$ for all $n\geq 0$ of the following spaces :
1) The rose with 2 petals (the "eight")
2) The torus $\mathbb{T}^2 :=[0;1]^2 /\mathcal{T}$ With the help of 1) and by using $X_1 := \mathbb{T}^{2} -\{(1/2;1/2)\}$, $X_2 = ]0,1[^{2}$ et $A=X_1 \cap X_2$.
I do not know how to begin. Thank you for any help.
Best Answer
For the rose with two petals, you might want to try $X_1 = $ the left petal plus part of the right, i.e., the shape of the letter $\alpha$; $X_2$ is the same, but the right petal with part of the left; then $A$ is a "$\times$" shape which is contractible, so all its reduced homology groups are 0. (Can you give a good reason for this? Your professor will want one.)
That tells you that in a typical part of the sequence, you have $$ 0 \to H_n(X_1) \times H_n(X_2) \to H_n(X) \to 0 $$ (where all groups should be reduced). That means that each homology group of $X$ is isomorphic to the product of the homology groups of the two $\alpha$s (Why?), which are both homotopy-equivalent to the circle (Can you prove this? Can you write a deformation retraction of $\alpha$ to the circle? Why would that suffice?), so their reduced homology groups (starting at $n = 0$) are $$ 0, \mathbb Z, 0, 0, \ldots $$ Can you now write down the groups for $X$?