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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.

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1 Answer 1

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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 $X$?

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  • $\begingroup$ Hugues Thank you, i am working on your questions... $\endgroup$
    – Far
    May 27, 2015 at 13:50
  • $\begingroup$ Thank you but your questions are a little bit hard for the moment. I am reading my course and i have to give an answer very soon. $\endgroup$
    – Far
    May 28, 2015 at 10:20
  • $\begingroup$ My questions are all things that you should know before trying to do the problem you gave us. So I'll reformulate my answer: you should read the material previous to this problem until you know the answer to all my questions, because only then will you be ready to proceed. In the words of an old friend, what you're saying is "I don't have time to stop for gas...I'm already late!" :( $\endgroup$ May 28, 2015 at 12:07
  • $\begingroup$ @JohnHugues Thank you, i will answer soon. Let me just a moment (it was my point) $\endgroup$
    – Far
    May 28, 2015 at 15:03
  • $\begingroup$ @JohnHugues I have woked on your questions. Sorry for the delay. So a contractible space is homotopic to a point and so it has an homology group of order 0 reduced to $\mathbb{Z}$, but its reduced homology group of the same order is ${0}$. Your exact sequence prove that we have the isomorphism. Now a deformation retract (which is a special case of homotopic-equivalent) induces an isomorphism on homology groups. So $H_0 (X_1)= H_1(X_2)=_{isom} \mathbb{Z}$, the others are reduced to $0$. Finally : $H_0 (X) = \mathbb{Z}$ ,$H_1 (X) = \mathbb{Z}\oplus \mathbb{Z}$ and $H_n (X) = {0}$ if $n\geq 2$. $\endgroup$
    – Far
    May 31, 2015 at 13:40

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