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By using the Mayer-Vietoris sequence in reduced homology :

I have to calculate the homology groups of :

The torus $\mathbb{T}^2 :=[0;1]^2 /\mathcal{T}$ by using the following decomposition $X_1 := \mathbb{T}^{2} -\{(1/2;1/2)\}$, $X_2 = ]0,1[^{2}$ et $A=X_1 \cap X_2$.

$\mathcal{T}$ is an equivalence relation meaning that we have identified the boudaries of the square $[0;1]^2$ : $(1,t)\sim (0,t)$ and $(t,1)\sim (t,0)$;$t\in[0;1]$. The point $(1/2;1/2)$ is the center of the square.

I have already done this work : the rose with 2 petals is a retract by deformation of $X_1$ so $\tilde{H}_0(X_1)=\mathbb{Z}$ $\tilde{H}_1(X_1)=\mathbb{Z\oplus\mathbb Z}$ and $\tilde{H}_n(X_1)=0$ for $n\geq 2$

$\tilde{H}_n(X_2)=0$ for all $n$.

But i am stuck here. How can i compute $H_1(\mathbb T^2)$ and $H_2(\mathbb T^2)$ ? Thanks for any help.

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  • $\begingroup$ Have you computed $\tilde H_*(A)$? And have you written out the Mayer-Vietoris sequence and substituted the values that you know for the various terms? $\endgroup$
    – Lee Mosher
    Commented May 31, 2015 at 17:03
  • $\begingroup$ @LeeMosher But how can i geometrically describe $A$ ? It is a square without boundary and a hole in the middle but how can i compute its homology ? $\endgroup$
    – Far
    Commented May 31, 2015 at 17:09
  • $\begingroup$ You compute the homology of $A$ the way you computed the homology of $X_1$, by figuring out what $A$ deformation retracts to. $\endgroup$
    – Lee Mosher
    Commented May 31, 2015 at 17:15
  • $\begingroup$ @LeeMosher That is my problem...i do not know how to retract $A$. $\endgroup$
    – Far
    Commented May 31, 2015 at 17:17
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    $\begingroup$ What is $\mathcal T$? What does $(1/2; 1/2)$ mean? One can guess, but it would be much better if you clarified this non-standard notation. $\endgroup$ Commented May 31, 2015 at 17:26

1 Answer 1

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From what you've worked out so far, the key to this problem is computation of the homomorphism $$\mathbb{Z} \approx H_1(\mathbb{S}^1) \xrightarrow{j_*} H_1(A) \xrightarrow{i_*} H_1(X_1) \approx \mathbb{Z}^2 $$ In this sequence, $i_*$ is the homomorphism induced by the inclusion $i : A \hookrightarrow X_1$, and $j_*$ is the isomorphism induced by a homotopy equivalence $j :\mathbb{S}^1 \to A$. Once you've computed that homomorphism, it will follow that $H_2(\mathbb{T}^2)$ is isomorphic to its kernel.

Let me give some further hints, without giving the whole thing away.

Your work will be aided by deriving a simple expression for a function $f : \mathbb{S}^1 \to X_1$ which is homotopic to the composition $\mathbb{S}^1 \xrightarrow{i} A \xrightarrow{j} X_1$.

If you understand how the boundary of a square is glued to a rose $X_1$ to form $\mathbb{T}^2$, then you should be able to use the description of that gluing to guess and to prove a formula for $f$.

And once you have a formula for $f : S^1 \to X_1$, you should be able to "abelianize" that formula to obtain a formula for the induced function $f_* = j_* \circ i_* : \mathbb{Z} \to \mathbb{Z}^2$.

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  • $\begingroup$ Thank you. The idea is clear but i am not able to write down a formula for $f$. I can glue the boundary of the square (the square is homeomorphic to a 2-cell) to the wedge of the circles ? $\endgroup$
    – Far
    Commented Jun 1, 2015 at 15:56
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    $\begingroup$ @Far: Yes, you can glue in that manner. Furthermore, the actual formula for that gluing is the formula for $f : \mathbb{S}^1 \to X_1$, up to homotopy. $\endgroup$
    – Lee Mosher
    Commented Jun 5, 2015 at 14:41

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