2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 May 31 '15 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 May 31 '15 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 May 31 '15 at 17:15
  • $\begingroup$ @LeeMosher That is my problem...i do not know how to retract $A$. $\endgroup$ – Far May 31 '15 at 17:17
  • 1
    $\begingroup$ 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. $\endgroup$ – Ayman Hourieh May 31 '15 at 17:26
1
$\begingroup$

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 $f_* = j_* \circ i_* : \mathbb{Z} \to \mathbb{Z}^2$.

$\endgroup$
  • $\begingroup$ 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 Jun 1 '15 at 15:56
  • 1
    $\begingroup$ @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 Jun 5 '15 at 14:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.